How do I sum tuples in a list where the first value is the same?

Question

I have a list of stocks and positions as tuples. Positive for buy, negative for sell. Example:

p = [(\'AAPL\', 50), (\'AAPL\', -50), (\'RY\', 100), (\'RY\',          


        

Answer 1

without any explicit loops:

>>> from itertools import groupby
>>> from operator import itemgetter
>>> p = [('AAPL', 50), ('AAPL', -50), ('RY', 100), ('RY', -43)]
>>> map(lambda g: (g[0], sum(map(itemgetter(1), g[1]))),
...     groupby(sorted(p), itemgetter(0)))
[('AAPL', 0), ('RY', 57)]

Answer 2

How about this? You can read about collections.defaultdict.

>>> from collections import defaultdict
>>> testDict = defaultdict(int)
>>> p = [('AAPL', 50), ('AAPL', -50), ('RY', 100), ('RY', -43)]
>>> for key, val in p:
        testDict[key] += val


>>> testDict.items()
[('AAPL', 0), ('RY', 57)]

Answer 3

Here is a solution that doesn't involve importing:

>>> p = [('AAPL', 50), ('AAPL', -50), ('RY', 100), ('RY', -43)]
>>> d = {x:0 for x,_ in p}
>>> for name,num in p: d[name] += num
...
>>> Result = map(tuple, d.items())
>>> Result
[('AAPL', 0), ('RY', 57)]
>>>

Note this is for Python 2.x. In 3.x, you'll need to do: Result = list(map(tuple, d.items())).

Answer 4

I would do this using collections.Counter:

In [2]: from collections import Counter

In [3]: c = Counter()

In [4]: for k, v in p:
   ...:     c[k] += v
   ...:     
In [5]: c
Out[5]: Counter({'AAPL': 0, 'RY': 57})

Then you can call the most_common method of Counter objects to get a list of tuples sorted by the values in descending order.

In [5]: c.most_common()
Out[5]: [('RY', 57), ('AAPL', 0)]

In case you need to sort tuples by their first elements, use sorted(c.items()):

In [6]: sorted(c.items())
Out[6]: [('AAPL', 0), ('RY', 57)]