使用索引数组（Python）将numpy数组划分为多个数组

Question

I have an array:

a = [1, 3, 5, 7, 29 ... 5030, 6000]

This array gets created from a previous process, and the length of the array could be different (it is depending on user input).

I also have an array:

b = [3, 15, 67, 78, 138]

(Which could also be completely different)

I want to use the array b to slice the array a into multiple arrays.

More specifically, I want the result arrays to be:

array1 = a[:3]
array2 = a[3:15]
...
arrayn = a[138:]

Where n = len(b) .

My first thought was to create a 2D array slices with dimension (len(b), something) . However we don't know this something beforehand so I assigned it the value len(a) as that is the maximum amount of numbers that it could contain.

I have this code:

 slices = np.zeros((len(b), len(a)))

 for i in range(1, len(b)):
     slices[i] = a[b[i-1]:b[i]]

But I get this error:

ValueError: could not broadcast input array from shape (518) into shape (2253412)

Answer 1

You can use numpy.split :

np.split(a, b)

Example :

np.split(np.arange(10), [3,5])
# [array([0, 1, 2]), array([3, 4]), array([5, 6, 7, 8, 9])]

Answer 2

b.insert(0,0)
result = []
for i in range(1,len(b)):
    sub_list = a[b[i-1]:b[i]]
    result.append(sub_list)
result.append(a[b[-1]:])

Answer 3

You are getting the error because you are attempting to create a ragged array. This is not allowed in numpy.

An improvement on @Bohdan's answer:

from itertools import zip_longest
result = [a[start:end] for start, end in zip_longest(np.r_[0, b], b)]

The trick here is that zip_longest makes the final slice go from b[-1] to None , which is equivalent to a[b[-1]:] , removing the need for special processing of the last element.

Please do not select this. This is just a thing I added for fun. The "correct" answer is @Psidom's answer.