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 need to split this string by the comma's. Though due to the comma's used in the numerical data fields the split doesn't work correctly. So I want to remove the comma's in the numbers before running splitting the string.
Thanks.
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.
"123,456,789" - then it will be "123.456,789"
999,999 inside the quotes..."999,999,999" at all - you would need something with lookaround, like: "(\d+),(?=\d\d\d(,\d\d\d)*") (which should be replaced with "\1)>>> 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']
io.StringIO: docs.python.org/py3k/library/…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')]
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.
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
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!
