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

Question

For python dict, I could use iteritems() to loop through key and value at the same time. But I cannot find such functionality for NumPy array. I have to manually track idx like this:

idx = 0  for j in theta:    some_function(idx,j,theta)    idx += 1 

Is there a better way to do this?

Answer 1

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