How to add list of lists in a list comprehension

Question

I have a list which is produced by a list comprehension and it sorts the data in stripped according to groups by finding which strings have a length of 3 and I want to merge them so that are in a single list separately from single length strings.

stripped = ['a,b', 'c,d', 'e', '', 'f,g', 'h', '', '']
lst = [[i.split(',')] if len(i) is 3 else i for i in stripped]
print(lst)
#[[['a', 'b']], [['c', 'd']], 'e', '', [['f', 'g']], 'h', '', '']

I want to produce [[['a', 'b'], ['c', 'd'],['f', 'g']], 'e', '','h', '', ''] instead

How can I achieve this with a Single list-comprehension if possible?

Edit:

accepted @HennyH's answer because of its high efficiency and simplicity

@GamesBrainiac just what I was afraid of... BTW I'll move on with that If I *must* — K DawG, Oct 27 '13 at 07:51
@hcwhsa It might use a whole lot of resources, my actual list is so damn large that running it over twice is expensive! — K DawG, Oct 27 '13 at 07:55

HennyH · Accepted Answer · 2013-10-27T08:39:17.463

4

l = [[]]
for e in stripped:
    (l[0] if len(e) == 3 else l).append(e)
>>> 
[['a,b', 'c,d', 'f,g'], 'e', '', 'h', '', '']

Or to match the OP's output for the 3 long strings:

for e in stripped:
    l[0].append(e.split(',')) if len(e) == 3 else l.append(e)
>>> 
[[['a', 'b'], ['c', 'd'], ['f', 'g']], 'e', '', 'h', '', '']

This way there is no additional concatenation of two lists A,B as Inbar's solution provided. You can also turn stripped into a generator expression so you don't need to hold the two lists in memory.

edited Oct 27 '13 at 08:39

answered Oct 27 '13 at 08:04

HennyH

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Good But actually it should be: `(l[0] if len(e) == 3 else l).append(e.split(','))` – Grijesh Chauhan Oct 27 '13 at 08:28
@HennyH but wait that will outputs `[[['a', 'b'], ['c', 'd'], ['f', 'g']], ['e'], [''], ['h'], [''], ['']]` that is not exactly same as OP wants – Grijesh Chauhan Oct 27 '13 at 08:31
@GrijeshChauhan fixed it up – HennyH Oct 27 '13 at 08:43
@HennyH Really Nice answer, Very Good :) – Grijesh Chauhan Oct 27 '13 at 19:21

score 2 · Answer 2 · answered Oct 27 '13 at 07:49

Use two list comprehensions:

>>> stripped = ['a,b', 'c,d', 'e', '', 'f,g', 'h', '', '']
>>> first = [x.split(',') for x in (item for item in stripped if len(item) == 3)]
>>> second = [item for item in stripped if len(item) != 3]
>>> [first] + second
[[['a', 'b'], ['c', 'd'], ['f', 'g']], 'e', '', 'h', '', '']

Inbar Rose · Answer 3 · 2013-10-27T08:02:22.060

2

Why require list comprehension? Better to do it in 1 pass.

stripped = ['a,b', 'c,d', 'e', '', 'f,g', 'h', '', '']
groups = []
final = [groups]
for item in stripped:
    if len(item) == 3:
        groups.append(item.split(','))
    else:
        final.append(item)

Result:

>>> print(final) 
[[['a', 'b'], ['c', 'd'], ['f', 'g']], 'e', '', 'h', '', '']

edited Oct 27 '13 at 08:02

answered Oct 27 '13 at 07:53

Inbar Rose

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You could even add with `B = [A]` before the loop, and fully meet the single-pass requirement. (The final `[A]+B` is technically a *pass* over `B`, although a very efficient one.) – user4815162342 Oct 27 '13 at 08:00
@user4815162342 That is true. – Inbar Rose Oct 27 '13 at 08:01

How to add list of lists in a list comprehension

Edit:

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