Sum numbers by letter in list of tuples

Question

I have a list of tuples:

[ (\'A\',100), (\'B\',50), (\'A\',50), (\'B\',20), (\'C\',10) ]

I am trying to sum up all numbers that have the s

Answer 1

How about this: (assuming a is the name of the tuple you have provided)

letters_to_numbers = {}
for i in a:
    if i[0] in letters_to_numbers:
        letters_to_numbers[i[0]] += i[1]
    else:
        letters_to_numbers[i[0]] = i[1]
b = letters_to_numbers.items()

The elements of the resulting tuple b will be in no particular order.

Answer 2

What is generating the list of tuples? Is it you? If so, why not try a defaultdict(list) to append the values to the right letter at the time of making the list of tuples. Then you can simply sum them. See example below.

>>> from collections import defaultdict
>>> val_store = defaultdict(list)
>>> # next lines are me simulating the creation of the tuple
>>> val_store['A'].append(10)
>>> val_store['B'].append(20)
>>> val_store['C'].append(30)
>>> val_store
defaultdict(<class 'list'>, {'C': [30], 'A': [10], 'B': [20]})
>>> val_store['A'].append(10)
>>> val_store['C'].append(30)
>>> val_store['B'].append(20)
>>> val_store
defaultdict(<class 'list'>, {'C': [30, 30], 'A': [10, 10], 'B': [20, 20]})

>>> for val in val_store:
...   print(val, sum(val_store[val]))
... 
C 60
A 20
B 40

Answer 3

Here is a one(and a half?)-liner: group by letter (for which you need to sort before), then take the sum of the second entries of your tuples.

from itertools import groupby
from operator import itemgetter

data = [('A', 100), ('B', 50), ('A', 50), ('B', 20), ('C', 10)]
res = [(k, sum(map(itemgetter(1), g)))
       for k, g in groupby(sorted(data, key=itemgetter(0)), key=itemgetter(0))]
print(res)
// => [('A', 150), ('B', 70), ('C', 10)]

The above is O(n log n) — sorting is the most expensive operation. If your input list is truly large, you might be better served by the following O(n) approach:

from collections import defaultdict

data = [('A', 100), ('B', 50), ('A', 50), ('B', 20), ('C', 10)]

d = defaultdict(int)
for letter, value in data:
    d[letter] += value
res = list(d.items())
print(res)
// => [('B', 70), ('C', 10), ('A', 150)]

Answer 4

A simpler approach

x = [('A',100),('B',50),('A',50),('B',20),('C',10)]
y = {}
for _tuple in x:
    if _tuple[0] in y:
        y[_tuple[0]] += _tuple[1]
    else:
        y[_tuple[0]] = _tuple[1]
print [(k,v) for k,v in y.iteritems()]

Answer 5

Try this:

a = [('A',100), ('B',50), ('A',50), ('B',20), ('C',10) ]

letters = set([s[0] for s in a])

new_a = []
for l in letters:
    nums = [s[1] for s in a if s[0] == l]
    new_a.append((l, sum(nums)))

print new_a    

Results:

[('A', 150), ('C', 10), ('B', 70)]

Answer 6

A one liner:

>>> x = [ ('A',100), ('B',50), ('A',50), ('B',20), ('C',10) ]
>>> {
...     k: reduce(lambda u, v: u + v, [y[1] for y in x if y[0] == k])
...     for k in [y[0] for y in x]
...     }.items()
[('A', 150), ('C', 10), ('B', 70)]