Replace string elements, using their index, by a list of strings

Question

The function I have to build is meant to replace digits in a string by (value of digit * next character).

So, foo = '2 hs4q q2w2 ' will become ' hsqqqq qww ' (mind the spaces)

Assumption - Digit can't be zero.

I fetched the (index,value) of digits and next char. Used that info to get the substrings I need to put back into the string:

foo = '2 hs4q q2w2 '
parameters=[(int(foo_list[b]),b+1) for b in range(len(foo_list)) if foo_list[b].isdigit()]
parameters # list of tuples (digit,charindex to be extended)
#[(2, 1), (4, 5), (2, 9), (2, 11)]
for p,i in parameters:
 hoo=p*foo[i]
 print (hoo,type(hoo))
 #Out
 <class 'str'> # Two spaces
qqqq <class 'str'>
ww <class 'str'>
 <class 'str'> # Two spaces

How can I use all this info in a loop that works with similar strings? I understand strings are immutable, hence a new str object has to be created for every insert/replace. Plus the index values change as the loop runs.

Comments after solution -

Thank you all for four different kinds of solutions, here is a reference for anyone who hasn't used yield from, yield - In practice, what are the main uses for the new "yield from" syntax in Python 3.3?

Answer 1

Single digit numbers

You can check if a character is a digit with str.isdigit, if it is then cast it to an int and multiply it with the next character. This logic can be written as a generator given to str.join.

Code

def expand_string(s):
 return ''.join([(int(c) - 1) * s[i+1] if c.isdigit() else c for i, c in enumerate(s)])

Example

foo = '2 hs4q q2w2 '
print(expand_string(foo)) # ' hsqqqq qww '

Although, the above fails for a string with multiple digit number such as f10o'.

Multiple digits numbers

If you also want to consider numbers with multiple digits, you can write a generator function that groups digits together using itertools.groupby.

Code

from itertools import groupby
def group_digits(s):
 for isdigit, group in groupby(s, str.isdigit):
 yield from [''.join(group)] if isdigit else group
def expand_string(s):
 s = list(group_digits(s))
 return ''.join((int(c) - 1) * s[i+1] if c.isdigit() else c for i, c in enumerate(s))

Example

foo = 'f10o'
print(expand_string(foo)) # 'foooooooooo'

Answer 2

You can use re.sub:

import re
foo = '2 hs4q q2w2 '
new_foo = re.sub('\d+[\w\W]{1}', lambda x:x.group()[-1]*int(x.group()[:-1]), foo)

Output:

' hsqqqq qww '

Answer 3

One idea is to iterate characters in your string pairwise using zip (or itertools.zip_longest) and yield an item depending on whether the character is a digit.

from itertools import zip_longest, islice
foo = '2 hs4q q2w2 '
def gen(x):
 for i, j in zip_longest(x, islice(x, 1, None), fillvalue=''):
 if i.isdigit():
 yield j * (int(i)-1)
 else:
 yield i
res = ''.join(list(gen(foo)))
# ' hsqqqq qww '

Answer 4

Be careful while using next index in the current iteration, It can give you error because if the string is something like foo = '2 hs4q q2w2 2' Use try and except

foo = '2 hs4q q2w2 '
splitted_data=list(foo)
for i,j in enumerate(splitted_data):
 try:
 if j.isdigit():
 splitted_data[i]=splitted_data[i+1]*int(j)
 except IndexError:
 pass
print("".join(splitted_data))

output:

 hsqqqqq qwww 

Answer 5

Such a function can be obtained using a while loop:

def new_string(string):
 # Create the variable new_string
 new_string = ""
 i = 0
 while i < len(string):
 char = string[i]
 # If char is a digit we multiply the next char by it
 if char.isdigit():
 new_string += int(char)*string[i+1]
 i += 2
 # If not we just concatenate char
 else:
 new_string += char
 i += 1
 return new_string

Hope it helps.