Increasing the speed of finding an index of an item in list

Question

I'm trying to speed up a program I've written, and after importing cProfile, I see that one function takes up a massive bit of computation time.

It's this, which finds an numpy.ndarray in a list:

    def locate(arr, l ):
        for i in range(len(l)):
            if np.all(l[i] == arr):
                return i
        return -1

As the list can't be ordered etc, I can't see any way to avoid scanning the entire list. I have read some pieces on vectorisation, and I wanted to know if that could be applied here, or if there's any other way to speed this up?

Thanks

Answer 1

You probably cannot avoid walking the list but you can speed up the comparison:

Set up example:

L  = list(np.floor(np.outer(*2*(np.linspace(1,10,1000),))))
arr = L[537]

Direct method for reference:

import itertools as it

next(it.chain((i for i, a in enumerate(L) if np.all(arr==a)), (-1,)))
# 537
timeit(lambda: next(it.chain((i for i, a in enumerate(L) if np.all(arr==a)), (-1,))), number=100)
# 0.27100146701559424

Approach 1: Use np.array_equal (slower)

next(it.chain((i for i, a in enumerate(L) if np.array_equal(arr, a)), (-1,)))
# 537
timeit(lambda: next(it.chain((i for i, a in enumerate(L) if np.array_equal(arr, a)), (-1,))), number=100)
# 0.2992244770284742

Approach 2: Use void view (faster)

arr_v = arr.reshape(-1).view(f'V{arr.itemsize*arr.size}')

next(it.chain((i for i, a in enumerate(L) if arr_v==a.reshape(-1).view(f'V{a.itemsize*a.size}')), (-1,)))
# 537
timeit(lambda: next(it.chain((i for i, a in enumerate(L) if arr_v==a.reshape(-1).view(f'V{a.itemsize*a.size}')), (-1,))), number=100)
# 0.11853155982680619

Answer 2

有一个名为index()的内置python函数，您可以通过将字符串作为值插入并在列表中查找其索引来使用它。

Answer 3

So are you looking for np.where

temp_list=np.array(temp_list)
np.where(temp_list==5)
(array([1, 3, 6, 8]),)

Answer 4

How to Check if a Matrix is in a List of Matrices Python

Here the accepted answer uses np.array_equal which first checks shape, then does the all(==) test.

Another SO: Check if 2d array exists in 3d array in Python?

Searching an array for a value faster than np.where(ar==value) using fortran and f2py

Why isn't "numpy.any" lazy (short-circuiting)