Why is the string split into characters in python

Question

The following is the code I wrote:

def comb(self, rows, cols):
    return [s+t for s in a for t in b]

If the values of rows and cols are

rows = ['abc','efg']
cols = ['123','456']

Expected output: ['abc123','abc456,'efg123','efg456']

Program output: ['a1', 'a2', 'a3', 'b1', 'b2', 'b3', 'c1', 'c2', 'c3']

I am new to Python programming. Could you help me understand what is happening? I already fixed the output but I would like to understand why this happened.

Answer 1

To understand what your list comprehension is doing, you can rewrite it like this:

results = []
for s in a:
    for t in b:
        results.append(s+t)

Presumably that's not what you want.

Answer 2

Try using the zip() function:

>>> rows = ['abc','efg']
>>> cols = ['123','456']
>>> def comb(rows, cols):
    return [r+c for r, c in zip(rows, cols)]

>>> comb(rows, cols)
['abc123', 'efg456']

The zip() function, in essence, pairs each value in rows with each value in cols .

>>> list(zip(rows, cols))
[('abc', '123'), ('efg', '456')]

On the other hand, [s+t for s in a for t in b] is a nested for loop, with iterations of a nested in iterations of b .

Answer 3

change it to this:

rows = ['abc','efg']
cols = ['123','456']

def comb(rows, cols):
    return [s+t for s in rows for t in cols]

print(comb(rows,cols))

output:

['abc123', 'abc456', 'efg123', 'efg456']

Answer 4

The itertools library is a popular Python standard library containing many tools that help you iterate over different ways of combining itrators (such as your given rows and cols lists).

import itertools

def comb(rows, cols):
    return map(lambda t: t[0] + t[1], itertools.product(rows, cols))

itertools.product will give you an iterator full of tuples which give you all the possible combinations of the entries in rows as the first element, and the entries in cols as the second element.

map uses the lambda function given to join the two strings in the tuple produced by itertools.product into one string, which is what you want.

To get the list that you mentioned you want, you can wrap the call to map in the function list() to evaluate the map -produced iterator into a list. Fyi, for large lists, this will be inefficient, as the result of combining the substrings will all live in memory, whereas if you use the iterator provided by the first def comb I gave you, when you iterate over those iterators, you'll use rules to generate each entry as you go.

Answer 5

It was an error in my code because instead of taking the string as it is I was using each string as an input for the comb function

def comb(self, rows, cols): return [s+t for s in a for t in b]

rows = ['abc','efg'] cols = ['123','456']

print [comb(rs, cs) for rs in rows for cs in cols]

so the output was ['a1', 'a2', 'a3', 'b1', 'b2', 'b3', 'c1', 'c2', 'c3']

in this function, it would take each character of the string and the output will be a combination of the characters instead of the string.

But thank you for trying to help. I really appreciate it.