How to extract each word consecutive to its own previous number in a string and sorting the result in Python

Question

Input : x3b4U5i2 Output : bbbbiiUUUUUxxx

How can i solve this problem in Python. I have to print the word next to it's number n times and sort it

Answer 1

One option, extract the character/digit(s) pairs with a regex, sort them by letter (ignoring case), multiply the letter by the number of repeats, join:

s = 'x3b4U5i2'

import re

out = ''.join([c*int(i) for c,i in 
               sorted(re.findall('(\D)(\d+)', s),
                      key=lambda x: x[0].casefold())
              ])

print(out)

Output: bbbbiiUUUUUxxx

If you want to handle multiple characters you can use '(\D+)(\d+)'

Answer 2

I'm assuming the formatting will always be <char><int> with <int> being in between 1 and 9...

input_ = "x3b4U5i2"

result_list = [input_[i]*int(input_[i+1]) for i in range(0, len(input_), 2)]
result_list.sort(key=str.lower)
result = ''.join(result_list)

There's probably a much more performance-oriented approach to solving this, it's just the first solution that came into my limited mind.

Edit

After the feedback in the comments I've tried to improve performance by sorting it first, but I have actually decreased performance in the following implementaiton:

input_ = "x3b4U5i2"

def sort_first(value):
    return value[0].lower()

tuple_construct = [(input_[i], int(input_[i+1])) for i in range(0, len(input_), 2)]
tuple_construct.sort(key=sort_first)
result = ''.join([tc[0] * tc[1] for tc in tuple_construct])

Execution time for 100,000 iterations on it:

1) The execution time is: 0.353036
2) The execution time is: 0.4361724

Answer 3

It wasn't clear if multiple digit counts or groups of letters should be handled. Here's a solution that does all of that:

import re

def main(inp):
    parts = re.split(r"(\d+)", inp)
    parts_map = {parts[i]:int(parts[i+1]) for i in range(0, len(parts)-1, 2)}
    print(''.join([c*parts_map[c] for c in sorted(parts_map.keys(),key=str.lower)]))

main("x3b4U5i2")
main("x3brx4U5i2")
main("x23b4U35i2")

Result:

bbbbiiUUUUUxxx
brxbrxbrxbrxiiUUUUUxxx
bbbbiiUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUxxxxxxxxxxxxxxxxxxxxxxx

Answer 4

No list comprehensions or generator expressions in sight. Just using re.sub with a lambda to expand the length encoding, then sorting that, and then joing that back into a string.

import re

s = "x3b4U5i2"

''.join(sorted(re.sub(r"(\D+)(\d+)", 
                      lambda m: m.group(1)*int(m.group(2)), 
                      s),
        key=lambda x: x[0].casefold()))
# 'bbbbiiUUUUUxxx'

---

If we use re.findall to extract a list of pairs of strings and multipliers:

import re

s = 'x3b4U5i2'

pairs = re.findall(r"(\D+)(\d+)", s)

Then we can use some functional style to sort that list before expanding it.

from operator import itemgetter

def compose(f, g): 
  return lambda x: f(g(x))

sorted(pairs, key=compose(str.lower, itemgetter(0)))
# [('b', '4'), ('i', '2'), ('U', '5'), ('x', '3')]