Numpy equivalent of list.index

Question

In a low-level function that is called many times, I need to do the equivalent of python\'s list.index, but with a numpy array. The function needs to return when it finds the f

Answer 1

The closest thing I could find to what you're asking for is nonzero. It may sound odd, but the documentation makes it look like it might have the desired result.

http://www.scipy.org/Numpy_Example_List_With_Doc#nonzero

Specifically this part:

a.nonzero()

Return the indices of the elements that are non-zero.

Refer to numpy.nonzero for full documentation.

See Also

numpy.nonzero : equivalent function

>>> from numpy import *
>>> y = array([1,3,5,7])
>>> indices = (y >= 5).nonzero()
>>> y[indices]
array([5, 7])
>>> nonzero(y)                                # function also exists
(array([0, 1, 2, 3]),)

Where (http://www.scipy.org/Numpy_Example_List_With_Doc#where) may also be of interest to you.

Answer 2

You can code it in Cython and just import from a Python script. There is no need to migrate your entire project into Cython.

# paste into: indexing.pyx
def index(long[:] lst, long value):
    cdef int i
    for i in range(len(lst)):
        if lst[i] == value:
            return i
    raise ValueError

# import in your .py code
import pyximport
pyximport.install()
from indexing import index

# example
from numpy import zeros
a = zeros(10**6, int)
a[-1] = 1

index(a, 1)
Wall time: 6.07 ms
999999

index(a, 0)
Wall time: 38.1 µs
0

Answer 3

The only time I've had this problem, it was sufficient to cast the numpy array as a list:

a = numpy.arange(3)
print(list(a).index(2))

>>> 2

Answer 4

See my comment on the OP's question for caveats, but in general, I would do the following:

import numpy as np
a = np.array([1, 2, 3])
np.min(np.nonzero(a == 2)[0])

if the value you are looking for is not in the array, you'll get a ValueError due to:

ValueError: zero-size array to ufunc.reduce without identity

because you are trying to take the min value of an empty array.

I would profile this code and see if it is an actual bottleneck, because in general when numpy searches through an entire array using a built-in function rather than an explicit python loop, it is relatively fast. An insistence on halting the search when it finds the first value may be functionally irrelevant.

Answer 5

NumPy's searchsorted is very similar to lists's index, except that it requires a sorted array and behaves more numerically. The big differences are that you don't need to have an exact match, and you can search starting from either the left or right sides. See the following examples to get an idea how it works:

import numpy as np
a = np.array([10, 20, 30])

a.searchsorted(-99) == a.searchsorted(0) == a.searchsorted(10)
# returns index 0 for value 10

a.searchsorted(20.1) == a.searchsorted(29.9) == a.searchsorted(30)
# returns index 2 for value 30

a.searchsorted(30.1) == a.searchsorted(99) == a.searchsorted(np.nan)
# returns index 3 for undefined value

With the last case, where an index of 3 is returned, you can handle this as you like. I gather from the name and intention of the function that it stops after finding the first suitable index.

Answer 6

If your numpy array is 1d array, maybe try like this:

a = np.array([1, 2, 3])
print a.tolist().index(2)
>>> 1

If is not 1d, you could search trough array like:

a = np.array([[1, 2, 3],[2,5,6],[0,0,2]])
print a[0,:].tolist().index(2)
>>> 1

print a[1,:].tolist().index(2)
>>> 0

print a[2,:].tolist().index(2)
>>> 2