Python: Replace ith occurence of x with ith element in list

Question

Suppose we have a string a = "01000111000011" with n=5 "1"s. The ith "1", I would like to replace with the ith character in "ORANGE". My result should look like:

b = "0O000RAN0000GE"

What could be the finest way to solve this problem in Python? Is it possible to bind an index to each substitution?

Many thanks! Helga

Answer 1

Tons of answers/ways to do it. Mine uses a fundamental assumption that your #of 1s is equal to the length of the word you are subsituting.

a = "01000111000011"
a = a.replace("1", "%s")
b = "ORANGE"
print a % tuple(b)

Or the pythonic 1 liner ;)

print "01000111000011".replace("1", "%s") % tuple("ORANGE")

Answer 2

a = '01000111000011'
for char in 'ORANGE':
  a = a.replace('1', char, 1)

Or:

b = iter('ORANGE')
a = ''.join(next(b) if i == '1' else i for i in '01000111000011')

Or:

import re
a = re.sub('1', lambda x, b=iter('ORANGE'): b.next(), '01000111000011')