python – Is there a NumPy function to return the first index of something in an array?

The Question :

494 people think this question is useful

I know there is a method for a Python list to return the first index of something:

>>> l = [1, 2, 3]
>>> l.index(2)
1

Is there something like that for NumPy arrays?

The Question Comments :

The Answer 1

561 people think this answer is useful

Yes, given an array, array, and a value, item to search for, you can use np.where as:

itemindex = numpy.where(array==item)

The result is a tuple with first all the row indices, then all the column indices.

For example, if an array is two dimensions and it contained your item at two locations then

array[itemindex[0][0]][itemindex[1][0]]

would be equal to your item and so would be:

array[itemindex[0][1]][itemindex[1][1]]

The Answer 2

77 people think this answer is useful

If you need the index of the first occurrence of only one value, you can use nonzero (or where, which amounts to the same thing in this case):

>>> t = array([1, 1, 1, 2, 2, 3, 8, 3, 8, 8])
>>> nonzero(t == 8)
(array([6, 8, 9]),)
>>> nonzero(t == 8)[0][0]
6

If you need the first index of each of many values, you could obviously do the same as above repeatedly, but there is a trick that may be faster. The following finds the indices of the first element of each subsequence:

>>> nonzero(r_[1, diff(t)[:-1]])
(array([0, 3, 5, 6, 7, 8]),)

Notice that it finds the beginning of both subsequence of 3s and both subsequences of 8s:

[1, 1, 1, 2, 2, 3, 8, 3, 8, 8]

So it’s slightly different than finding the first occurrence of each value. In your program, you may be able to work with a sorted version of t to get what you want:

>>> st = sorted(t)
>>> nonzero(r_[1, diff(st)[:-1]])
(array([0, 3, 5, 7]),)

The Answer 3

55 people think this answer is useful

You can also convert a NumPy array to list in the air and get its index. For example,

l = [1,2,3,4,5] # Python list
a = numpy.array(l) # NumPy array
i = a.tolist().index(2) # i will return index of 2
print i

It will print 1.

The Answer 4

20 people think this answer is useful

Just to add a very performant and handy numba alternative based on np.ndenumerate to find the first index:

from numba import njit
import numpy as np

@njit
def index(array, item):
    for idx, val in np.ndenumerate(array):
        if val == item:
            return idx
    # If no item was found return None, other return types might be a problem due to
    # numbas type inference.

This is pretty fast and deals naturally with multidimensional arrays:

>>> arr1 = np.ones((100, 100, 100))
>>> arr1[2, 2, 2] = 2

>>> index(arr1, 2)
(2, 2, 2)

>>> arr2 = np.ones(20)
>>> arr2[5] = 2

>>> index(arr2, 2)
(5,)

This can be much faster (because it’s short-circuiting the operation) than any approach using np.where or np.nonzero.


However np.argwhere could also deal gracefully with multidimensional arrays (you would need to manually cast it to a tuple and it’s not short-circuited) but it would fail if no match is found:

>>> tuple(np.argwhere(arr1 == 2)[0])
(2, 2, 2)
>>> tuple(np.argwhere(arr2 == 2)[0])
(5,)

The Answer 5

14 people think this answer is useful

If you’re going to use this as an index into something else, you can use boolean indices if the arrays are broadcastable; you don’t need explicit indices. The absolute simplest way to do this is to simply index based on a truth value.

other_array[first_array == item]

Any boolean operation works:

a = numpy.arange(100)
other_array[first_array > 50]

The nonzero method takes booleans, too:

index = numpy.nonzero(first_array == item)[0][0]

The two zeros are for the tuple of indices (assuming first_array is 1D) and then the first item in the array of indices.

The Answer 6

12 people think this answer is useful

l.index(x) returns the smallest i such that i is the index of the first occurrence of x in the list.

One can safely assume that the index() function in Python is implemented so that it stops after finding the first match, and this results in an optimal average performance.

For finding an element stopping after the first match in a NumPy array use an iterator (ndenumerate).

In [67]: l=range(100)

In [68]: l.index(2)
Out[68]: 2

NumPy array:

In [69]: a = np.arange(100)

In [70]: next((idx for idx, val in np.ndenumerate(a) if val==2))
Out[70]: (2L,)

Note that both methods index() and next return an error if the element is not found. With next, one can use a second argument to return a special value in case the element is not found, e.g.

In [77]: next((idx for idx, val in np.ndenumerate(a) if val==400),None)

There are other functions in NumPy (argmax, where, and nonzero) that can be used to find an element in an array, but they all have the drawback of going through the whole array looking for all occurrences, thus not being optimized for finding the first element. Note also that where and nonzero return arrays, so you need to select the first element to get the index.

In [71]: np.argmax(a==2)
Out[71]: 2

In [72]: np.where(a==2)
Out[72]: (array([2], dtype=int64),)

In [73]: np.nonzero(a==2)
Out[73]: (array([2], dtype=int64),)

Time comparison

Just checking that for large arrays the solution using an iterator is faster when the searched item is at the beginning of the array (using %timeit in the IPython shell):

In [285]: a = np.arange(100000)

In [286]: %timeit next((idx for idx, val in np.ndenumerate(a) if val==0))
100000 loops, best of 3: 17.6 µs per loop

In [287]: %timeit np.argmax(a==0)
1000 loops, best of 3: 254 µs per loop

In [288]: %timeit np.where(a==0)[0][0]
1000 loops, best of 3: 314 µs per loop


This is an open NumPy GitHub issue.

See also: Numpy: find first index of value fast

The Answer 7

10 people think this answer is useful

For one-dimensional sorted arrays, it would be much more simpler and efficient O(log(n)) to use numpy.searchsorted which returns a NumPy integer (position). For example,

arr = np.array([1, 1, 1, 2, 3, 3, 4])
i = np.searchsorted(arr, 3)

Just make sure the array is already sorted

Also check if returned index i actually contains the searched element, since searchsorted’s main objective is to find indices where elements should be inserted to maintain order.

if arr[i] == 3:
    print("present")
else:
    print("not present")

The Answer 8

6 people think this answer is useful

To index on any criteria, you can so something like the following:

In [1]: from numpy import *
In [2]: x = arange(125).reshape((5,5,5))
In [3]: y = indices(x.shape)
In [4]: locs = y[:,x >= 120] # put whatever you want in place of x >= 120
In [5]: pts = hsplit(locs, len(locs[0]))
In [6]: for pt in pts:
   .....:         print(', '.join(str(p[0]) for p in pt))
4, 4, 0
4, 4, 1
4, 4, 2
4, 4, 3
4, 4, 4

And here’s a quick function to do what list.index() does, except doesn’t raise an exception if it’s not found. Beware — this is probably very slow on large arrays. You can probably monkey patch this on to arrays if you’d rather use it as a method.

def ndindex(ndarray, item):
    if len(ndarray.shape) == 1:
        try:
            return [ndarray.tolist().index(item)]
        except:
            pass
    else:
        for i, subarray in enumerate(ndarray):
            try:
                return [i] + ndindex(subarray, item)
            except:
                pass

In [1]: ndindex(x, 103)
Out[1]: [4, 0, 3]

The Answer 9

5 people think this answer is useful

For 1D arrays, I’d recommend np.flatnonzero(array == value)[0], which is equivalent to both np.nonzero(array == value)[0][0] and np.where(array == value)[0][0] but avoids the ugliness of unboxing a 1-element tuple.

The Answer 10

4 people think this answer is useful

An alternative to selecting the first element from np.where() is to use a generator expression together with enumerate, such as:

>>> import numpy as np
>>> x = np.arange(100)   # x = array([0, 1, 2, 3, ... 99])
>>> next(i for i, x_i in enumerate(x) if x_i == 2)
2

For a two dimensional array one would do:

>>> x = np.arange(100).reshape(10,10)   # x = array([[0, 1, 2,... 9], [10,..19],])
>>> next((i,j) for i, x_i in enumerate(x) 
...            for j, x_ij in enumerate(x_i) if x_ij == 2)
(0, 2)

The advantage of this approach is that it stops checking the elements of the array after the first match is found, whereas np.where checks all elements for a match. A generator expression would be faster if there’s match early in the array.

The Answer 11

4 people think this answer is useful

There are lots of operations in NumPy that could perhaps be put together to accomplish this. This will return indices of elements equal to item:

numpy.nonzero(array - item)

You could then take the first elements of the lists to get a single element.

The Answer 12

3 people think this answer is useful

The numpy_indexed package (disclaimer, I am its author) contains a vectorized equivalent of list.index for numpy.ndarray; that is:

sequence_of_arrays = [[0, 1], [1, 2], [-5, 0]]
arrays_to_query = [[-5, 0], [1, 0]]

import numpy_indexed as npi
idx = npi.indices(sequence_of_arrays, arrays_to_query, missing=-1)
print(idx)   # [2, -1]

This solution has vectorized performance, generalizes to ndarrays, and has various ways of dealing with missing values.

The Answer 13

-1 people think this answer is useful

Note: this is for python 2.7 version

You can use a lambda function to deal with the problem, and it works both on NumPy array and list.

your_list = [11, 22, 23, 44, 55]
result = filter(lambda x:your_list[x]>30, range(len(your_list)))
#result: [3, 4]

import numpy as np
your_numpy_array = np.array([11, 22, 23, 44, 55])
result = filter(lambda x:your_numpy_array [x]>30, range(len(your_list)))
#result: [3, 4]

And you can use

result[0]

to get the first index of the filtered elements.

For python 3.6, use

list(result)

instead of

result

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