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groupingpython
hoju
hoju

Reputation: 29452

Grouping Python tuple list

I have a list 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?

Upvotes: 31

Views: 38703

Answers (8)

Md Shahbaz
Md Shahbaz

Reputation: 1

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

Upvotes: 0

Thomas Wouters
Thomas Wouters

Reputation: 133425

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)]

Upvotes: 44

Shameem
Shameem

Reputation: 2814

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()).

Upvotes: 1

Anton Suslov
Anton Suslov

Reputation: 71

my version without itertools
[(k, sum([y for (x,y) in l if x == k])) for k in dict(l).keys()]

Upvotes: 4

Paul Kenjora
Paul Kenjora

Reputation: 2004

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)

Upvotes: 0

John La Rooy
John La Rooy

Reputation: 304167

>>> 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)]

Upvotes: 5

ghostdog74
ghostdog74

Reputation: 342373

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)

Upvotes: 6

cobbal
cobbal

Reputation: 70733

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).

Upvotes: 8

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