How would you iterate through a list of lists, such as:
[[1,2,3,4], [5,6], [7,8,9]]
and construct a new list by grabbing the first item of each list, then the second, etc. So the above becomes this:
[1, 5, 7, 2, 6, 8, 3, 9, 4]
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5 Answers
You can use a list comprehension along with itertools.izip_longest (or zip_longest in Python 3)
from itertools import izip_longest
a = [[1,2,3,4], [5,6], [7,8,9]]
[i for sublist in izip_longest(*a) for i in sublist if i is not None]
# [1, 5, 7, 2, 6, 8, 3, 9, 4]
8 Comments
Padraic Cunningham
You will lose any 0's, you should check for None
9000
Your example would be even better if you renamed
j in a more readable way, like result_with_holes or something.
pp_
In Python 3 the name has changed to
zip_longest.
Julien Spronck
@9000 Changed it. Thanks :-)
Paul
If you want to be fancy, this is equivalent to
list(filter(lambda x: x is not None, chain(*zip_longest(*a)))) |
As long as you know that None will not appear, you can take advantage of how map() works with no function:
outlist = [y for sub in map(None, *inlist) for y in sub if not y is None]
Comments
DRY: itertools has a recipe that sounds right for this task: roundrobin
from itertools import cycle, islice
def roundrobin(*iterables):
"roundrobin('ABC', 'D', 'EF') --> A D E B F C"
# Recipe credited to George Sakkis
pending = len(iterables)
nexts = cycle(iter(it).next for it in iterables)
while pending:
try:
for next in nexts:
yield next()
except StopIteration:
pending -= 1
nexts = cycle(islice(nexts, pending))
l = [[1,2,3,4], [5,6], [7,8,9]]
print (list(roundrobin(*l)))
#[1, 5, 7, 2, 6, 8, 3, 9, 4]
4 Comments
chapelo
You are right, @pp_, python 2 uses the attribute
next and python 3 uses __next__.Paul
I think
next(iter(it)) would work for both, right?
Kevin
@Paul: Yes, but then you'd need to use
functools.partial or a lambda, in this code.You can do it manually without using any import if you don't want to.
Keep N different counters where n is the number of lists in your list and increment them whenever you add a new element from that sublists. While adding them you should control that respective sets counter must be the minimum of all other counters if all lists have remaining elements.When you add the last element of list you can increment its counter to something like 9999 to protect our minimum rule.
This will be harder to implement and I do not suggest you to implement in this way but this is also a possibility if you do not want to import anything or want to have a programming challenge for your level.
Comments
Solutions that filter out None won't work if None is a value from the sublists that you would like to keep. To fix this, you can do a list comprehension that iterates over the indices of the longest sublist, and only adds sublist items if the index is in range.
a = [[1,2,3,4],[5,6,None],[7,8,9]]
range_longest = range(max(map(len, a)))
[sublist[i] for i in range_longest for sublist in a if i < len(sublist)]
# [1, 5, 7, 2, 6, 8, 3, None, 9, 4]
