[re.finditer] Getting all occurences in one go?

[Gilles Ganault]

Hello

I'd like to make sure there isn't an easier way to extract all the
occurences found with re.finditer:

=======================
req = urllib2.Request(url, None, headers)
response = urllib2.urlopen(req).read()
matches = re.compile("(\d+).html").finditer(response)
# ----------- BEGIN
for match in matches:
if mytable[item]=="":
mytable[item]= match.group(1)
else:
mytable[item]= mytable[item] + "," + match.group(1) #
----------- END
=======================

Can the lines between BEGIN/END be simplified so I can copy all the
items into the mytable[] dictionary in one go, instead of getting each
one in a row, and append them with a comma, eg.

# I want this : mytable[123] = 6548,8457,etc."
# DOESN' T WORK
# mytable[item] = matches.group(0)
mytable[item] = matches.getall()

Thank you.

[Gilles Ganault]

On Sun, 22 Apr 2007 23:28:23 +0200, Gilles Ganault <no****@nospam.com>
wrote:

>I'd like to make sure there isn't an easier way to extract all the
occurences found with re.finditer:

Oops, s/match/item/:

req = urllib2.Request(url, None, headers)
response = urllib2.urlopen(req).read()
matches = re.compile("(\d+).html").finditer(response)
# ----------- BEGIN
for item in matches:
if mytable[item]=="":
mytable[item]= match.group(1)
else:
mytable[item]= mytable[item] + "," + match.group(1) #
----------- END

[Paul Rubin]

Gilles Ganault <no****@nospam.comwrites:

for match in matches:
if mytable[item]=="":
mytable[item]= match.group(1)
else:
mytable[item]= mytable[item] + "," + match.group(1) #
----------- END
=======================

Can the lines between BEGIN/END be simplified so I can copy all the
items into the mytable[] dictionary in one go, instead of getting each
one in a row, and append them with a comma, eg.

Yes, look at the string.join method and generator expressions. You'd
say:

mytable[item] = ','.join(m.group(1) for m in matches)

[Gilles Ganault]

On 22 Apr 2007 15:33:37 -0700, Paul Rubin
<http://ph****@NOSPAM.invalidwrote:

mytable[item] = ','.join(m.group(1) for m in matches)

Thanks, that did it.