How to get for loop to move onto the next inner list?

Question

I'm trying to count the number of zeros within a nested list. My code outputs 2 because it only goes through the first inner list.

This is what I have currently:

def e(xlst):
    cnt = 0
    for numbers in xlst:
        for numbers2 in numbers:
            if numbers2 == 0:
                cnt += 1
        return cnt
xlst = [[1,0,0],[1,1,1],[1,1,0]]
e(xlst)

Answer 1

您放错了return ，它应该与外部循环匹配

Answer 2

The return statement is inside the first for , so after the first row it will return and exit the function

also, list has a built-in function count , that can be used to make the code a little simples and faster:

from time import time

def e(xlst):
    cnt = 0
    for numbers in xlst:
        for numbers2 in numbers:
            if numbers2 == 0:
                cnt += 1
    return cnt


def with_count(xlst):
    cnt = 0
    for numbers in xlst:
        cnt += numbers.count(0)
    return cnt

# Creates a 2D list with 300 rows of [1,0,0]
xlst = [[1,0,0]]*300

t0=time()
print(f'\nfirst function found: {e(xlst)}')
print(f'time: {time() - t0}\n')

t0=time()
print(f'second function found: {with_count(xlst)}')
print(f'time: {time() - t0}\n')

first function found: 600

time: 0.0003502368927001953

second function found: 600

time: 0.000194549560546875

Answer 3

Correct the return indent

Code should be like this

def e(xlst):
    cnt = 0
    for numbers in xlst:
        for numbers2 in numbers:
            if numbers2 == 0:
                cnt += 1
    return cnt
xlst = [[1,0,0],[1,1,1],[1,1,0]]
print(e(xlst))

It prints 3

Answer 4

Using global variables is bad practice, but I am too tired to figure out how to do it without, without using two functions

def e(xlst):
    cnt = 0
    if (type(xlst) == list):
        for x in xlst:
            e(x)
    elif (xlst == 0):
        cnt +=1

Here we use recursion, calling the function on each nested list. This works for infinitely nested lists.

Answer 5

You could do it as a one liner.

from itertools import chain

def e(xlst):
    return len([i for i in chain.from_iterable(xlst) if i == 0])

xlst = [[1,0,0],[1,1,1],[1,1,0]]
print(e(xlst))

or without the temporary list.

def e(xlst):
    return sum(1 for i in chain.from_iterable(xlst) if i == 0)