How to get one number specific times in an array python

Question

I have one array like A = [1,2,3] and another array B = [4,5,6]. Now, I need another array C so that the elements in C should be the same elements of B having occurrence in the order of element A. Like, C = [4, 5, 5, 6, 6, 6]

score 17 · Accepted Answer · edited May 23 '17 at 11:43

17

A = [1,2,3]
B = [4,5,6]
C = [b_item for a_item, b_item in zip(A,B) for _ in range(a_item)]
print C

Result:

[4, 5, 5, 6, 6, 6]

This is a one-line equivalent to:

C = []
for a_item, b_item in zip(A,B):
    for _ in range(a_item):
        C.append(b_item)

... Which is roughly equivalent to

C = []
for i in range(min(len(A), len(B))):
    a_item = A[i]
    b_item = B[i]
    for _ in range(a_item):
        C.append(b_item)

(N.B. Don't get tripped up by the underscore. It's an ordinary variable. It is conventionally used when you don't actually have to refer to the variable's value, as in this example)

edited May 23 '17 at 11:43

Community

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answered May 20 '15 at 19:02

Kevin

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Could you explain this for the.. less pythonic inclined people like me? I was thinking 2 for loops lol – Sterling Archer May 20 '15 at 19:03
@SterlingArcher, sure. In a way, I _am_ using two for loops. I just crammed both of them into one expression. – Kevin May 20 '15 at 19:09
Thanks. But as Sterling Archer told .... is there any simple logic to do this? Although, this works. – sparrow May 20 '15 at 19:11

score 6 · Answer 2 · answered May 20 '15 at 19:15

I will provide another way of doing it, using a different idea (not claiming he should use this, but for didactic reasons)

I find pretty neat to be able to replicate elements in python using

[0]*3 # it means we create 3 elements of zero: [0,0,0]

if I simply do this:

[[x]*i for i,x in zip(A, B)]

I get groups of elements:

[[4], [5, 5], [6, 6, 6]]

Then, we could use itertools.chain to get back to a single list:

from itertools import chain
list(chain(*[[x]*i for i,x in zip(A, B)]))
[4, 5, 5, 6, 6, 6]

score 5 · Answer 3 · answered May 20 '15 at 19:51

Here's another solution using numpy:

import numpy

a = numpy.array = [1,2,3]
b = numpy.array = [4,5,6]

c = numpy.repeat(b,a)

Result:

array([4, 5, 5, 6, 6, 6])

Here more information about using numpy.repeat : http://docs.scipy.org/doc/numpy/reference/generated/numpy.repeat.html

score 2 · Answer 4 · answered May 20 '15 at 19:12

2

My solution:

C = sum([[B[i]]*A[i] for i in range(len(A))], [])
print C

Explaination:

[B[i]]*A[i] will create a list with A[i] items of B[i]

[B[i]]*A[i] for i in range(len(A))] will give you a list of lists, eg [[4], [5, 5], [6, 6, 6]]

sum(C, []) will convert list of lists into list

answered May 20 '15 at 19:12

Tuan Anh Hoang-Vu

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Yep, this works. `sum(sequence, [])` is a little inefficient memory-wise since it has to create multiple intermediary lists, but that's not a big deal if C is reasonably small. – Kevin May 20 '15 at 19:14

score 1 · Answer 5 · answered May 20 '15 at 20:06

Another way:

C = []
for a, b in zip(A, B):
    C.extend([b] * a)

But I'd prefer the list comprehension, just without the annoying _item, i.e.,

C = [b for a, b in zip(A, B) for _ in range(a)]

Or the shorter (but for large cases slow) sum:

C = sum(([b] * a for a, b in zip(A, B)), [])

How to get one number specific times in an array python

5 Answers5