I have a list of tuples like this:
[
('a', 1),
('a', 2),
('a', 3),
('b', 1),
('b', 2),
('c', 1),
]
I want to iterate through this keying by the first item, so, for example, I could print something like this:
a 1 2 3 b 1 2 c 1
How would I go about doing this without keeping an item to track whether the first item is the same as I loop around the tuples? This feels rather messy (plus I have to sort the list to start with)…
Answers:
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Method 1
l = [
('a', 1),
('a', 2),
('a', 3),
('b', 1),
('b', 2),
('c', 1),
]
d = {}
for x, y in l:
d.setdefault(x, []).append(y)
print d
produces:
{'a': [1, 2, 3], 'c': [1], 'b': [1, 2]}
Method 2
Slightly simpler…
from collections import defaultdict
fq = defaultdict(list)
for n, v in myList:
fq[n].append(v)
print(fq) # defaultdict(<type 'list'>, {'a': [1, 2, 3], 'c': [1], 'b': [1, 2]})
Method 3
A solution using groupby
from itertools import groupby
l = [('a',1), ('a', 2),('a', 3),('b', 1),('b', 2),('c', 1),]
[(label, [v for l,v in value]) for (label, value) in groupby(l, lambda x:x[0])]
Output:
[('a', [1, 2, 3]), ('b', [1, 2]), ('c', [1])]
groupby(l, lambda x:x[0]) gives you an iterator that contains
['a', [('a', 1), ...], c, [('c', 1)], ...]
Method 4
I would just do the basic
answer = {}
for key, value in list_of_tuples:
if key in answer:
answer[key].append(value)
else:
answer[key] = [value]
If it’s this short, why use anything complicated. Of course if you don’t mind using setdefault that’s okay too.
Method 5
Print list of tuples grouping by the first item
This answer is based on the @gommen one.
#!/usr/bin/env python
from itertools import groupby
from operator import itemgetter
L = [
('a', 1),
('a', 2),
('a', 3),
('b', 1),
('b', 2),
('c', 1),
]
key = itemgetter(0)
L.sort(key=key) #NOTE: use `L.sort()` if you'd like second items to be sorted too
for k, group in groupby(L, key=key):
print k, ' '.join(str(item[1]) for item in group)
Output:
a 1 2 3 b 1 2 c 1
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0