Python has shortened a list by a specific word
I have a list like this:
[["tab1", None], ["val1", 10], ["val2", "test"], ["val3", 20], ["tab2", None], ["val4", "test"], ["val5", 30]]
and I am looking for a method that slices n lists, if the word "tab" is found, the result could be this:
list1 = [["val1", 10], ["val2", "test"], ["val3", 20]]
list2 = [["val4", "test"], ["val5", 30]]
I try with some for a loop but nothing is done.
but I have no idea how this is possible with python. Anyone have an idea?
Thank you in advance
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I would prefer a simple loop for
for this:
In [56]: new_l = []
In [57]: for i in l:
...: if 'tab' in i[0]:
...: new_l.append([])
...: else:
...: new_l[-1].append(i)
...:
In [58]: new_l
Out[58]:
[[['val1', 10], ['val2', 'test'], ['val3', 20]],
[['val4', 'test'], ['val5', 30]]]
This is probably a shorter solution, but I doubt it would be better.
Edit: Found a shorter version with itertools.groupby
(still prefer loop):
In [66]: [list(v) for _, v in filter(lambda x: x[0], itertools.groupby(l, key=lambda x: 'tab' not in x[0] ))]
Out[66]:
[[['val1', 10], ['val2', 'test'], ['val3', 20]],
[['val4', 'test'], ['val5', 30]]]
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Here's a pythonic way using itertools.groupby()
:
In [10]: from itertools import groupby
In [12]: delimiters = {'tab1', 'tab2'}
In [13]: [list(g) for k, g in groupby(lst, delimiters.intersection) if not k]
Out[13]:
[[['val1', 10], ['val2', 'test'], ['val3', 20]],
[['val4', 'test'], ['val5', 30]]]
A more general approach that does not require delimiters to be specified uses a lambda function as a key function groupby
(but with less performance):
In [14]: [list(g) for k, g in groupby(lst, lambda x: x[0].startswith('tab')) if not k]
Out[14]:
[[['val1', 10], ['val2', 'test'], ['val3', 20]],
[['val4', 'test'], ['val5', 30]]]
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