Summing values in nested list when item changes

Question

I have some problem which has stumped my beginner-knowledge of python and I hope someone out there can point me in the correct direction.

I generated a nested list, each

Answer 1

Here's a variation on m170897017's technique:

a = [[1, 0],[1, 2],[2, 9],[3, 0],[3, 8],[3, 1]]

result = {}
for day, val in a:
    if day not in result:
       result[day] = 0
    result[day] += val

print result

#Convert back into a list
print [list(t) for t in result.items()]

output

{1: 2, 2: 9, 3: 9}
[[1, 2], [2, 9], [3, 9]]

If you're using Python 2.7 or later, you could also use a Counter.

Another possibility is to use a defaultdict, which has been available since Python 2.5.

from collections import defaultdict

a = [[1, 0],[1, 2],[2, 9],[3, 0],[3, 8],[3, 1]]

result = defaultdict(int)
for day, val in a:
    result[day] += val

print [list(t) for t in result.items()]

output

[[1, 2], [2, 9], [3, 9]]

Answer 2

You can put the key in a dictionary and store the counts in the values. Like this:

#!/usr/bin/python
# -*- coding: utf-8 -*-

a = [[1, 0],[1, 2],[2, 9],[3, 0],[3, 8],[3, 1]]
res = {}
for i in a:
    if i[0] in res:
        res[i[0]] += i[1]
    else:
        res[i[0]] = i[1]

print res

OUTPUT:

{1: 2, 2: 9, 3: 9}

This output is in dictionary format. You can turn it to list format as you like.

Answer 3

You can use a collections.OrderedDict mapping items to list:

l = [[1, 0],[1, 2],[2, 9],[3, 0],[3, 8],[3, 1]]

from collections import OrderedDict
d = OrderedDict()

for a, b in l:
    d.setdefault(a, 0)
    d[a] += b
print(map(list,d.iteritems()))

[[1, 2], [2, 9], [3, 9]]

Answer 4

Problem's like this calls for itertools.groupby. Particularly groupby, groups consecutive values with same keys, where keys can be specified by user. Nevertheless, in this case, key is trivial enough as indexing a particular element of a list, so that should be a reason enough to use operator.itemgetter. Finally, you can wrap up either as a functional (using map/imap) or as a generator expression based on your taste and choice.

>>> from itertools import groupby, imap
>>> from operator import itemgetter
>>> lst=[[1, 0],[1, 2],[2, 9],[3, 0],[3, 8],[3, 1]]
>>> [[k, sum(imap(itemgetter(1), v))]
     for k, v in groupby(lst,key = itemgetter(0))]
[[1, 2], [2, 9], [3, 9]]

Answer 5

You can try using defaultdict ...

from collections import defaultdict

dat = [[1, 0],[1, 2],[2, 9],[3, 0],[3, 8],[3, 1]]

d = defaultdict(int)
for k,v in dat: d[k] += v