I wrote a function to create a nested list.
For example:
input= ['a','b','c','','d','e','f','g','','d','s','d','a','']
I want to create a sublist before ''
As a return I want a nested list like:
[['a','b','c'],['d','e','f','g'],['d','s','d','a']]
Try the following implementation
>>> def foo(inlist, delim = ''):
start = 0
try:
while True:
stop = inlist.index(delim, start)
yield inlist[start:stop]
start = stop + 1
except ValueError:
# if '' may not be the end delimiter
if start < len(inlist):
yield inlist[start:]
return
>>> list(foo(inlist))
[['a', 'b', 'c'], ['d', 'e', 'f', 'g'], ['d', 's', 'd', 'a']]
Another possible implementation could be by itertools.groupby. But then you have to filter the result to remove the ['']. But though it might look to be one-liner yet the above implementation is more pythonic as its intuitive and readable
>>> from itertools import ifilter, groupby
>>> list(ifilter(lambda e: '' not in e,
(list(v) for k,v in groupby(inlist, key = lambda e:e == ''))))
[['a', 'b', 'c'], ['d', 'e', 'f', 'g'], ['d', 's', 'd', 'a']]
I'd use itertools.groupby:
l = ['a','b','c','','d','e','f','g','','d','s','d','a','']
from itertools import groupby
[list(g) for k, g in groupby(l, bool) if k]
gives
[['a', 'b', 'c'], ['d', 'e', 'f', 'g'], ['d', 's', 'd', 'a']]
def nester(nput):
out = [[]]
for n in nput:
if n == '':
out.append([])
else:
out[-1].append(n)
if out[-1] == []:
out = out[:-1]
return out
edited to add check for empty list at end
if not out[-1]: out.pop() before the return