reduce array of dictionaries

Question

I have an array of dictionaries that I want to reduce

foo = [{'value': 45}, {'value': 2}, {'value': 3}, {'value': 0}]
reduce(lambda x, y: x['value']+y['value'] , foo)

gives me the following error

TypeError: 'int' object has no attribute '__getitem__'

what I ended up doing was to first create a simple array using a comprehension.

reduce(lambda x, y: x+y, [x['value'] for x in foo])

For some reason when the number of dicts exceeds 2, `reduce()` does not work. I am not sure why yet. — Y2H, Jun 21 '17 at 14:11

Tomalak · Answer 1 · 2017-06-21T13:21:54.930

3

You are using reduce wrong.

from functools import reduce

foo = [{'value': 45}, {'value': 2}, {'value': 3}, {'value': 0}]
reduce(lambda acc, item: acc + item['value'], foo, 0)

Once reduces list items against an accumulator, not list items against each other.

edited Jun 21 '17 at 13:21

answered Jun 21 '17 at 13:16

Tomalak

332,285
67
532
628

score 3 · Accepted Answer · answered Jun 21 '17 at 13:22

reduce(function, [a, b, c]) is equivalent to function(function(a, b), c). The result of function must therefore be suitable to act as the first argument of function.

To illustrate, this works:

>>> reduce(lambda x, y: {'value': x['value']+y['value']} , foo)
{'value': 50}

score 0 · Answer 3 · answered Jun 21 '17 at 13:20

0

If you're trying to sum the integer values

You could just reduce using a builtin from the operator library:

from functools import reduce
from operator import add, itemgetter

reduce(add, map(itemgetter('value'), foo))

answered Jun 21 '17 at 13:20

Jack Evans

1,697
3
17
33

reduce array of dictionaries

3 Answers3