6

Perhaps this has been raised and addressed somewhere else but I haven't found it. Suppose we have a numpy array: 

a = np.arange(100).reshape(10,10)
b = np.zeros(a.shape)
start = np.array([1,4,7])   # can be arbitrary but valid values
end = np.array([3,6,9])     # can be arbitrary but valid values

start and end both have valid values so that each slicing is also valid for a. I wanted to copy value of subarrays in a to corresponding spots in in b:

b[:, start:end] = a[:, start:end]   #error

this syntax doesn't work, but it's equivalent to: 

b[:, start[0]:end[0]] = a[:, start[0]:end[0]]
b[:, start[1]:end[1]] = a[:, start[1]:end[1]]
b[:, start[2]:end[2]] = a[:, start[2]:end[2]]

I wonder if there is a better way of doing this instead of an explicit for-loop over the start and end arrays.

Thanks!

galactica
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  • Would the start and end pairs always have a constant difference (2 here)? – Divakar Oct 13 '17 at 16:20
  • not necessarily, this is just an example, but can assume the indices are valid ones to index with – galactica Oct 13 '17 at 16:22
  • Does this answer your question? [How to slice numpy rows using start and end index](https://stackoverflow.com/questions/49188161/how-to-slice-numpy-rows-using-start-and-end-index) – Ta946 Oct 23 '21 at 13:13

1 Answers1

6

We can use broadcasting to create a mask of places to be edited with two sets of comparisons against start and end arrays and then simply assign with boolean-indexing for a vectorized solution -

# Range array for the length of columns
r = np.arange(b.shape[1])

# Broadcasting magic to give us the mask of places
mask = (start[:,None] <= r) & (end[:,None] >= r)

# Boolean-index to select and assign 
b[:len(mask)][mask] = a[:len(mask)][mask]

Sample run -

In [222]: a = np.arange(50).reshape(5,10)
     ...: b = np.zeros(a.shape,dtype=int)
     ...: start = np.array([1,4,7])
     ...: end = np.array([5,6,9]) # different from sample for variety
     ...: 

# Mask of places to be edited
In [223]: mask = (start[:,None] <= r) & (end[:,None] >= r)

In [225]: print mask
[[False  True  True  True  True  True False False False False]
 [False False False False  True  True  True False False False]
 [False False False False False False False  True  True  True]]

In [226]: b[:len(mask)][mask] = a[:len(mask)][mask]

In [227]: a
Out[227]: 
array([[ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9],
       [10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
       [20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
       [30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
       [40, 41, 42, 43, 44, 45, 46, 47, 48, 49]])

In [228]: b
Out[228]: 
array([[ 0,  1,  2,  3,  4,  5,  0,  0,  0,  0],
       [ 0,  0,  0,  0, 14, 15, 16,  0,  0,  0],
       [ 0,  0,  0,  0,  0,  0,  0, 27, 28, 29],
       [ 0,  0,  0,  0,  0,  0,  0,  0,  0,  0],
       [ 0,  0,  0,  0,  0,  0,  0,  0,  0,  0]])
Divakar
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  • Thanks! This mask-trick looks neat! But does it cause any efficiency issues, i.e., making copy of the whole ndarray in this example? In the practical scenario, I have a pretty big ndarray, e.g, a.shape=(128, 32x32x32) and such value assignment would be repeated multiple times, depending on different start/end values at each iteration – galactica Oct 13 '17 at 16:43
  • @galactica Which copy are we referring to? Which step are you referring to? – Divakar Oct 13 '17 at 16:44
  • @galactica If you are talking about the copy at : `a[:len(mask)][mask]`, that can't be avoided with masking and you can't slice the whole thing in a vectorized manner. So, if memory efficiency is a concern, i guess keep the loop and use something like : `b[0, start[0]:end[0]] = a[0, start[0]:end[0]]` and so on. – Divakar Oct 13 '17 at 17:36
  • sounds great! Thanks for the detailed explanations! Let's wait for a bit to see if there are other answers – galactica Oct 13 '17 at 18:00
  • What is 'r' in this example? – Captain Trojan Sep 06 '20 at 11:42
  • @CaptainTrojan As listed in the code `r = np.arange(b.shape[1])`? – Divakar Sep 06 '20 at 13:38
  • @Divakar My bad, I'm blind apparently. – Captain Trojan Sep 07 '20 at 10:18