Numpy provides various methods for searching different kinds of numerical values, in this article, we will cover two important ones.
- numpy.where()
- numpy.searchsorted()
1. numpy.where:() It returns the indices of elements in an input array where the given condition is satisfied.
Syntax: numpy.where(condition[, x, y])
Parameters:
- condition : When True, yield x, otherwise yield y.
- x, y : Values from which to choose. x, y and condition need to be broadcastable to some shape.
Returns:
out : [ndarray or tuple of ndarrays] If both x and y are specified, the output array contains elements of x where condition is True, and elements from y elsewhere.If only condition is given, return the tuple condition.nonzero(), the indices where condition is True.
The following example demonstrates how to search using where().
Python3
import
numpy as np
arr
=
np.array([
10
,
32
,
30
,
50
,
20
,
82
,
91
,
45
])
print
(
"arr = {}"
.
format
(arr))
i
=
np.where(arr
=
=
30
)
print
(
"i = {}"
.
format
(i))
Output:
arr = [10 32 30 50 20 82 91 45] i = (array([2], dtype=int64),)
As you can see variable i is an iterable with the index of our searched value as the first element. We can make it look better by replacing the last print statement with
print("i = {}".format(i[0]))
This will change the final output to
arr = [10 32 30 50 20 82 91 45] i = [2]
2. numpy.searchsorted(): The function is used to find the indices into a sorted array arr such that, if elements are inserted before the indices, the order of arr would be still preserved. Here, a binary search is used to find the required insertion indices.
Syntax : numpy.searchsorted(arr, num, side=’left’, sorter=None)
Parameters :
- arr : [array_like] Input array. If sorter is None, then it must be sorted in ascending order, otherwise sorter must be an array of indices that sort it.
- num : [array_like]The Values which we want to insert into arr.
- side : [‘left’, ‘right’], optional.If ‘left’, the index of the first suitable location found is given. If ‘right’, return the last such index. If there is no suitable index, return either 0 or N (where N is the length of a).
- num : [array_like, Optional] array of integer indices that sort array a into ascending order. They are typically the result of argsort.
Return : [indices], Array of insertion points with the same shape as num.
The following example explains the use of searchsorted().
Python3
import
numpy as np
arr
=
[
1
,
2
,
2
,
3
,
3
,
3
,
4
,
5
,
6
,
6
]
print
(
"arr = {}"
.
format
(arr))
print
(
"left-most index = {}"
.
format
(np.searchsorted(arr,
3
, side
=
"left"
)))
print
(
"right-most index = {}"
.
format
(np.searchsorted(arr,
3
, side
=
"right"
)))
Output:
arr = [1, 2, 2, 3, 3, 3, 4, 5, 6, 6] left-most index = 3 right-most index = 6
Last Updated :
01 Oct, 2020
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ndarrays
can be indexed using the standard Python
x[obj]
syntax, where x is the array and obj the selection.
There are different kinds of indexing available depending on obj:
basic indexing, advanced indexing and field access.
Most of the following examples show the use of indexing when
referencing data in an array. The examples work just as well
when assigning to an array. See Assigning values to indexed arrays for
specific examples and explanations on how assignments work.
Note that in Python, x[(exp1, exp2, ..., expN)]
is equivalent to
x[exp1, exp2, ..., expN]
; the latter is just syntactic sugar
for the former.
Basic indexing#
Single element indexing#
Single element indexing works
exactly like that for other standard Python sequences. It is 0-based,
and accepts negative indices for indexing from the end of the array.
>>> x = np.arange(10) >>> x[2] 2 >>> x[-2] 8
It is not necessary to
separate each dimension’s index into its own set of square brackets.
>>> x.shape = (2, 5) # now x is 2-dimensional >>> x[1, 3] 8 >>> x[1, -1] 9
Note that if one indexes a multidimensional array with fewer indices
than dimensions, one gets a subdimensional array. For example:
>>> x[0] array([0, 1, 2, 3, 4])
That is, each index specified selects the array corresponding to the
rest of the dimensions selected. In the above example, choosing 0
means that the remaining dimension of length 5 is being left unspecified,
and that what is returned is an array of that dimensionality and size.
It must be noted that the returned array is a view, i.e., it is not a
copy of the original, but points to the same values in memory as does the
original array.
In this case, the 1-D array at the first position (0) is returned.
So using a single index on the returned array, results in a single
element being returned. That is:
So note that x[0, 2] == x[0][2]
though the second case is more
inefficient as a new temporary array is created after the first index
that is subsequently indexed by 2.
Note
NumPy uses C-order indexing. That means that the last
index usually represents the most rapidly changing memory location,
unlike Fortran or IDL, where the first index represents the most
rapidly changing location in memory. This difference represents a
great potential for confusion.
Slicing and striding#
Basic slicing extends Python’s basic concept of slicing to N
dimensions. Basic slicing occurs when obj is a slice
object
(constructed by start:stop:step
notation inside of brackets), an
integer, or a tuple of slice objects and integers. Ellipsis
and newaxis
objects can be interspersed with these as
well.
The simplest case of indexing with N integers returns an array
scalar representing the corresponding item. As in
Python, all indices are zero-based: for the i-th index (n_i),
the valid range is (0 le n_i < d_i) where (d_i) is the
i-th element of the shape of the array. Negative indices are
interpreted as counting from the end of the array (i.e., if
(n_i < 0), it means (n_i + d_i)).
All arrays generated by basic slicing are always views
of the original array.
Note
NumPy slicing creates a view instead of a copy as in the case of
built-in Python sequences such as string, tuple and list.
Care must be taken when extracting
a small portion from a large array which becomes useless after the
extraction, because the small portion extracted contains a reference
to the large original array whose memory will not be released until
all arrays derived from it are garbage-collected. In such cases an
explicit copy()
is recommended.
The standard rules of sequence slicing apply to basic slicing on a
per-dimension basis (including using a step index). Some useful
concepts to remember include:
-
The basic slice syntax is
i:j:k
where i is the starting index,
j is the stopping index, and k is the step ((kneq0)).
This selects the m elements (in the corresponding dimension) with
index values i, i + k, …, i + (m — 1) k where
(m = q + (rneq0)) and q and r are the quotient and remainder
obtained by dividing j — i by k: j — i = q k + r, so that
i + (m — 1) k < j.
For example:>>> x = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]) >>> x[1:7:2] array([1, 3, 5])
-
Negative i and j are interpreted as n + i and n + j where
n is the number of elements in the corresponding dimension.
Negative k makes stepping go towards smaller indices.
From the above example:>>> x[-2:10] array([8, 9]) >>> x[-3:3:-1] array([7, 6, 5, 4])
-
Assume n is the number of elements in the dimension being
sliced. Then, if i is not given it defaults to 0 for k > 0 and
n — 1 for k < 0 . If j is not given it defaults to n for k > 0
and -n-1 for k < 0 . If k is not given it defaults to 1. Note that
::
is the same as:
and means select all indices along this
axis.
From the above example:>>> x[5:] array([5, 6, 7, 8, 9])
-
If the number of objects in the selection tuple is less than
N, then:
is assumed for any subsequent dimensions.
For example:>>> x = np.array([[[1],[2],[3]], [[4],[5],[6]]]) >>> x.shape (2, 3, 1) >>> x[1:2] array([[[4], [5], [6]]])
-
An integer, i, returns the same values as
i:i+1
except the dimensionality of the returned object is reduced by
1. In particular, a selection tuple with the p-th
element an integer (and all other entries:
) returns the
corresponding sub-array with dimension N — 1. If N = 1
then the returned object is an array scalar. These objects are
explained in Scalars. -
If the selection tuple has all entries
:
except the
p-th entry which is a slice objecti:j:k
,
then the returned array has dimension N formed by
concatenating the sub-arrays returned by integer indexing of
elements i, i+k, …, i + (m — 1) k < j, -
Basic slicing with more than one non-
:
entry in the slicing
tuple, acts like repeated application of slicing using a single
non-:
entry, where the non-:
entries are successively taken
(with all other non-:
entries replaced by:
). Thus,
x[ind1, ..., ind2,:]
acts likex[ind1][..., ind2, :]
under basic
slicing.Warning
The above is not true for advanced indexing.
-
You may use slicing to set values in the array, but (unlike lists) you
can never grow the array. The size of the value to be set in
x[obj] = value
must be (broadcastable to) the same shape as
x[obj]
. -
A slicing tuple can always be constructed as obj
and used in thex[obj]
notation. Slice objects can be used in
the construction in place of the[start:stop:step]
notation. For example,x[1:10:5, ::-1]
can also be implemented
asobj = (slice(1, 10, 5), slice(None, None, -1)); x[obj]
. This
can be useful for constructing generic code that works on arrays
of arbitrary dimensions. See Dealing with variable numbers of indices within programs
for more information.
Dimensional indexing tools#
There are some tools to facilitate the easy matching of array shapes with
expressions and in assignments.
Ellipsis
expands to the number of :
objects needed for the
selection tuple to index all dimensions. In most cases, this means that the
length of the expanded selection tuple is x.ndim
. There may only be a
single ellipsis present.
From the above example:
>>> x[..., 0] array([[1, 2, 3], [4, 5, 6]])
This is equivalent to:
>>> x[:, :, 0] array([[1, 2, 3], [4, 5, 6]])
Each newaxis
object in the selection tuple serves to expand
the dimensions of the resulting selection by one unit-length
dimension. The added dimension is the position of the newaxis
object in the selection tuple. newaxis
is an alias for
None
, and None
can be used in place of this with the same result.
From the above example:
>>> x[:, np.newaxis, :, :].shape (2, 1, 3, 1) >>> x[:, None, :, :].shape (2, 1, 3, 1)
This can be handy to combine two
arrays in a way that otherwise would require explicit reshaping
operations. For example:
>>> x = np.arange(5) >>> x[:, np.newaxis] + x[np.newaxis, :] array([[0, 1, 2, 3, 4], [1, 2, 3, 4, 5], [2, 3, 4, 5, 6], [3, 4, 5, 6, 7], [4, 5, 6, 7, 8]])
Advanced indexing#
Advanced indexing is triggered when the selection object, obj, is a
non-tuple sequence object, an ndarray
(of data type integer or bool),
or a tuple with at least one sequence object or ndarray (of data type
integer or bool). There are two types of advanced indexing: integer
and Boolean.
Advanced indexing always returns a copy of the data (contrast with
basic slicing that returns a view).
Warning
The definition of advanced indexing means that x[(1, 2, 3),]
is
fundamentally different than x[(1, 2, 3)]
. The latter is
equivalent to x[1, 2, 3]
which will trigger basic selection while
the former will trigger advanced indexing. Be sure to understand
why this occurs.
Also recognize that x[[1, 2, 3]]
will trigger advanced indexing,
whereas due to the deprecated Numeric compatibility mentioned above,
x[[1, 2, slice(None)]]
will trigger basic slicing.
Integer array indexing#
Integer array indexing allows selection of arbitrary items in the array
based on their N-dimensional index. Each integer array represents a number
of indices into that dimension.
Negative values are permitted in the index arrays and work as they do with
single indices or slices:
>>> x = np.arange(10, 1, -1) >>> x array([10, 9, 8, 7, 6, 5, 4, 3, 2]) >>> x[np.array([3, 3, 1, 8])] array([7, 7, 9, 2]) >>> x[np.array([3, 3, -3, 8])] array([7, 7, 4, 2])
If the index values are out of bounds then an IndexError
is thrown:
>>> x = np.array([[1, 2], [3, 4], [5, 6]]) >>> x[np.array([1, -1])] array([[3, 4], [5, 6]]) >>> x[np.array([3, 4])] Traceback (most recent call last): ... IndexError: index 3 is out of bounds for axis 0 with size 3
When the index consists of as many integer arrays as dimensions of the array
being indexed, the indexing is straightforward, but different from slicing.
Advanced indices always are broadcast and
iterated as one:
result[i_1, ..., i_M] == x[ind_1[i_1, ..., i_M], ind_2[i_1, ..., i_M], ..., ind_N[i_1, ..., i_M]]
Note that the resulting shape is identical to the (broadcast) indexing array
shapes ind_1, ..., ind_N
. If the indices cannot be broadcast to the
same shape, an exception IndexError: shape mismatch: indexing arrays could
is raised.
not be broadcast together with shapes...
Indexing with multidimensional index arrays tend
to be more unusual uses, but they are permitted, and they are useful for some
problems. We’ll start with the simplest multidimensional case:
>>> y = np.arange(35).reshape(5, 7) >>> y array([[ 0, 1, 2, 3, 4, 5, 6], [ 7, 8, 9, 10, 11, 12, 13], [14, 15, 16, 17, 18, 19, 20], [21, 22, 23, 24, 25, 26, 27], [28, 29, 30, 31, 32, 33, 34]]) >>> y[np.array([0, 2, 4]), np.array([0, 1, 2])] array([ 0, 15, 30])
In this case, if the index arrays have a matching shape, and there is an
index array for each dimension of the array being indexed, the resultant
array has the same shape as the index arrays, and the values correspond
to the index set for each position in the index arrays. In this example,
the first index value is 0 for both index arrays, and thus the first value
of the resultant array is y[0, 0]
. The next value is y[2, 1]
, and
the last is y[4, 2]
.
If the index arrays do not have the same shape, there is an attempt to
broadcast them to the same shape. If they cannot be broadcast to the same
shape, an exception is raised:
>>> y[np.array([0, 2, 4]), np.array([0, 1])] Traceback (most recent call last): ... IndexError: shape mismatch: indexing arrays could not be broadcast together with shapes (3,) (2,)
The broadcasting mechanism permits index arrays to be combined with
scalars for other indices. The effect is that the scalar value is used
for all the corresponding values of the index arrays:
>>> y[np.array([0, 2, 4]), 1] array([ 1, 15, 29])
Jumping to the next level of complexity, it is possible to only partially
index an array with index arrays. It takes a bit of thought to understand
what happens in such cases. For example if we just use one index array
with y:
>>> y[np.array([0, 2, 4])] array([[ 0, 1, 2, 3, 4, 5, 6], [14, 15, 16, 17, 18, 19, 20], [28, 29, 30, 31, 32, 33, 34]])
It results in the construction of a new array where each value of the
index array selects one row from the array being indexed and the resultant
array has the resulting shape (number of index elements, size of row).
In general, the shape of the resultant array will be the concatenation of
the shape of the index array (or the shape that all the index arrays were
broadcast to) with the shape of any unused dimensions (those not indexed)
in the array being indexed.
Example
From each row, a specific element should be selected. The row index is just
[0, 1, 2]
and the column index specifies the element to choose for the
corresponding row, here [0, 1, 0]
. Using both together the task
can be solved using advanced indexing:
>>> x = np.array([[1, 2], [3, 4], [5, 6]]) >>> x[[0, 1, 2], [0, 1, 0]] array([1, 4, 5])
To achieve a behaviour similar to the basic slicing above, broadcasting can be
used. The function ix_
can help with this broadcasting. This is best
understood with an example.
Example
From a 4×3 array the corner elements should be selected using advanced
indexing. Thus all elements for which the column is one of [0, 2]
and
the row is one of [0, 3]
need to be selected. To use advanced indexing
one needs to select all elements explicitly. Using the method explained
previously one could write:
>>> x = np.array([[ 0, 1, 2], ... [ 3, 4, 5], ... [ 6, 7, 8], ... [ 9, 10, 11]]) >>> rows = np.array([[0, 0], ... [3, 3]], dtype=np.intp) >>> columns = np.array([[0, 2], ... [0, 2]], dtype=np.intp) >>> x[rows, columns] array([[ 0, 2], [ 9, 11]])
However, since the indexing arrays above just repeat themselves,
broadcasting can be used (compare operations such as
rows[:, np.newaxis] + columns
) to simplify this:
>>> rows = np.array([0, 3], dtype=np.intp) >>> columns = np.array([0, 2], dtype=np.intp) >>> rows[:, np.newaxis] array([[0], [3]]) >>> x[rows[:, np.newaxis], columns] array([[ 0, 2], [ 9, 11]])
This broadcasting can also be achieved using the function ix_
:
>>> x[np.ix_(rows, columns)] array([[ 0, 2], [ 9, 11]])
Note that without the np.ix_
call, only the diagonal elements would
be selected:
>>> x[rows, columns] array([ 0, 11])
This difference is the most important thing to remember about
indexing with multiple advanced indices.
Example
A real-life example of where advanced indexing may be useful is for a color
lookup table where we want to map the values of an image into RGB triples for
display. The lookup table could have a shape (nlookup, 3). Indexing
such an array with an image with shape (ny, nx) with dtype=np.uint8
(or any integer type so long as values are with the bounds of the
lookup table) will result in an array of shape (ny, nx, 3) where a
triple of RGB values is associated with each pixel location.
Boolean array indexing#
This advanced indexing occurs when obj is an array object of Boolean
type, such as may be returned from comparison operators. A single
boolean index array is practically identical to x[obj.nonzero()]
where,
as described above, obj.nonzero()
returns a
tuple (of length obj.ndim
) of integer index
arrays showing the True
elements of obj. However, it is
faster when obj.shape == x.shape
.
If obj.ndim == x.ndim
, x[obj]
returns a 1-dimensional array
filled with the elements of x corresponding to the True
values of obj. The search order will be row-major,
C-style. If obj has True
values at entries that are outside
of the bounds of x, then an index error will be raised. If obj is
smaller than x it is identical to filling it with False
.
A common use case for this is filtering for desired element values.
For example, one may wish to select all entries from an array which
are not NaN
:
>>> x = np.array([[1., 2.], [np.nan, 3.], [np.nan, np.nan]]) >>> x[~np.isnan(x)] array([1., 2., 3.])
Or wish to add a constant to all negative elements:
>>> x = np.array([1., -1., -2., 3]) >>> x[x < 0] += 20 >>> x array([ 1., 19., 18., 3.])
In general if an index includes a Boolean array, the result will be
identical to inserting obj.nonzero()
into the same position
and using the integer array indexing mechanism described above.
x[ind_1, boolean_array, ind_2]
is equivalent to
x[(ind_1,) + boolean_array.nonzero() + (ind_2,)]
.
If there is only one Boolean array and no integer indexing array present,
this is straightforward. Care must only be taken to make sure that the
boolean index has exactly as many dimensions as it is supposed to work
with.
In general, when the boolean array has fewer dimensions than the array being
indexed, this is equivalent to x[b, ...]
, which means x is indexed by b
followed by as many :
as are needed to fill out the rank of x. Thus the
shape of the result is one dimension containing the number of True elements of
the boolean array, followed by the remaining dimensions of the array being
indexed:
>>> x = np.arange(35).reshape(5, 7) >>> b = x > 20 >>> b[:, 5] array([False, False, False, True, True]) >>> x[b[:, 5]] array([[21, 22, 23, 24, 25, 26, 27], [28, 29, 30, 31, 32, 33, 34]])
Here the 4th and 5th rows are selected from the indexed array and
combined to make a 2-D array.
Example
From an array, select all rows which sum up to less or equal two:
>>> x = np.array([[0, 1], [1, 1], [2, 2]]) >>> rowsum = x.sum(-1) >>> x[rowsum <= 2, :] array([[0, 1], [1, 1]])
Combining multiple Boolean indexing arrays or a Boolean with an integer
indexing array can best be understood with the
obj.nonzero()
analogy. The function ix_
also supports boolean arrays and will work without any surprises.
Example
Use boolean indexing to select all rows adding up to an even
number. At the same time columns 0 and 2 should be selected with an
advanced integer index. Using the ix_
function this can be done
with:
>>> x = np.array([[ 0, 1, 2], ... [ 3, 4, 5], ... [ 6, 7, 8], ... [ 9, 10, 11]]) >>> rows = (x.sum(-1) % 2) == 0 >>> rows array([False, True, False, True]) >>> columns = [0, 2] >>> x[np.ix_(rows, columns)] array([[ 3, 5], [ 9, 11]])
Without the np.ix_
call, only the diagonal elements would be
selected.
Or without np.ix_
(compare the integer array examples):
>>> rows = rows.nonzero()[0] >>> x[rows[:, np.newaxis], columns] array([[ 3, 5], [ 9, 11]])
Example
Use a 2-D boolean array of shape (2, 3)
with four True elements to select rows from a 3-D array of shape
(2, 3, 5) results in a 2-D result of shape (4, 5):
>>> x = np.arange(30).reshape(2, 3, 5) >>> x array([[[ 0, 1, 2, 3, 4], [ 5, 6, 7, 8, 9], [10, 11, 12, 13, 14]], [[15, 16, 17, 18, 19], [20, 21, 22, 23, 24], [25, 26, 27, 28, 29]]]) >>> b = np.array([[True, True, False], [False, True, True]]) >>> x[b] array([[ 0, 1, 2, 3, 4], [ 5, 6, 7, 8, 9], [20, 21, 22, 23, 24], [25, 26, 27, 28, 29]])
Combining advanced and basic indexing#
When there is at least one slice (:
), ellipsis (...
) or newaxis
in the index (or the array has more dimensions than there are advanced indices),
then the behaviour can be more complicated. It is like concatenating the
indexing result for each advanced index element.
In the simplest case, there is only a single advanced index combined with
a slice. For example:
>>> y = np.arange(35).reshape(5,7) >>> y[np.array([0, 2, 4]), 1:3] array([[ 1, 2], [15, 16], [29, 30]])
In effect, the slice and index array operation are independent. The slice
operation extracts columns with index 1 and 2, (i.e. the 2nd and 3rd columns),
followed by the index array operation which extracts rows with index 0, 2 and 4
(i.e the first, third and fifth rows). This is equivalent to:
>>> y[:, 1:3][np.array([0, 2, 4]), :] array([[ 1, 2], [15, 16], [29, 30]])
A single advanced index can, for example, replace a slice and the result array
will be the same. However, it is a copy and may have a different memory layout.
A slice is preferable when it is possible.
For example:
>>> x = np.array([[ 0, 1, 2], ... [ 3, 4, 5], ... [ 6, 7, 8], ... [ 9, 10, 11]]) >>> x[1:2, 1:3] array([[4, 5]]) >>> x[1:2, [1, 2]] array([[4, 5]])
The easiest way to understand a combination of multiple advanced indices may
be to think in terms of the resulting shape. There are two parts to the indexing
operation, the subspace defined by the basic indexing (excluding integers) and
the subspace from the advanced indexing part. Two cases of index combination
need to be distinguished:
-
The advanced indices are separated by a slice,
Ellipsis
or
newaxis
. For examplex[arr1, :, arr2]
. -
The advanced indices are all next to each other.
For examplex[..., arr1, arr2, :]
but notx[arr1, :, 1]
since1
is an advanced index in this regard.
In the first case, the dimensions resulting from the advanced indexing
operation come first in the result array, and the subspace dimensions after
that.
In the second case, the dimensions from the advanced indexing operations
are inserted into the result array at the same spot as they were in the
initial array (the latter logic is what makes simple advanced indexing
behave just like slicing).
Example
Suppose x.shape
is (10, 20, 30) and ind
is a (2, 3, 4)-shaped
indexing intp
array, then result = x[..., ind, :]
has
shape (10, 2, 3, 4, 30) because the (20,)-shaped subspace has been
replaced with a (2, 3, 4)-shaped broadcasted indexing subspace. If
we let i, j, k loop over the (2, 3, 4)-shaped subspace then
result[..., i, j, k, :] = x[..., ind[i, j, k], :]
. This example
produces the same result as x.take(ind, axis=-2)
.
Example
Let x.shape
be (10, 20, 30, 40, 50) and suppose ind_1
and ind_2
can be broadcast to the shape (2, 3, 4). Then
x[:, ind_1, ind_2]
has shape (10, 2, 3, 4, 40, 50) because the
(20, 30)-shaped subspace from X has been replaced with the
(2, 3, 4) subspace from the indices. However,
x[:, ind_1, :, ind_2]
has shape (2, 3, 4, 10, 30, 50) because there
is no unambiguous place to drop in the indexing subspace, thus
it is tacked-on to the beginning. It is always possible to use
.transpose()
to move the subspace
anywhere desired. Note that this example cannot be replicated
using take
.
Example
Slicing can be combined with broadcasted boolean indices:
>>> x = np.arange(35).reshape(5, 7) >>> b = x > 20 >>> b array([[False, False, False, False, False, False, False], [False, False, False, False, False, False, False], [False, False, False, False, False, False, False], [ True, True, True, True, True, True, True], [ True, True, True, True, True, True, True]]) >>> x[b[:, 5], 1:3] array([[22, 23], [29, 30]])
Field access#
If the ndarray
object is a structured array the fields
of the array can be accessed by indexing the array with strings,
dictionary-like.
Indexing x['field-name']
returns a new view to the array,
which is of the same shape as x (except when the field is a
sub-array) but of data type x.dtype['field-name']
and contains
only the part of the data in the specified field. Also,
record array scalars can be “indexed” this way.
Indexing into a structured array can also be done with a list of field names,
e.g. x[['field-name1', 'field-name2']]
. As of NumPy 1.16, this returns a
view containing only those fields. In older versions of NumPy, it returned a
copy. See the user guide section on Structured arrays for more
information on multifield indexing.
If the accessed field is a sub-array, the dimensions of the sub-array
are appended to the shape of the result.
For example:
>>> x = np.zeros((2, 2), dtype=[('a', np.int32), ('b', np.float64, (3, 3))]) >>> x['a'].shape (2, 2) >>> x['a'].dtype dtype('int32') >>> x['b'].shape (2, 2, 3, 3) >>> x['b'].dtype dtype('float64')
Flat Iterator indexing#
x.flat
returns an iterator that will iterate
over the entire array (in C-contiguous style with the last index
varying the fastest). This iterator object can also be indexed using
basic slicing or advanced indexing as long as the selection object is
not a tuple. This should be clear from the fact that x.flat
is a 1-dimensional view. It can be used for integer
indexing with 1-dimensional C-style-flat indices. The shape of any
returned array is therefore the shape of the integer indexing object.
Assigning values to indexed arrays#
As mentioned, one can select a subset of an array to assign to using
a single index, slices, and index and mask arrays. The value being
assigned to the indexed array must be shape consistent (the same shape
or broadcastable to the shape the index produces). For example, it is
permitted to assign a constant to a slice:
>>> x = np.arange(10) >>> x[2:7] = 1
or an array of the right size:
>>> x[2:7] = np.arange(5)
Note that assignments may result in changes if assigning
higher types to lower types (like floats to ints) or even
exceptions (assigning complex to floats or ints):
>>> x[1] = 1.2 >>> x[1] 1 >>> x[1] = 1.2j Traceback (most recent call last): ... TypeError: can't convert complex to int
Unlike some of the references (such as array and mask indices)
assignments are always made to the original data in the array
(indeed, nothing else would make sense!). Note though, that some
actions may not work as one may naively expect. This particular
example is often surprising to people:
>>> x = np.arange(0, 50, 10) >>> x array([ 0, 10, 20, 30, 40]) >>> x[np.array([1, 1, 3, 1])] += 1 >>> x array([ 0, 11, 20, 31, 40])
Where people expect that the 1st location will be incremented by 3.
In fact, it will only be incremented by 1. The reason is that
a new array is extracted from the original (as a temporary) containing
the values at 1, 1, 3, 1, then the value 1 is added to the temporary,
and then the temporary is assigned back to the original array. Thus
the value of the array at x[1] + 1
is assigned to x[1]
three times,
rather than being incremented 3 times.
Dealing with variable numbers of indices within programs#
The indexing syntax is very powerful but limiting when dealing with
a variable number of indices. For example, if you want to write
a function that can handle arguments with various numbers of
dimensions without having to write special case code for each
number of possible dimensions, how can that be done? If one
supplies to the index a tuple, the tuple will be interpreted
as a list of indices. For example:
>>> z = np.arange(81).reshape(3, 3, 3, 3) >>> indices = (1, 1, 1, 1) >>> z[indices] 40
So one can use code to construct tuples of any number of indices
and then use these within an index.
Slices can be specified within programs by using the slice() function
in Python. For example:
>>> indices = (1, 1, 1, slice(0, 2)) # same as [1, 1, 1, 0:2] >>> z[indices] array([39, 40])
Likewise, ellipsis can be specified by code by using the Ellipsis
object:
>>> indices = (1, Ellipsis, 1) # same as [1, ..., 1] >>> z[indices] array([[28, 31, 34], [37, 40, 43], [46, 49, 52]])
For this reason, it is possible to use the output from the
np.nonzero()
function directly as an index since
it always returns a tuple of index arrays.
Because of the special treatment of tuples, they are not automatically
converted to an array as a list would be. As an example:
>>> z[[1, 1, 1, 1]] # produces a large array array([[[[27, 28, 29], [30, 31, 32], ... >>> z[(1, 1, 1, 1)] # returns a single value 40
Detailed notes#
These are some detailed notes, which are not of importance for day to day
indexing (in no particular order):
-
The native NumPy indexing type is
intp
and may differ from the
default integer array type.intp
is the smallest data type
sufficient to safely index any array; for advanced indexing it may be
faster than other types. -
For advanced assignments, there is in general no guarantee for the
iteration order. This means that if an element is set more than once,
it is not possible to predict the final result. -
An empty (tuple) index is a full scalar index into a zero-dimensional array.
x[()]
returns a scalar ifx
is zero-dimensional and a view
otherwise. On the other hand,x[...]
always returns a view. -
If a zero-dimensional array is present in the index and it is a full
integer index the result will be a scalar and not a zero-dimensional array.
(Advanced indexing is not triggered.) -
When an ellipsis (
...
) is present but has no size (i.e. replaces zero
:
) the result will still always be an array. A view if no advanced index
is present, otherwise a copy. -
The
nonzero
equivalence for Boolean arrays does not hold for zero
dimensional boolean arrays. -
When the result of an advanced indexing operation has no elements but an
individual index is out of bounds, whether or not anIndexError
is
raised is undefined (e.g.x[[], [123]]
with123
being out of bounds). -
When a casting error occurs during assignment (for example updating a
numerical array using a sequence of strings), the array being assigned
to may end up in an unpredictable partially updated state.
However, if any other error (such as an out of bounds index) occurs, the
array will remain unchanged. -
The memory layout of an advanced indexing result is optimized for each
indexing operation and no particular memory order can be assumed. -
When using a subclass (especially one which manipulates its shape), the
defaultndarray.__setitem__
behaviour will call__getitem__
for
basic indexing but not for advanced indexing. For such a subclass it may
be preferable to callndarray.__setitem__
with a base class ndarray
view on the data. This must be done if the subclasses__getitem__
does
not return views.
In this tutorial, we will look at how to find the index of an element in a numpy array.
You can use the numpy’s where() function to get the index of an element inside the array. The following example illustrates the usage.
np.where(arr==i)
Here, arr
is the numpy array and i
is the element for which you want to get the index. Inside the function, we pass arr==i which is a vectorized operation on the array arr to compare each of its elements with the value in i and result in a numpy array of boolean True
and False
values.
Now, np.where()
gives you all the indices where the element occurs in the array. That is the indices of all the elements for which arr==i evaluates to True. This method works for both one-dimensional and multidimensional arrays. See the examples below:
1. Index of element in 1D array
Let’s apply the above syntax on a one-dimensional numpy array and find all the indices where a particular element occurs. First, let’s create a 1D array and print it out.
import numpy as np # create a numpy array arr = np.array([7, 5, 8, 6, 3, 9, 5, 2, 3, 5]) # print the original array print("Original array:", arr)
Output:
Original array: [7 5 8 6 3 9 5 2 3 5]
Now that we have a 1D numpy array, let’s find the indexes where the element 5
occurs inside the array:
# find index of 5 result = np.where(arr==5) # print the result print("Index of 5:", result)
Output:
Index of 5: (array([1, 6, 9], dtype=int64),)
We get a tuple of numpy arrays as an output. Note that this tuple only has one numpy array storing the indices of occurrence of the element 5
inside the array.
If you were to use the np.where()
function on a multidimensional numpy array, the returned tuple would have multiple numpy arrays, one for each axis.
What if the element is not present in the array?
If the element is not present in the array we get an empty array with np.where()
. For example, let’s use it to find the index of 1
, an element that is not present in the above array arr
.
# index of 1 print(np.where(arr==1))
Output:
(array([], dtype=int64),)
You can see that the returned tuple contains an empty numpy array.
Index of the first occurrence of element
Since np.where()
returns all the indexes of the occurrence of an element. You can use it to find the index of the first occurrence. For example, let’s find the index of the first occurrence of 5
in the above array.
# the first index of 5 print("First index of 5:", np.where(arr==5)[0][0])
Output:
First index of 5: 1
From the previous examples, we know that 5 is present at indexes 1, 6, and 9 in the array arr. Here we get its first occurrence which is at index 1.
2. Index of element in 2D array
We can also use the np.where()
function to find the position/index of occurrences of elements in a two-dimensional or multidimensional array. For a 2D array, the returned tuple will contain two numpy arrays one for the rows and the other for the columns.
First, let’s create a two-dimensional numpy array.
# create a 2D numpy array arr = np.array([[21, 17, 19], [15, 23, 17], [17, 11, 16]]) # print the original array print("Original array:n", arr)
Output:
Original array: [[21 17 19] [15 23 17] [17 11 16]]
Here we created a 2D numpy array with three rows and three columns. Let’s find the indexes where 17 occurs inside this array.
# find index of 17 result = np.where(arr == 17) # print the result print("Index of 17:", result)
Output:
Index of 17: (array([0, 1, 2], dtype=int64), array([1, 2, 0], dtype=int64))
The returned tuple from np.where()
contains two numpy arrays. The first array values tell the row indexes whereas the second array values tell the column indexes of the occurrences of the element inside the 2D array.
Let’s make these indexes more readable by showing the (row, column) index tuples for each occurrence.
# get the position in the 2-d array print(list(zip(result[0], result[1])))
Output:
[(0, 1), (1, 2), (2, 0)]
The result shows that 17 occurs at the following locations – row 0 column 1, row 1 column 2, and row 2 column 0 with the index for rows and columns starting from 0.
What if the element is not present in the array?
If the element is not present in the 2D array. The returned tuple from np.where() will have two empty numpy arrays. For example, if we check for the index of 13, an element that is not present in the 2D array above –
print("Index of 13:", np.where(arr==13))
Output:
Index of 13: (array([], dtype=int64), array([], dtype=int64))
We get a tuple of two empty numpy arrays.
For more on the numpy where function, refer to its documentation.
With this, we come to the end of this tutorial. The code examples and results presented in this tutorial have been implemented in a Jupyter Notebook with a python (version 3.8.3) kernel having numpy version 1.18.5
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Tutorials on numpy arrays –
- How to sort a Numpy Array?
- Create Pandas DataFrame from a Numpy Array
- Different ways to Create NumPy Arrays
- Convert Numpy array to a List – With Examples
- Append Values to a Numpy Array
- Find Index of Element in Numpy Array
- Read CSV file as NumPy Array
- Filter a Numpy Array – With Examples
- Python – Randomly select value from a list
- Numpy – Sum of Values in Array
- Numpy – Elementwise sum of two arrays
- Numpy – Elementwise multiplication of two arrays
- Using the numpy linspace() method
- Using numpy vstack() to vertically stack arrays
- Numpy logspace() – Usage and Examples
- Using the numpy arange() method
- Using numpy hstack() to horizontally stack arrays
- Trim zeros from a numpy array in Python
- Get unique values and counts in a numpy array
- Horizontally split numpy array with hsplit()
-
Piyush is a data professional passionate about using data to understand things better and make informed decisions. He has experience working as a Data Scientist in the consulting domain and holds an engineering degree from IIT Roorkee. His hobbies include watching cricket, reading, and working on side projects.
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17 авг. 2022 г.
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Вы можете использовать следующие методы, чтобы найти положение индекса определенных значений в массиве NumPy:
Метод 1: найти все ценные позиции в индексе
np.where (x== value )
Метод 2: найти первую позицию индекса значения
np.where (x== value )[0][0]
Метод 3: найти первую позицию индекса нескольких значений
#define values of interest
vals = np.array([ value1 , value2 , value3 ])
#find index location of first occurrence of each value of interest
sorter = np.argsort (x)
sorter[np.searchsorted (x, vals, sorter=sorter)]
В следующих примерах показано, как использовать каждый метод на практике.
Метод 1: найти все ценные позиции в индексе
В следующем коде показано, как найти каждую позицию индекса, которая равна определенному значению в массиве NumPy:
import numpy as np
#define array of values
x = np.array([4, 7, 7, 7, 8, 8, 8])
#find all index positions where x is equal to 8
np.where (x== 8 )
(array([4, 5, 6]),)
Из вывода мы видим, что позиции индекса 4, 5 и 6 равны значению 8 .
Метод 2: найти первую позицию индекса значения
В следующем коде показано, как найти первую позицию индекса, равную определенному значению в массиве NumPy:
import numpy as np
#define array of values
x = np.array([4, 7, 7, 7, 8, 8, 8])
#find first index position where x is equal to 8
np.where (x== 8 )[0][0]
4
Из вывода мы видим, что значение 8 сначала встречается в позиции индекса 4.
Метод 3: найти первую позицию индекса нескольких значений
В следующем коде показано, как найти первую позицию индекса нескольких значений в массиве NumPy:
import numpy as np
#define array of values
x = np.array([4, 7, 7, 7, 8, 8, 8])
#define values of interest
vals = np.array([4, 7, 8])
#find index location of first occurrence of each value of interest
sorter = np.argsort (x)
sorter[np.searchsorted (x, vals, sorter=sorter)]
array([0, 1, 4])
Из вывода мы видим:
- Значение 4 сначала встречается в индексной позиции 0.
- Значение 7 сначала встречается в индексной позиции 1.
- Значение 8 сначала встречается в индексной позиции 4.
Дополнительные ресурсы
В следующих руководствах объясняется, как выполнять другие распространенные операции в NumPy:
Как сопоставить функцию с массивом NumPy
Как преобразовать массив NumPy в список в Python
Как рассчитать величину вектора с помощью NumPy
Use np.where
to get the indices where a given condition is True
.
Examples:
For a 2D np.ndarray
called a
:
i, j = np.where(a == value) # when comparing arrays of integers
i, j = np.where(np.isclose(a, value)) # when comparing floating-point arrays
For a 1D array:
i, = np.where(a == value) # integers
i, = np.where(np.isclose(a, value)) # floating-point
Note that this also works for conditions like >=
, <=
, !=
and so forth…
You can also create a subclass of np.ndarray
with an index()
method:
class myarray(np.ndarray):
def __new__(cls, *args, **kwargs):
return np.array(*args, **kwargs).view(myarray)
def index(self, value):
return np.where(self == value)
Testing:
a = myarray([1,2,3,4,4,4,5,6,4,4,4])
a.index(4)
#(array([ 3, 4, 5, 8, 9, 10]),)