CODEWITHSUNDEEP
python
Dylan Drake
Dylan Drake

Reputation: 21

Return the Row Index with the minimum sum in a two dimensional list Python

I am using a two dimensional list in python, I need to find the sum of every item in the list and then return the index with the lowest value.

It should return like this:

Row([[1000, 30], [12, 23], [560, 30, 1]])
1
Row([[10, 9], [20, 1, 2], [1, 3]])
2

This is what I have so far in the shell:

def Row(S):
count=0
for r in range(len(S[0])):
    for c in range(len(S)):
        count += sum([r])
        return r

I can't get the code in the shell to return the indice in a two dimensional list with the lowest sum. Any ideas?

Upvotes: 2

Views: 1585

Answers (3)

mgilson
mgilson

Reputation: 309881

You can do it in 1 line:

minidx = min(enumerate(data), key=lambda x:sum(x[1]))[0]

Although I would prefer to do the unpacking (instead of indexing the result):

minidx, minlist = min(enumerate(data), key=lambda x:sum(x[1]))

This has the side benefit that you get a quick handle on the list which has the minimum sum ... Although that's easy enough to get once you have the index ...

Upvotes: 1

Anton Beloglazov
Anton Beloglazov

Reputation: 5089

def row(data):
    sums = [sum(x) for x in data]
    return sums.index(min(sums))

print row([[1000, 30], [12, 23], [560, 30, 1]])

Upvotes: 0

nneonneo
nneonneo

Reputation: 179392

You can specify a "key" to min to get the minimum of a list according to some arbitrary criteria. Using this feature, we can do it in just one line:

def min_sum(data):
    return min(xrange(len(data)), key=lambda i: sum(data[i]))

Upvotes: 1

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