Group a list by word length

Question

For example, I have a list, say

list = ['sight', 'first', 'love', 'was', 'at', 'It']

I want to group this list by word length, say

newlist = [['sight', 'first'],['love'], ['was'], ['at', 'It']]

Please help me on it. Appreciation!

Answer 1

Use itertools.groupby:

>>> from itertools import groupby
>>> lis = ['sight', 'first', 'love', 'was', 'at', 'It']
>>> [list(g) for k, g in groupby(lis, key=len)]
[['sight', 'first'], ['love'], ['was'], ['at', 'It']]

Note that for itertools.groupby to work properly all the items must be sorted by length, otherwise use collections.defaultdict(O(N)) or sort the list first and then use itertools.groupby(O(NlogN)). :

>>> from collections import defaultdict
>>> d = defaultdict(list)
>>> lis = ['sight', 'first', 'foo', 'love', 'at', 'was', 'at', 'It']
>>> for x in lis:
...     d[len(x)].append(x)
...     
>>> d.values()
[['at', 'at', 'It'], ['foo', 'was'], ['love'], ['sight', 'first']]

If you want the final output list to be sorted too then better sort the list items by length and apply itertools.groupby to it.

Answer 2

You can use a temp dictionary then sort by length:

li=['sight', 'first', 'love', 'was', 'at', 'It']

d={}
for word in li:
    d.setdefault(len(word), []).append(word)

result=[d[n] for n in sorted(d, reverse=True)] 

print result  
# [['sight', 'first'], ['love'], ['was'], ['at', 'It']]

You can use defaultdict:

from collections import defaultdict
d=defaultdict(list)
for word in li:
    d[len(word)].append(word)

result=[d[n] for n in sorted(d, reverse=True)] 
print result

or use __missing__ like so:

class Dicto(dict):
    def __missing__(self, key):
        self[key]=[]
        return self[key]

d=Dicto()
for word in li:
    d[len(word)].append(word)

result=[d[n] for n in sorted(d, reverse=True)] 
print result