Counting the number of elements in a list that contain more 0's than 1's

Question

i'm new to python programming and my task is to tell how many binary values of the list there are where the number of 0's is grater than 1. The data for this task is in a text file, I've opened the file and put every line of text into separet value in list.

binary = list()
file = 'liczby.txt'
with open(file) as fin:
    for line in fin:
        binary.append(line)
print(*binary, sep = "\n")

And now im stuck.

Answer 1

more_zeros = 0
file = 'liczby.txt'
with open(file) as fin:
    for line in fin:
        if line.count('0') > line.count('1'):
            more_zeros += 1
print(more_zeros)

Out[1]: 6 # based on the 17 lines you gave me in your comment above

Answer 2

def count(fname):
    cnt = 0
    with open(fname, newline='') as f:
        for line in f:
            if line.count('0') > line.count('1'):
                cnt += 1
    return cnt

print(count('/tmp/g.data'))

Read help(str) , there are many useful function.

EDIT:
If you like minimalist notation, you can use;-)
Including Nicolas Gervais len() trick – it is awesome.

def count(fname):
    with open(fname, newline='') as f:
        return sum(line.count('0') > len(line) // 2 for line in f)

EDIT2: Misunderstanding question. I've updated to count only lines contains more zeros.