grouping list elements in python

Question

list = [('a5', 1), 1, ('a1', 1), 0, 0]

I want to group the elements of the list into 3, if the second or third element is missing in the list 'None' has to appended in the corresponding location.

exepected_output = [[('a5', 1), 1,None],[('a1', 1), 0, 0]]

Is there a pythonic way for this? New to this, any suggestions would be helpful.

Here is how you split a [list into chunks](http://stackoverflow.com/questions/312443/how-do-you-split-a-list-into-evenly-sized-chunks) — Brendan Abel, Sep 23 '16 at 16:29
As a useful aside - where do you get this data *from*? It's highly likely that there's a better way to get your data out of its native state into the state that you're looking for. — Wayne Werner, Sep 23 '16 at 16:31
Why you shouldn't use `list` as a name is because it shadows the built-in class `list`. This prevents you from accessing the class and can lead to unwanted results or errors later on in your program. — Ted Klein Bergman, Sep 23 '16 at 16:52

score 2 · Answer 1 · answered Sep 23 '16 at 17:02

Here's a slightly different approach from the other answers, doing a comparison on the type of each element and then breaking the original list into chunks.

li = [('a5', 1), 1, ('a1', 1), 0, 0]

for i in range(0, len(li), 3):
    if type(li[i]) is not tuple:
        li.insert(i, None)
    if type(li[i+1]) is not int:
        li.insert(i+1, None)
    if type(li[i+2]) is not int:
        li.insert(i+2, None)

print [li[i:i + 3] for i in range(0, len(li), 3)]

trelltron · Accepted Answer · 2016-09-23T16:48:38.120

1

As far as I am aware, the only way to get the result you want is to loop through your list and detect when you encounter tuples.

Example which should work:

temp = None
result = []
for item in this_list:
    if type(item) == tuple:
        if temp is not None:
            while len(temp) < 3:
                temp.append(None)
            result.append(temp)
        temp = []
    temp.append(item)

Edit: As someone correctly commented, don't name a variable list, you'd be overwriting the built in list function. Changed name in example.

edited Sep 23 '16 at 16:48

answered Sep 23 '16 at 16:39

trelltron

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1

last temp never gets added to result. In the rare case that the OP actually gives you a good test case, use it! Preferably before you post the answer. :) expected_output = [[('a5', 1), 1,None],[('a1', 1), 0, 0]] assert(result == expected_output) – Kenny Ostrom Sep 23 '16 at 16:53

grouping list elements in python

2 Answers2