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

If you need a regex solution, this should do:

r"(\d+),(?=\d\d\d)"

then replace with:

"\1"

It will replace any comma-delimited numbers anywhere in your string with their number-only equivalent, thus turning this:

Place,08/09/2010,"15,531","548,122,909",650

into this:

Place,08/09/2010,"15531","548122909",650

I'm sure there are a few holes to be found and places you don't want this done, and that's why you should use a parser!

Good luck!

Answer 2

Another way of doing it using regex directly:

>>> import re
>>> data = "Place,08/09/2010,\"15,531\",\"2,909\",650"
>>> res = re.findall(r"(\w+),(\d{2}/\d{2}/\d{4}),\"([\d,]+)\",\"([\d,]+)\",(\d+)", data)
>>> res
[('Place', '08/09/2010', '15,531', '2,909', '650')]

Answer 3

new_string = re.sub(r'"(\d+),(\d+)"', r'\1.\2', original_string)

This will substitute the , inside the quotes with a . and you can now just use the strings split method.

Answer 4

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.

Answer 5

>>> from StringIO import StringIO
>>> import csv
>>> r = csv.reader(StringIO('Place,08/09/2010,"15,531","2,909",650'))
>>> r.next()
['Place', '08/09/2010', '15,531', '2,909', '650']

Answer 6

a = """Place,08/09/2010,"15,531","2,909",650""".split(',')
result = []
i=0
while i<len(a):
    if not "\"" in a[i]:
        result.append(a[i])
    else:
        string = a[i]
        i+=1
        while True:
            string += ","+a[i]
            if "\"" in a[i]:
                break
            i+=1
        result.append(string)
    i+=1
print result

Result:
['Place', '08/09/2010', '"15,531"', '"2,909"', '650']
Not a big fan of regular expressions unless you absolutely need them