How do I sum the numbers of a string vertically?-CodePudding

I'm trying to add all the numbers in a string in the Python language. For example,

s="""11 9 5
     6 6 8
     4 6 4"""

If we considered this string in 3 lines, and each line separated with "Enter" and some space between them, how could we this output: 21,21,17 (11 6 4 = 21) or (9 6 6=21) or(5 8 4=17). Can you help me complete it?

CodePudding user response：

You can use zip for vertically operations like that. and map for evaluating sum function to all the elements in your list:

s="""11 9 5
 6 6 8
 4 6 4""" 
print(list(map(sum, zip(*[map(int, i.strip().split()) for i in s.split("\n")]))))


# [21, 21, 17]

Explanation: to be honest one-linear codes are always hard to read and not maintainable :) let's break the code to these parts for better understanding:

list_with_string_numbers = [i.strip().split() for i in s.split("\n")]
# returns => [['11', '9', '5'], ['6', '6', '8'], ['4', '6', '4']]


map_all_str_elements_to_int = [map(int, i.strip().split()) for i in s.split("\n")]
# returns => [<map object at 0x7f991725f6a0>, <map object at 0x7f991725f730>, <map object at 0x7f9917b293a0>]

join_vertically = list(zip(*[map(int, i.strip().split()) for i in s.split("\n")]))
# returns [(11, 6, 4), (9, 6, 6), (5, 8, 4)]

and finally sum do the aggregation to all tuples in the list.

list(map(sum, zip(*[map(int, i.strip().split()) for i in s.split("\n")])))
# [21, 21, 17]

CodePudding user response：

This is not a one-line, but it is readable

v = list(map(int, s.split()))
# col_size = int(len(v)**0.5)
col_size = len(s.split("\n")[0].split())

res = [0]*col_size
for i in range(0, len(v)):
    res[i%col_size]  = v[i]

CodePudding user response：

Take into account that there may be cases of unequal lengths, such as this.

"""
1 2 3
2 3
3 4 5
"""

So, based on @Mojtaba Kamyabi's answer, I made a modification, replacing the function zip with itertools.zip_longest.

Make an iterator that aggregates elements from each of the iterables. If the iterables are of uneven length, missing values are filled-in with fillvalue. Iteration continues until the longest iterable is exhausted.

from itertools import zip_longest

s="""1 2 3
 2 3
 3 4 5"""
print(list(map(sum, zip_longest(*[map(int, i.strip().split()) for i in s.split("\n")], fillvalue=0))))

# [6, 9, 8]

CodePudding user response：

This is an one-line method:

s = """11 9 5 6 6 8 4 6 4"""
out = ','.join(map(str, map(sum, zip(*zip(*[iter(map(int, s.split()))]*3)))))
print(out) # 21,21,17