CODEWITHSUNDEEP
pythonpython-3.xlist
New to Java Script
New to Java Script

Reputation: 83

Sum total from multiple lists in one dictionary

very new to python and hope can get some help here. I'm trying to sum up the total from different lists in dictionary

{'1': [0, 2, 2, 0, 0], '2': [0, 1, 1, 0, 0], '3': [2, 4, 2, 0, 2]}

I have been trying to find a way to sum up the total as follow and append in a new list:

'1': [0, 2, 2, 0, 0]
'2': [0, 1, 1, 0, 0]
'3': [2, 4, 2, 0, 2]

0+0+2 = 2
2+1+4 = 7
2+1+2 = 5

[2, 7, 5, 0, 2]

I was able to get partial results for one row if I do this way but wasn't able ti figure out a way to get the output I want.

total_all = list()
    for x, result_total in result_all.items():
        new_total = (result_total[1])
        total_all.append((new_total))
    print(sum(total_all))

output 7

Any suggestions and help would be greatly appreciated

Upvotes: 0

Views: 171

Answers (2)

Andrej Kesely
Andrej Kesely

Reputation: 195408

Using zip() (doc) function to transpose dictionary values and sum() to sum them inside list comprehension:

d = {'1': [0, 2, 2, 0, 0], '2': [0, 1, 1, 0, 0], '3': [2, 4, 2, 0, 2]}

out = [sum(i) for i in zip(*d.values())]
print(out)

Prints:

[2, 7, 5, 0, 2]

EDIT (little explanation):

The star-expression * inside zip() effectively unpacks dict values into this:

out = [sum(i) for i in zip([0, 2, 2, 0, 0], [0, 1, 1, 0, 0], [2, 4, 2, 0, 2])]

zip() iterates over each of it's argument: 

1. iteration -> (0, 0, 2)    
2. iteration -> (2, 1, 4)
...

sum() does sum of these tuples:

1. iteration -> sum( (0, 0, 2) ) -> 2
2. iteration -> sum( (2, 1, 4) ) -> 7
...

Upvotes: 2

Rahul K P
Rahul K P

Reputation: 16081

You can do like this,

In [6]: list(map(sum,zip(*d.values())))                                                                                                                                                                     
Out[6]: [2, 7, 5, 0, 2]

Upvotes: 0

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