Как найти сумму строки матрицы python

lista = [[1,2,3],[4,5,6],[7,8,9]]

print(lista)

def filas(lista):

    res=[]
    for elemento in lista:
        x = sum(lista[elemento])
        res.append(x)
    print(res)

I need to sum the numbers of each row, and then print it as a list. It seems the problem I have is that I try to sum the sublists, instead of the numbers of each row.

Do you want like this?

lista = [[1,2,3],[4,5,6],[7,8,9]]

print(lista)

def filas(lista):
    summed_list = [sum(i) for i in lista]
    print(summed_list)
filas(lista)

answered Jun 13, 2018 at 16:11

You can do this easly, with dot product. This is much faster than any loop.

lista = np.float32([[1,2,3],[4,5,6],[7,8,9]])
vet_one = np.ones(len(lista))
vet_sum = lista.dot(vet_one)

answered Jul 18, 2020 at 4:53

Is this what you want to do?

def filas(lista):
    res=[]
    for elemento in lista:
        x = sum(elemento) # <- change this line.
        res.append(x)
    print(res)

answered May 22, 2017 at 1:15

I want to sum a 2 dimensional array in python:

Here is what I have:

def sum1(input):
    sum = 0
    for row in range (len(input)-1):
        for col in range(len(input[0])-1):
            sum = sum + input[row][col]

    return sum


print sum1([[1, 2],[3, 4],[5, 6]])

It displays 4 instead of 21 (1+2+3+4+5+6 = 21). Where is my mistake?

Better still, forget the index counters and just iterate over the items themselves:

def sum1(input):
    my_sum = 0
    for row in input:
        my_sum += sum(row)
    return my_sum

print sum1([[1, 2],[3, 4],[5, 6]])

One of the nice (and idiomatic) features of Python is letting it do the counting for you. sum() is a built-in and you should not use names of built-ins for your own identifiers.

answered May 23, 2012 at 3:59

This is the issue

for row in range (len(input)-1):
    for col in range(len(input[0])-1):

try

for row in range (len(input)):
    for col in range(len(input[0])):

Python’s range(x) goes from 0..x-1 already

range(…)
range([start,] stop[, step]) -> list of integers

Return a list containing an arithmetic progression of integers.
range(i, j) returns [i, i+1, i+2, ..., j-1]; start (!) defaults to 0.
When step is given, it specifies the increment (or decrement).
For example, range(4) returns [0, 1, 2, 3].  The end point is omitted!
These are exactly the valid indices for a list of 4 elements.

answered May 23, 2012 at 3:45

range() in python excludes the last element. In other words, range(1, 5) is [1, 5) or [1, 4]. So you should just use len(input) to iterate over the rows/columns.

def sum1(input):
    sum = 0
    for row in range (len(input)):
        for col in range(len(input[0])):
            sum = sum + input[row][col]

    return sum

answered May 23, 2012 at 3:45

Don’t put -1 in range(len(input)-1) instead use:

range(len(input))

range automatically returns a list one less than the argument value so no need of explicitly giving -1

answered May 23, 2012 at 3:46

def sum1(input):
    return sum([sum(x) for x in input])

answered Sep 13, 2018 at 22:49

In Python 3.7

import numpy as np
x = np.array([ [1,2], [3,4] ])
sum(sum(x))

outputs

10

answered Jan 21, 2019 at 14:51

It seems like a general consensus is that numpy is a complicated solution. In comparison to simpler algorithms. But for the sake of the answer being present:

import numpy as np


def addarrays(arr):

    b = np.sum(arr)
    return sum(b)


array_1 = [
  [1, 2],
  [3, 4],
  [5, 6]
]
print(addarrays(array_1))

This appears to be the preferred solution:

x=[[1, 2],[3, 4],[5, 6]]                                                   
sum(sum(x,[]))                                                             

answered Sep 26, 2019 at 0:14

def sum1(input):
    sum = 0
    for row in input:
        for col in row:
            sum += col
    return sum
print(sum1([[1, 2],[3, 4],[5, 6]]))