0

I have a list of list like below: 

[[1,345,0,304],[0,345,678,946,90],[0,23,3,56,3,5,9,0]]

Now I want to append a new number (same number) to each of internal list at the end.

so result should be like: 

[[1,345,0,304,90],[0,345,678,946,90,90],[0,23,3,56,3,5,9,0,90]]

I want to use a list comprehension as I don't want to do it the normal way by iterating over each internal list in main list and then use a temporary list to add to it.

Any help?

Improve this question
1
  • 1
    You can go a long way in improving this question by clarifying your intent. You claim that you want to append to each list using a list comprehension, but you've selected an answer which creates a copy, appending to those lists instead of the lists you mention. If that's what you want, please edit your question to make that clear. Commented Sep 16, 2017 at 3:33

5 Answers 5

Reset to default
3

you can just use list addition to accomplish this:

[l + [90] for l in lists]
Improve this answer
Sign up to request clarification or add additional context in comments.

2 Comments

This is tremendously wasteful because it creates copies of the list. It doesn't, in fact, append each list. It creates brand new lists. Imagine the asymptotic case of a 1000x1000 list of lists.
@AryaMcCarthy yes it's wasteful, but if you read OP's question it's what he/she wanted.
1

Approaches

Native Python (@acushner presented this first):

lsts = [[1,345,0,304], [0,345,678,946,90], [0,23,3,56,3,5,9,0]]
[lst + [90] for lst in lsts]

Alternatively, with itertools.chain:

import itertools as it

[list(it.chain(lst, [90])) for lst in lsts]

For fun, a third-party library more-itertools (pip install more_itertools):

import more_itertools as mit

[list(mit.padded(lst, fillvalue=90, n=len(lst)+1)) for lst in lsts]

Caveat

Some answers attempt to mutate a list while iterating. While those options give equivalent results, are possibly more memory efficient and may even be faster for larger data for this specific problem, they are arguably non-pythonic and not recommended practices.

From the Python docs:

It is sometimes tempting to change a list while you are looping over it; however, it is often simpler and safer to create a new list instead.

This is especially true when removing or inserting elements from a list while iterating it. The latter approaches adopt this convention of creating a new list. However, for certain innocuous circumstances, a compromise may be iterating over a copy of the nested list:

lists = [[1,345,0,304], [0,345,678,946,90], [0,23,3,56,3,5,9,0]]
for lst in lists[:]:
    lst.append(90)
lists

Otherwise, default to @acushner's approach, which is the next performant option discussed here.

Improve this answer

2 Comments

This is tremendously wasteful because it creates copies of the list. It doesn't, in fact, append each list. It creates brand new lists. Imagine the asymptotic case of a 1000x1000 list of lists.
No copies are made. New lists are created. This is intentional and a common convention with list comprehensions in Python. What alternative do you suggest?
0

This also works

lists = [[1,345,0,304],[0,345,678,946,90],[0,23,3,56,3,5,9,0]]
[l.append(90) for l in lists]
print(lists)

[[1, 345, 0, 304, 90], [0, 345, 678, 946, 90, 90], [0, 23, 3, 56, 3, 5, 9, 0, 90]]

Improve this answer

6 Comments

This solution is best because it mutates the data structure rather than creating a new one. That being said, this is not how you should use list comprehensions. The question is best answered with a for-loop.
Could you explain why list comprehensions shouldn't be used for this?
@Serjik This is not correct. It won't work since append returns None
@Baktaawar It absolutely works, just print the result, None is because append does not return a value
@Arya McCarthy : as your comment the best, why any other answers get an upvote and me getting the down ???
|
0

And another one:

lists = [[1,345,0,304],[0,345,678,946,90],[0,23,3,56,3,5,9,0]]
for _ in map(lambda x: x.append(90), lists):
    pass
Improve this answer

Comments

0

Another option

lists = [[1,345,0,304], [0,345,678,946,90], [0,23,3,56,3,5,9,0]]

for lst in lists:
    lst.append(90)

print lists
[[1, 345, 0, 304, 90],[0, 345, 678, 946, 90, 90],[0, 23, 3, 56, 3, 5, 9, 0, 90]]
Improve this answer

4 Comments

Consider mutating a copy of the nested list (see post for rationale). Also for clarity, I would swap the names of lst and lists. Otherwise, this is a good answer.
Mutating the same list you are iterating is not a best-practice. In this case, it is harmless since you are not removing items, but a safer habit is to iterate a copy, e.g. lists[:]. There are several posts of this kind stackoverflow.com/questions/1637807/…. Just substitute for lst in lists[:]:.
@pylang he is not asking to iterate a copy
I was suggesting an idiom to improve your answer. See the cavaet I posted for more information. Thanks for your post.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.