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 1.14.3

np.random.seed(0)

arr = np.random.random(10**6)

def enumerater(arr):
    for index, value in enumerate(arr):
        index, value
        pass

def ranger(arr):
    for index in range(len(arr)):
        index, arr[index]
        pass

def 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

As I've become more familiar with the NumPy and general scientific array processing ecosystem, it's become clearer to me that iteration is usually outright wrong due to how slow it is in comparison, and writing to use a vectorized operation is best whenever possible. Though the style is not as obvious/Pythonic at first, I've (anecdotally) gained ridiculous speedups with vectorized operations; more than 1000x in a case of swapping out a form like some row iteration .apply(lambda)

@MSeifert's answer much better provides this and will be significantly more performant on a dataset of any real size

More general Answer by @cs95 covering and comparing alternatives to iteration in Pandas

Original Answer

You can iterate through the values in your array with numpy.ndenumerate to get the indices of the values in your array.

Using the documentation above:

A = np.array([[1,2,3],[4,5,6],[7,8,9]])
for index, values in np.ndenumerate(A):
    print(index, values)  # operate here

I think you're looking for the ndenumerate.

>>> import numpy as np
>>> a = np.array([[1,2],[3,4],[5,6]])
>>> for (i, j), value in np.ndenumerate(a):
...     print(i, j)
... 
0 0
0 1
1 0
1 1
2 0
2 1

Regarding the performance. It is a bit slower than a list comprehension.

X = np.zeros((100, 100, 100))

%timeit list([((i, j, k), X[i, j, k]) for i in range(X.shape[0]) for j in range(X.shape[1]) for k in range(X.shape[2])])
1 loop, best of 3: 376 ms per loop

%timeit list(np.ndenumerate(X))
1 loop, best of 3: 570 ms per loop

If you are worried about the performance you could optimise a bit further by looking at the implementation of ndenumerate, which does 2 things, converting to an array and looping. If you know you have an array, you can call the .coords attribute of the flat iterator.

a = X.flat
%timeit list([(a.coords, x) for x in a.flat])
1 loop, best of 3: 305 ms per loop