Python Regex replace

Question

Hey I\'m trying to figure out a regular expression to do the following.

Here is my string

Place,08/09/2010,\"15,531\",\"2,909\",650

I

Answer 1

You could parse a string of that format using pyparsing:

import pyparsing as pp
import datetime as dt

st='Place,08/09/2010,"15,531","2,909",650'

def line_grammar():
    integer=pp.Word(pp.nums).setParseAction(lambda s,l,t: [int(t[0])])
    sep=pp.Suppress('/')
    date=(integer+sep+integer+sep+integer).setParseAction(
              lambda s,l,t: dt.date(t[2],t[1],t[0]))
    comma=pp.Suppress(',')
    quoted=pp.Regex(r'("|\').*?\1').setParseAction(
              lambda s,l,t: [int(e) for e in t[0].strip('\'"').split(',')])
    line=pp.Word(pp.alphas)+comma+date+comma+quoted+comma+quoted+comma+integer
    return line

line=line_grammar()
print(line.parseString(st))
# ['Place', datetime.date(2010, 9, 8), 15, 531, 2, 909, 650]

The advantage is you parse, convert, and validate in a few lines. Note that the ints are all converted to ints and the date to a datetime structure.