Numpy Dynamic Slicing Per Row

How do I dynamically slice each row given a starting and ending index without using a for loop. I can do it with loop listed below, but it is way too slow for something where the x

Solution 1:

You can work with masked arrays:

import numpy as np

np.random.seed(100)

x = np.arange(0, 100)
x = x.reshape(20, 5)
s_idx = np.random.randint(0, 3, x.shape[0])
e_idx = np.random.randint(3, 6, x.shape[0])

# This is optional, reduce x to the minimum possible block
first_col, last_col = s_idx.min(), e_idx.max()
x = x[:, first_col:last_col]
s_idx -= first_col
e_idx -= first_col

col_idx = np.arange(x.shape[1])
# Mask elements out of range
mask = (col_idx < s_idx[:, np.newaxis]) | (col_idx >= e_idx[:, np.newaxis])
x_masked = np.ma.array(x, mask=mask)
print(x_masked)

Output:

[[0 1 2 3 --][5 6 7 8 9][10 11 12 13 14][-- -- 17 -- --][-- -- 22 -- --][25 26 27 28 --][-- -- 32 33 --][-- 36 37 38 --][-- -- 42 -- --][-- -- 47 -- --][-- -- 52 53 --][-- -- 57 58 --][-- 61 62 63 --][65 66 67 68 69][70 71 72 -- --][75 76 77 78 79][80 81 82 83 --][-- -- 87 88 --][90 91 92 93 94][-- 96 97 98 99]]

You can do most NumPy operations with a masked array, but if you still want the list of arrays you could do something like:

list_arrays = [row[~m] for row, m inzip(x, x_masked.mask)]
print(list_arrays)

Output:

[array([0, 1, 2, 3]),
 array([5, 6, 7, 8, 9]),
 array([10, 11, 12, 13, 14]),
 array([17]),
 array([22]),
 array([25, 26, 27, 28]),
 array([32, 33]),
 array([36, 37, 38]),
 array([42]),
 array([47]),
 array([52, 53]),
 array([57, 58]),
 array([61, 62, 63]),
 array([65, 66, 67, 68, 69]),
 array([70, 71, 72]),
 array([75, 76, 77, 78, 79]),
 array([80, 81, 82, 83]),
 array([87, 88]),
 array([90, 91, 92, 93, 94]),
 array([96, 97, 98, 99])]

Although in this case obviously you do not need to construct the intermediate masked array, you can just iterate through the rows of x and mask.