Python split for lists

If we have a list of strings in python and want to create sublists based on some special string how should we do?

For instance:

l = ["data","more data","","data 2","more data 2","danger","","date3","lll"]
p = split_special(l,"")

would generate:

p = [["data","more data"],["data 2","more data 2","danger"],["date3","lll"]]

Answers:

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 1

itertools.groupby is one approach (as it often is):

>>> l = ["data","more data","","data 2","more data 2","danger","","date3","lll"]
>>> from itertools import groupby
>>> groupby(l, lambda x: x == "")
<itertools.groupby object at 0x9ce06bc>
>>> [list(group) for k, group in groupby(l, lambda x: x == "") if not k]
[['data', 'more data'], ['data 2', 'more data 2', 'danger'], ['date3', 'lll']]

We can even cheat a little because of this particular case:

>>> [list(group) for k, group in groupby(l, bool) if k]
[['data', 'more data'], ['data 2', 'more data 2', 'danger'], ['date3', 'lll']]

Method 2

One possible implementation using itertools

>>> l
['data', 'more data', '', 'data 2', 'more data 2', 'danger', '', 'date3', 'lll']
>>> it_l = iter(l)
>>> from itertools import takewhile, dropwhile
>>> [[e] + list(takewhile(lambda e: e != "", it_l)) for e in it_l if e != ""]
[['data', 'more data'], ['data 2', 'more data 2', 'danger'], ['date3', 'lll']]

Note*

This is as fast as using groupby

>>> stmt_dsm = """
[list(group) for k, group in groupby(l, lambda x: x == "") if not k]
"""
>>> stmt_ab = """
it_l = iter(l)
[[e] + list(takewhile(lambda e: e != "", it_l)) for e in it_l if e != ""]
"""
>>> t_ab = timeit.Timer(stmt = stmt_ab, setup = "from __main__ import l, dropwhile, takewhile")
>>> t_dsm = timeit.Timer(stmt = stmt_dsm, setup = "from __main__ import l, groupby")
>>> t_ab.timeit(100000)
1.6863486541265047
>>> t_dsm.timeit(100000)
1.5298066765462863
>>> t_ab.timeit(100000)
1.735611326163962
>>>

Method 3

reduce comes to mind:

def split(iterable, where):
    def splitter(acc, item, where=where):
        if item == where:
            acc.append([])
        else:
            acc[-1].append(item)
        return acc
    return reduce(splitter, iterable, [[]])


data = ["data","more data","","data 2","more data 2","danger","","date3","lll"]
print split(data, '')

Result:

[['data', 'more data'], ['data 2', 'more data 2', 'danger'], ['date3', 'lll']]

Method 4

I’m not sure wether this is the most “pythonic” way of solving it.

def split_seq(seq, sep):
    start = 0
    while start < len(seq):
        try:
           stop = start + seq[start:].index(sep)
           yield seq[start:stop]
           start = stop + 1
        except ValueError:
           yield seq[start:]
           break

ll = ["data","more data","","data 2","more data 2","danger","","date3","lll"]
p = [i for i in split_seq(ll,"")]

Method 5

Heres one idea. 🙂

def spec_split(seq,sep):
    # Ideally this separator will never be in your list
    odd_sep = "<a href="https://getridbug.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="597819">[email protected]</a>#$%^&*()"

    # Join all the items with the odd separator and split
    # anywhere the odd separator + separator + odd seperator meet
    # This makes a list of items broken by the separator
    jumble = odd_sep.join(seq).split(odd_sep+sep+odd_sep)

    # split the remaining items broken by odd separators into sublists
    return [item.split(odd_sep) for item in jumble]

Method 6

    lst = ["data","more data","","data 2","more data 2","danger","","date3","lll"]
    join_list = ",".join(lst)
    split_list = join_list.split(",,")
    result = [i.split() for i in split_list]
    #result =[['data,more', 'data'], ['data', '2,more', 'data', '2,danger'],  ['date3,lll']]

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0