Python regex - faster search

Question

I need a way to optimize by regex, here is the string I am working with:

rr='JA=3262SGF432643;KL=ASDF43TQ;ME=FQEWF43344;JA=4355FF;PE=FDSDFHSDF;EB=SFGDASDSD;JA=THISONE;IH=42DFG43;'

and i want to take only JA=4355FF which is before JA=THISONE, so i did it this way:

aa='.*JA=([^.]*)JA=THISONE[^.]*'
aa=re.compile(aa)
print (re.findall(aa,rr))

and i get:

['4355FF;PE=FDSDFHSDF;EB=SFGDASDSD;']

My first problem is slow searching apropriete part of string (becouse the string which i want to search is too large and usually JA=THISONE is at the end of string)

And second problem is i dont get 4355FF but all string until JA=THISONE.

Can someone help me optimize my regex? Thank you!

orip · Accepted Answer · 2013-10-28 13:41:22Z

3

I. Consider using string search instead of regexes:

thisone_pos = rr.find('JA=THISONE')
range_start = rr.rfind("JA=", 0, thisone_pos) + 3
range_end = rr.find(';', range_start)
print rr[range_start:range_end]

II. Consider flipping the string and constructing your regex in reverse:

re.findall(pattern, rr[::-1])

edited Oct 28, 2013 at 13:41

answered Oct 28, 2013 at 13:06

orip

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1 Comment

Jovan Over a year ago

Can you give me solution with using string search on this example, key thing here for me is that i only take JA='4355FF' which is before JA=THISONE

Tomasz Nguyen · Accepted Answer · 2013-10-28 13:26:07Z

1

You could consider the following solution:

import re

rr='JA=3262SGF432643;KL=ASDF43TQ;ME=FQEWF43344;JA=4355FF;PE=FDSDFHSDF;EB=SFGDASDSD;JA=THISONE;IH=42DFG43;'

m = re.findall( r"(JA=[^;]+;)", rr )

# Print all hits
print m

# Print the hit preceding "JA=THISONE;"
print m[ m.index( "JA=THISONE;" ) - 1]

First, you look for all instances starting with "JA;" and then, you pick the last instance located before "JA=THISONE;".

answered Oct 28, 2013 at 13:26

Tomasz Nguyen

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