Question

Is it possible to use a back reference to specify the number of replications in a regular expression?

foo= ‘ADCKAL+2AG.+2AG.+2AG.+2AGGG+.G+3AGGa.’

The substrings that start with ‘+[0-9]’ followed by ‘[A-z]{n}.’ need to be replaced with simply ‘+’ where the variable n is the digit from earlier in the substring. Can that n be back referenced? For example (doesn’t work) ‘+([0-9])[A-z]{/1}.’ is the pattern I want replaced with “+” (that last dot can be any character and represents a quality score) so that foo should come out to ADCKAL+++G.G+.

 import re
 foo = 'ADCKAL+2AG.+2AG.+2AG.+2AGGG+.+G+3AGGa.'
 indelpatt = re.compile('\+([0-9])')
 while indelpatt.search(foo):
     indelsize=int(indelpatt.search(foo).group(1))
     new_regex = '\+%s[ACGTNacgtn]{%s}.' % (indelsize,indelsize)
     newpatt=re.compile(new_regex)
     foo = newpatt.sub("+", foo)

I’m probably missing an easier way to parse the string.

Answer 1

No, you cannot use back-references as quantifiers. A workaround is to construct a regular expression that can handle each of the cases in an alternation.

import re

foo = 'ADCKAL+2AG.+2AG.+2AG.+2AGGG^+.+G+3AGGa4.'
pattern = '|'.join('\+%s[ACGTNacgtn]{%s}.' % (i, i) for i in range(1, 10))
regex = re.compile(pattern)
foo = regex.sub("+", foo)
print foo

Result:

ADCKAL++++G^+.+G+4.

Note also that your code contains an error that causes it to enter an infinite loop on the input you gave.