Use list comprehension to replace nested for loop

Question

This nested for loop accomplishes my goal but I would like to do it using list comprehension and/or lambda func

new_a=[]
new_b=[]
a=[1,2,3]
b=[1,2,3]

for num_a in a:
    for num_b in b:
        new_a.append(num_a)
        new_b.append(num_b)

print(new_a)
print(new_b)

output:

new_a=[1, 1, 1, 2, 2, 2, 3, 3, 3]

new_b=[1, 2, 3, 1, 2, 3, 1, 2, 3]

I can get new_a with [num for num in a for num in b]

but can't figure out how to get new_b using listcomp or lambda

Agree with @MarkMeyer; sometimes you don’t need comprehension/loop/etc. Similar for new_a, if you wanted to use numpy: np.repeat([1,2,3], 3). — s3dev
– s3dev, Commented Sep 7, 2020 at 20:07
@S3DEV I'd be hesitant to include something as heavy weight as numpy for this unless it was already in the project, but generally agree. — flakes
– flakes, Commented Sep 7, 2020 at 20:10

flakes · Accepted Answer · 2020-09-07 20:01:13Z

1

What you really want to do is just use the length of the one array to perform the operation on the other. You could use for _ in range(len(old)), but just iterating over the list and ignoring the output performs the same thing. The trick here is the arrangement of for loops

old_a = [1, 2, 3]
old_b = [1, 2, 3]

new_a = [a for a in old_a for _ in old_b]
new_b = [b for _ in old_a for b in old_b]

print(new_a)
print(new_b)

Output

[1, 1, 1, 2, 2, 2, 3, 3, 3]
[1, 2, 3, 1, 2, 3, 1, 2, 3]

answered Sep 7, 2020 at 20:01

flakes

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Comments

Vasilis G. · Accepted Answer · 2020-09-07 20:03:15Z

1

You can use zip along with map function to do that:

new_a, new_b = map(list,zip(*[[num_a, num_b] for num_a in a for num_b in b]))

print(new_a)
print(new_b)

This will return:

[1, 1, 1, 2, 2, 2, 3, 3, 3]
[1, 2, 3, 1, 2, 3, 1, 2, 3]

answered Sep 7, 2020 at 20:03

Vasilis G.

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Comments

Georgina Skibinski · Accepted Answer · 2020-09-07 20:14:33Z

0

How about:

a=[1,2,3]
b=[1,2,3]

new_a=list(el for el1 in zip(* [a]*len(b)) for el in el1)
new_b=b*len(a)

Outputs:

>>> new_a
[1, 1, 1, 2, 2, 2, 3, 3, 3] 

>>> new_b
[1, 2, 3, 1, 2, 3, 1, 2, 3]

answered Sep 7, 2020 at 20:14

Georgina Skibinski

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