How to iterate 1d NumPy array with index and value [duplicate]

There are a few alternatives. The below assumes you are iterating over a 1d NumPy array.

Iterate with range

for j in range(theta.shape[0]):  # or range(len(theta))   some_function(j, theta[j], theta)

Note this is the only of the 3 solutions which will work with numba. This is noteworthy since iterating over a NumPy array explicitly is usually only efficient when combined with numba or another means of pre-compilation.

Iterate with enumerate

for idx, j in enumerate(theta):   some_function(idx, j, theta)

The most efficient of the 3 solutions for 1d arrays. See benchmarking below.

Iterate with np.ndenumerate

for idx, j in np.ndenumerate(theta):   some_function(idx[0], j, theta)

Notice the additional indexing step in idx[0]. This is necessary since the index (like shape) of a 1d NumPy array is given as a singleton tuple. For a 1d array, np.ndenumerate is inefficient; its benefits only show for multi-dimensional arrays.

Performance benchmarking

# Python 3.7, NumPy 1.14.3np.random.seed(0)arr = np.random.random(10**6)def enumerater(arr):    for index, value in enumerate(arr):        index, value        passdef ranger(arr):    for index in range(len(arr)):        index, arr[index]        passdef ndenumerater(arr):    for index, value in np.ndenumerate(arr):        index[0], value        pass%timeit enumerater(arr)    # 131 ms%timeit ranger(arr)        # 171 ms%timeit ndenumerater(arr)  # 579 ms

You can use numpy.ndenumerate for example

import numpy as nptest_array = np.arange(2, 3, 0.1)for index, value in np.ndenumerate(test_array):    print(index[0], value)

For more information refer to https://docs.scipy.org/doc/numpy/reference/generated/numpy.ndenumerate.html