How to get the height for each collumn in Python?-CodePudding

This is a nested list:

matrix = [['0', '0', '0', '0', '0', '1', '1', '0', '0', '0'],
          ['0', '1', '1', '1', '1', '1', '0', '0', '0', '0'],
          ['0', '0', '1', '1', '0', '1', '0', '1', '1', '0'],
          ['0', '1', '0', '0', '0', '0', '1', '0', '0', '1'],
          ['0', '0', '0', '1', '0', '0', '1', '0', '1', '0'],
          ['0', '0', '0', '0', '1', '0', '1', '1', '1', '0'],
          ['0', '1', '1', '1', '1', '0', '1', '1', '1', '1'],
          ['1', '1', '1', '1', '1', '1', '1', '1', '1', '1']]
#output should be [1,2,2,2,3,1,5,3,4,2]

Height = 8 Width = 10 I want to know how I can get the height of each collumn. And then each height, I want to put it in a list. We counting 1's and they only count for the height if they are adjoint 1's. We start with counting below and then go up. #output should be [1,2,2,2,3,1,5,3,4,1]

I only want to use build in Python functions.

I tried with a for loop and if, else statements.

For loop iterate through the list.
like if i == '1' add 1 to counter.
if i == '0' reset counter and add the last value from counter to counter1, but only if counter is greater then counter1.

CodePudding user response：

You can use itertools.takewhile on the reversed, zipped matrix:

from itertools import takewhile
out = [len(list(takewhile(lambda x: x=='1', reversed(l)))) for l in zip(*matrix)]

output:

[1, 2, 2, 2, 3, 1, 5, 3, 4, 2]

If you don't want to import takewhile, use the recipe:

def takewhile(predicate, iterable):
    # takewhile(lambda x: x<5, [1,4,6,4,1]) --> 1 4
    for x in iterable:
        if predicate(x):
            yield x
        else:
            break

How it works:

zip rotates the matrix:

>>> list(zip(*matrix))

[('0', '0', '0', '0', '0', '0', '0', '1'),
 ('0', '1', '0', '1', '0', '0', '1', '1'),
 ('0', '1', '1', '0', '0', '0', '1', '1'),
 ('0', '1', '1', '0', '1', '0', '1', '1'),
 ('0', '1', '0', '0', '0', '1', '1', '1'),
 ('1', '1', '1', '0', '0', '0', '0', '1'),
 ('1', '0', '0', '1', '1', '1', '1', '1'),
 ('0', '0', '1', '0', '0', '1', '1', '1'),
 ('0', '0', '1', '0', '1', '1', '1', '1'),
 ('0', '0', '0', '1', '0', '0', '1', '1')]

The list comprehension with reversed rotates the other way around (actually inverses each row):

>>> [list(reversed(l)) for l in zip(*matrix)]

[['1', '0', '0', '0', '0', '0', '0', '0'],
 ['1', '1', '0', '0', '1', '0', '1', '0'],
 ['1', '1', '0', '0', '0', '1', '1', '0'],
 ['1', '1', '0', '1', '0', '1', '1', '0'],
 ['1', '1', '1', '0', '0', '0', '1', '0'],
 ['1', '0', '0', '0', '0', '1', '1', '1'],
 ['1', '1', '1', '1', '1', '0', '0', '1'],
 ['1', '1', '1', '0', '0', '1', '0', '0'],
 ['1', '1', '1', '1', '0', '1', '0', '0'],
 ['1', '1', '0', '0', '1', '0', '0', '0']]

takewhile keepd the elements while the condition is True, here while the items are '1' (lambda x: x=='1'), and len gets the length of the output:

>>> l = ['1', '1', '1', '0', '0', '0', '1', '0']
>>> list(takewhile(lambda x: x=='1', l))

['1', '1', '1']

>>> len(list(takewhile(lambda x: x=='1', l)))
3

NB. functions like zip, reversed, takewhile are generators, they don't produce output unless something consumes it, that's why I used list(generator(…)) in the exammples

solution with classical python loops:

out = []
for l in zip(*matrix):
    counter = 0
    for elem in reversed(l):
        if elem == '1':
            counter  =1
        else:
            break
    out.append(counter)