Python - How can I create a new list based on the values of another list?

Question

Newbie question: I have 2 lists:

# The first one is a list of lists:
list_a = [[1,5,3],[4,2],[2,3,3,5],[2,3,1]]
# The second one is a list of strings:
list_b = ['a','b','c','d','e']

The question is: How can I create a new list list_c, that looks like this?

list_c = [['a','e','c'], ['d','b'], ['b','c','c','e'],['b','c','a']]

So basically, what I need to do is some sort of mapping out the values of list_a which represents the indices of list_b.

I tried doing nested for loops, but I'm really confused.

Python indexes lists (etc) starting from 0, so it would be better if you worked in zero-based indices if possible, rather than having to subtract 1 every time. e.g. list_a = [[0,4,2],... I am not suggesting that you edit the question now, as this would break answers that have already been posted, but something to consider in future. — alani
– alani, Commented Jul 5, 2020 at 22:16

Dhaval Taunk · Accepted Answer · 2020-06-25 17:18:35Z

2

You can try this:-

list_c = [[list_b[i-1] for i in j] for j in list_a]
print(list_c)

Output:-

[['a', 'e', 'c'], ['d', 'b'], ['b', 'c', 'c', 'e'], ['b', 'c', 'a']]

answered Jun 25, 2020 at 17:18

Dhaval Taunk

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Brian61354270 · Accepted Answer · 2020-06-25 17:18:42Z

0

If you're not concerned about index errors, you can express this concisely using a nested list comprehension:

list_c = [[list_b[i-1] for i in sublist] for sublist in list_a]

answered Jun 25, 2020 at 17:18

Brian61354270

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Balaji Ambresh · Accepted Answer · 2020-06-25 17:20:18Z

0

Here you go:

list_c = [
    [list_b[item - 1] for item in inner_list] for inner_list in list_a
]
print(list_c)

Output:

[['a', 'e', 'c'], ['d', 'b'], ['b', 'c', 'c', 'e'], ['b', 'c', 'a']]

answered Jun 25, 2020 at 17:20

Balaji Ambresh

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Leonardo Gasparini · Accepted Answer · 2020-06-25 17:31:07Z

0

You need to do something like that:

for index, item in enumerate(list_a):
 for list_index, list_item in enumerate(list_a[index]):
    list_a[index][list_index] = None
    list_a[index][list_index] = list_b[list_item]

print(list_a)

Output:

[['b', 'f', 'd'], ['e', 'c'], ['c', 'd', 'd', 'f'], ['c', 'd', 'b']]

answered Jun 25, 2020 at 17:31

Leonardo Gasparini

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Red · Accepted Answer · 2020-07-05 21:35:30Z

There's a module for that: operator. You can use operator.itemgetter():

from operator import itemgetter as ig

list_a = [[1,5,3],[4,2],[2,3,3,5],[2,3,1]]
list_b = ['a','b','c','d','e']
list_c = [ig(*[i-1 for i in l])(list_b) for l in list_a]

print(list_c)

Output:

[('a', 'e', 'c'), ('d', 'b'), ('b', 'c', 'c', 'e'), ('b', 'c', 'a')]

It results in a list of tuples, which can be adjusted by adding list() to ig(*[i-1 for i in l])(list_b).

This would of course be more neat if the indices in list_a were not shifted by one unit:

from operator import itemgetter as ig

list_a = [[0,4,2],[3,1],[1,2,2,4],[1,2,0]]
list_b = ['a','b','c','d','e']
list_c = [ig(*l)(list_b) for l in list_a]

print(list_c)

Output:

[('a', 'e', 'c'), ('d', 'b'), ('b', 'c', 'c', 'e'), ('b', 'c', 'a')]

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