Vikram K wrote:
Suppose i have this nested list:
>>> x
[['NM100', 3, 4, 5, 6, 7], ['NM100', 10, 11, 12, 13], ['NM200', 15, 16, 17]]
>>> for i in x:
... print i
...
['NM100', 3, 4, 5, 6, 7]
['NM100', 10, 11, 12, 13]
['NM200', 15, 16, 17]
>>>
how do i obtain from the above the following nested list:
>>> z
[['NM100', 3, 4, 5, 6, 7, 10, 11, 12, 13], ['NM200', 15, 16, 17]]
>>> for i in z:
... print i
...
['NM100', 3, 4, 5, 6, 7, 10, 11, 12, 13]
['NM200', 15, 16, 17]
>>>
z = [['NM100', 3, 4, 5, 6, 7], ['NM100', 10, 11, 12, 13], ['NM200', 15,
16, 17]]
from itertools import groupby
# assumes that the data is already sorted by the first element as it is here
# if not then you must sort, eg. with the same function you group by:
# z.sort(key=lambda X: X[0])
# also assumes that there are no empty lists
def merge(inlist):
for k, g in groupby(inlist, lambda X: X[0]):
k = [k]
for item in g:
k.extend(item[1:])
yield k
print list(merge(z))
[['NM100', 3, 4, 5, 6, 7, 10, 11, 12, 13], ['NM200', 15, 16, 17]]
def merge(inlist):
for k, g in groupby(inlist, lambda X: X[0]):
newlist = []
for item in g:
newlist.extend(item[1:])
yield k, newlist
print dict(merge(z))
{'NM100': [3, 4, 5, 6, 7, 10, 11, 12, 13], 'NM200': [15, 16, 17]}
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