Grouping Python tuple list
I have a l开发者_Python百科ist of (label, count) tuples like this:
[('grape', 100), ('grape', 3), ('apple', 15), ('apple', 10), ('apple', 4), ('banana', 3)]
From that I want to sum all values with the same label (same labels always adjacent) and return a list in the same label order:
[('grape', 103), ('apple', 29), ('banana', 3)]
I know I could solve it with something like:
def group(l):
result = []
if l:
this_label = l[0][0]
this_count = 0
for label, count in l:
if label != this_label:
result.append((this_label, this_count))
this_label = label
this_count = 0
this_count += count
result.append((this_label, this_count))
return result
But is there a more Pythonic / elegant / efficient way to do this?
itertools.groupby
can do what you want:
import itertools
import operator
L = [('grape', 100), ('grape', 3), ('apple', 15), ('apple', 10),
('apple', 4), ('banana', 3)]
def accumulate(l):
it = itertools.groupby(l, operator.itemgetter(0))
for key, subiter in it:
yield key, sum(item[1] for item in subiter)
print(list(accumulate(L)))
# [('grape', 103), ('apple', 29), ('banana', 3)]
using itertools and list comprehensions
import itertools
[(key, sum(num for _, num in value))
for key, value in itertools.groupby(l, lambda x: x[0])]
Edit: as gnibbler pointed out: if l
isn't already sorted replace it with sorted(l)
.
import collections
d=collections.defaultdict(int)
a=[]
alist=[('grape', 100), ('banana', 3), ('apple', 10), ('apple', 4), ('grape', 3), ('apple', 15)]
for fruit,number in alist:
if not fruit in a: a.append(fruit)
d[fruit]+=number
for f in a:
print (f,d[f])
output
$ ./python.py
('grape', 103)
('banana', 3)
('apple', 29)
>>> from itertools import groupby
>>> from operator import itemgetter
>>> L=[('grape', 100), ('grape', 3), ('apple', 15), ('apple', 10), ('apple', 4), ('banana', 3)]
>>> [(x,sum(map(itemgetter(1),y))) for x,y in groupby(L, itemgetter(0))]
[('grape', 103), ('apple', 29), ('banana', 3)]
my version without itertools
[(k, sum([y for (x,y) in l if x == k])) for k in dict(l).keys()]
Method
def group_by(my_list):
result = {}
for k, v in my_list:
result[k] = v if k not in result else result[k] + v
return result
Usage
my_list = [
('grape', 100), ('grape', 3), ('apple', 15),
('apple', 10), ('apple', 4), ('banana', 3)
]
group_by(my_list)
# Output: {'grape': 103, 'apple': 29, 'banana': 3}
You Convert to List of tuples like list(group_by(my_list).items())
.
Or a simpler more readable answer ( without itertools ):
pairs = [('foo',1),('bar',2),('foo',2),('bar',3)]
def sum_pairs(pairs):
sums = {}
for pair in pairs:
sums.setdefault(pair[0], 0)
sums[pair[0]] += pair[1]
return sums.items()
print sum_pairs(pairs)
Simpler answer without any third-party libraries:
dct={}
for key,value in alist:
if key not in dct:
dct[key]=value
else:
dct[key]+=value
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