Loop a list to output each element twice with different prefix

Question

A simple list, that I want to loop through to print each element twice. Each time to add a different prefix. The output will be appended into a new list.

List1 = ["9016","6416","9613"]

The ideal result is:

['AB9016', 'CD9016', 'AB6416', 'CD6416', AB9613', 'CD9613']

I have tried below and but the output is:

new_list = []

for x in List1:
    for _ in [0,1]:
        new_list.append("AB" + x)
        new_list.append("CD" + x)

print (new_list)

['AB9016', 'CD9016', 'AB9016', 'CD9016', 'AB6416', 'CD6416', 'AB6416', 'CD6416', 'AB9613', 'CD9613', 'AB9613', 'CD9613']

And I can't use:

new_list.append("AB" + x).append("CD" + x)

What's the proper way to do it? Thank you.

kol · Accepted Answer · 2020-07-20T13:57:37.640

4

The problem is caused by the inner loop: the two appends will be called twice. Fixed code:

new_list = []

for x in List1:
    new_list.append("AB" + x)
    new_list.append("CD" + x)

Regarding chaining append calls: It would work if append returned the list (with the new item appended to it), but this is not the case, the append method returns None (doc).

edited Jul 20 '20 at 13:57

answered Jul 20 '20 at 13:45

kol

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thank you for pointing out the problem. Glad to choose yours as the answer for its simplicity. – Mark K Jul 20 '20 at 14:02

score 4 · Answer 2 · answered Jul 20 '20 at 13:48

4

Also could try an easy comprehension:

List1 = ["9016","6416","9613"]
result = [j+i for i in List1 for j in ('AB','CD')]
# ['AB9016', 'CD9016', 'AB6416', 'CD6416', 'AB9613', 'CD9613']

answered Jul 20 '20 at 13:48

jizhihaoSAMA

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score 2 · Answer 3 · answered Jul 20 '20 at 13:42

The fastest method is to use the list comprehension. We are using 2 list comprehension to create the desired_list. Note that I also used the f string so I could easily ad the 'ABandCD` prefix.

list1 = ["9016","6416","9613"]
desired_list = [f'AB{x}' for x in list1] + [f'CD{x}' for x in list1]
print(desired_list)

score 2 · Answer 4 · answered Jul 20 '20 at 13:52

I would use itertools.product for that task following way

import itertools
list1 = ["9016","6416","9613"]
prefixes = ["AB","CD"]
result = [x+y for y,x in itertools.product(list1,prefixes)]
print(result)

Output:

['AB9016', 'CD9016', 'AB6416', 'CD6416', 'AB9613', 'CD9613']

score 2 · Answer 5 · answered Jul 20 '20 at 13:52

2

Here is a solution using itertools.product:

from itertools import product

lst1 = ['9016', '6416', '9613']
lst2 = ['AB', 'CD']

result = list(map(''.join, map(reversed, product(lst1, lst2))))

answered Jul 20 '20 at 13:52

Riccardo Bucco

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score 1 · Answer 6 · answered Jul 20 '20 at 14:09

1

we can use sum too:

In [25]: sum([[f'AB{i}',f'CD{i}'] for i in List1],[])
Out[25]: ['AB9016', 'CD9016', 'AB6416', 'CD6416', 'AB9613', 'CD9613']

answered Jul 20 '20 at 14:09

Akhilesh_IN

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Loop a list to output each element twice with different prefix

6 Answers6