CODEWITHSUNDEEP
pythonlistsplit
MLSC
MLSC

Reputation: 5972

Split python list values

Imagin I have three lists like:

l = ['2g_', '3k', '3p']
p = ['3f', '2y_', '4k', 'p']
s = ['12g', 'k_', '3p']

So:

>>> ''.join(i[1]*int(i[0])+i[2:] if i[0].isdigit() else i for i in l)
'gg_kkkppp'
>>> ''.join(i[1]*int(i[0])+i[2:] if i[0].isdigit() else i for i in p)
'fffyy_kkkkp'
>>> ''.join(i[1]*int(i[0])+i[2:] if i[0].isdigit() else i for i in s)
'2gk_ppp'

But what in list s: 2gk_ppp must be ggggggggggggk_ppp

Upvotes: 1

Views: 437

Answers (2)

tobias_k
tobias_k

Reputation: 82949

You can use a nested list comprehension / generator expression with regular expressions.

def join(lst):
    return ''.join((int(n or 1) * c + r 
                    for (n, c, r) 
                    in (re.search(r"(\d*)(\w)(.*)", x).groups() for x in lst)))

First, use re.search with (\d*)(\w)(.*) to get the (optional) number, the character, and the (optional) rest for each string.

[re.search(r"(\d*)(\w)(.*)", x).groups() for x in lst]

For your second example, this gives [('3', 'f', ''), ('2', 'y', '_'), ('4', 'k', ''), ('', 'p', '')]. Now, in the outer generator, you can use or to provide a "default-value" in case the number is '' (or use a ternary expression if you prefer: int(n) if n else 1):

[int(n or '1') * c + r 
 for (n, c, r) 
 in (re.search(r"(\d*)(\w)(.*)", x).groups() for x in lst)]

This gives ['fff', 'yy_', 'kkkk', 'p']. Finally, join to get fffyy_kkkkp.

Upvotes: 4

MarengoHue
MarengoHue

Reputation: 1825

This variant is a little bit more imperative and simple to follow:

import re as regex

def process_template(template):
    result = regex.match('(\d*)(.)(.*)', template)
    if (result.group(1)):
        return int(result.group(1)) * result.group(2) + result.group(3)
    else:
        return result.group(2) + result.group(3)

''.join([process_template(template) for template in ['3f', '2y_', '4k', 'p']])

Upvotes: 0

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