Python replace, using patterns in array -

i need replace things in string using array, can this:

array = [3, "$x" , "$y", "$hi_buddy"] #the first number number of things in array string = "$xena here $x , $y." 

i've got array things replace things, let's called rep_array.

rep_array = [3, "a", "b", "c"] 

for replacement use this:

for x in range (1, array[0] + 1):   string = string.replace(array[x], rep_array[x]) 

but result is:

string = "aena here , b." 

but need lonely $x not $x in word. result should this:

string = "$xena here , b." 

note that:

• all patterns in array start $.
• a pattern matches if matches whole word after $; $xena doesn't match $x, foo$x would match.
• $ can escaped @ , should not matched (for example $x not match @$x)

use regular expression wraps source text whitespace look-behind , \b anchor; make sure include start of string too:

import re  pattern, replacement in zip(array[1:], rep_array[1:]):     pattern = r'{}\b'.format(re.escape(pattern))     string = re.sub(pattern, replacement, string) 

this uses re.escape() ensure regular expression meta characters in pattern escaped first. zip() used pair patterns , replacement values; more pythonic alternative range() loop.

\b matches @ position word character followed non-word character (or vice versa), word boundary. patterns end in word character, makes sure patterns match if next character not word character, blocking $x matching inside $xena.

demo:

>>> import re >>> array = [3, "$x" , "$y", "$hi_buddy"] >>> rep_array = [3, "a", "b", "c"] >>> string = "$xena here $x , $y. foo$x matches too!" >>> pattern, replacement in zip(array[1:], rep_array[1:]): ...     pattern = r'{}\b'.format(re.escape(pattern)) ...     string = re.sub(pattern, replacement, string) ...  >>> print string $xena here , b. fooa matches too!