I've got a list of strings and I'm trying to make a list of lists of strings by string length.
i.e.
['a', 'b', 'ab', 'abc']
becomes
[['a', 'b'], ['ab'], ['abc']]
I've accomplished this like so:
lst = ['a', 'b', 'ab', 'abc']
lsts = []
for num in set(len(i) for i in lst):
lsts.append([w for w in lst if len(w) == num])
I'm fine with that code, but I'm trying to wrap my head around comprehensions. I want to use nested comprehensions to do the same thing, but I can't figure out how.
>>> [[w for w in L if len(w) == num] for num in set(len(i) for i in L)]
[['a', 'b'], ['ab'], ['abc']]
Also, itertools is likely to be a little more efficient.
lst = ['a', 'b', 'ab', 'abc']
lst.sort(key=len) # does not make any change on this data,but
# all strings of given length must occur together
from itertools import groupby
lst = [list(grp) for i,grp in groupby(lst, key=len)]
results in
[['a', 'b'], ['ab'], ['abc']]
key=len (also on groupby)
L=['a','b','ab','abc']
result = [ [ w for w in L if len(w) == n] for n in set(len(i) for i in L)]
That is for all lengths from 1 to maximum (some of lists will be empty if there are no strings of that length in the a list):
>>> a = ['a', 'b', 'ab', 'abc']
>>> m = max(len(x) for x in a)
>>> print [[x for x in a if len(x) == i + 1] for i in range(m)]
[['a', 'b'], ['ab'], ['abc']]
But if you want to have only lists for the lengths that are in a you must use set(len(i) for i in lst) instead of range.
>>> print [[x for x in a if len(x) == i] for i in set(len(k) for k in a)]
[['a', 'b'], ['ab'], ['abc']]
There is no difference for the list ['a', 'b', 'ab', 'abc']. But if you change it a little bit, e.g so: [['a', 'b'], ['ab'], ['abcd']], you will see the difference:
>>> a = ['a', 'b', 'ab', 'abcd']
>>> print [[x for x in a if len(x) == i] for i in set(len(k) for k in a)]
[['a', 'b'], ['ab'], ['abcd']]
>>> print [[x for x in a if len(x) == i + 1] for i in range(max(len(x) for x in a))]
[['a', 'b'], ['ab'], [], ['abcd']]
from itertools import groupby
mylist = ['a', 'b', 'ab', 'abc']
[list(vals) for key, vals in groupby(mylist, lambda L: len(L))]
note that since groupby only works on adjacent elements - you may need to force a sort on mylist with key=len)
the outside list becomes built from the above
