嵌套列表理解示例

Question

Problem

I have next list:

a = (('_a_1', '_a_2'), [0.3, 0.6]) 
b = (('_a_3', '_a_4'), [0.15, 0.56]) 
c = (('_a_5', '_a_6'), [0.22, 0.6])

l = [a, b, c]

What I wan't: choose first tuples of list and remove _a_ labels. Result:

[['1','2'],['3','4'],['5','6']]

---

What I tried

I know I can do it using list comprehension . But I can't comprehend it :D What I'm trying:

[ el.replace('_a_','') for m in l for el in m[0]]

result:

['1', '2', '3', '4', '5', '6']

I think it is closed to right solution. Just add something like:

[ [el.replace('_a_','') for m in l] for el in m[0]]

But it doesn't work...

Answer 1

Here is one way using str.split :

a = (('_a_1', '_a_2'), [0.3, 0.6]) 
b = (('_a_3', '_a_4'), [0.15, 0.56]) 
c = (('_a_5', '_a_6'), [0.22, 0.6])

res = [[i.split('_')[-1], j.split('_')[-1]] for (i, j), k in [a, b, c]]

# [['1', '2'], ['3', '4'], ['5', '6']]

You can also use str.replace :

res = [[i.replace('_a_', ''), j.replace('_a_', '')] for (i, j), k in [a, b, c]]

As suggested by @Ev.Kounis, you can also write this functionally:

res = [list(map(lambda x: x.rsplit('_', 1)[-1], k)) for k, _ in [a, b, c]]

Nested list comprehensions require specific ordering , intuitive to some but not to others. The above solutions avoid them altogether.

Answer 2

You were close:

a = (('_a_1', '_a_2'), [0.3, 0.6]) 
b = (('_a_3', '_a_4'), [0.15, 0.56]) 
c = (('_a_5', '_a_6'), [0.22, 0.6])

l = [a, b, c]

[ [el.replace('_a_','') for el in tup[0]] for tup in l]
# [['1', '2'], ['3', '4'], ['5', '6']]

Answer 3

res=[[j.replace("_a_","") for j in i[0]]  for i in l]
#OUT
[['1', '2'], ['3', '4'], ['5', '6']]

Answer 4

Another way of doing it would be to use the zip function:

a = (('_a_1', '_a_2'), [0.3, 0.6]) 
b = (('_a_3', '_a_4'), [0.15, 0.56]) 
c = (('_a_5', '_a_6'), [0.22, 0.6])

l = [[elem[0][-1],elem[1][-1]] for elem in list(zip(a,b,c))[0]]

print(l)

Output:

[['1', '2'], ['3', '4'], ['5', '6']]

Answer 5

You can go with regex without splitting it :

import re
pattern=r'[0-9]+'

print([tuple(map(lambda x:re.search(pattern,x).group(),i[0])) for i in l])

output:

[('1', '2'), ('3', '4'), ('5', '6')]