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]
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5 Answers
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)
3 Comments
Sterling Archer
Could you explain this for the.. less pythonic inclined people like me? I was thinking 2 for loops lol
Kevin
@SterlingArcher, sure. In a way, I am using two for loops. I just crammed both of them into one expression.
sparrow
Thanks. But as Sterling Archer told .... is there any simple logic to do this? Although, this works.
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]
Comments
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
Comments
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
1 Comment
Kevin
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.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)), [])
