Reputation: 163
If for instance I have a list
['My', 'Is', 'Name', 'Hello', 'William']
How can I manipulate it such that I can create a new list
[['My', 'Is'], ['Name'], ['Hello'], ['William']]
Upvotes: 2
Views: 662
Reputation: 22420
You could use itertools.groupby:
>>> from itertools import groupby
>>> l = ['My', 'Is', 'Name', 'Hello', 'William']
>>> [list(g) for k, g in groupby(l, key=len)]
[['My', 'Is'], ['Name'], ['Hello'], ['William']]
If however the list is not already sorted by length you will need to sort it first as @recnac mentions in the comments below:
>>> l2 = ['My', 'Name', 'Hello', 'Is', 'William']
>>> [list(g) for k, g in groupby(sorted(l2, key=len), key=len)]
[['My', 'Is'], ['Name'], ['Hello'], ['William']]
Upvotes: 7
Reputation: 163
Hi guys I have also found a solution, while it is not the most concise I thought it would be worth sharing
data = ['My', 'Is', 'Name', 'Hello', 'William']
dict0 = {}
for index in data:
if len(index) not in dict0:
dict0[len(index)] = [index]
elif len(index) in dict0:
dict0[len(index)] += [index]
list0 = []
for i in dict0:
list0.append(dict0[i])
print(list0)
Upvotes: 2
Reputation: 106881
You can build a dict that maps word lengths to a list of matching words, and then get the list of the dict's values:
l = ['My', 'Is', 'Name', 'Hello', 'William']
d = {}
for w in l:
d.setdefault(len(w), []).append(w)
print(list(d.values()))
This outputs:
[['My', 'Is'], ['Name'], ['Hello'], ['William']]
Upvotes: 2
Reputation: 3744
you can use dict
to record the string group by length, defaultdict
is used for convenient here.
from collections import defaultdict
str_list = ['My', 'Is', 'Name', 'Hello', 'William']
group_by_len = defaultdict(list)
for s in str_list:
group_by_len[len(s)].append(s)
result = list(group_by_len.values())
output:
[['My', 'Is'], ['Name'], ['Hello'], ['William']]
Hope that will help you, and comment if you have further questions. : )
Upvotes: 1