Re: Cascading ifs
tbl = [(my_regex, doSomething), (my_regex2, doSomething2), (my_regex3, doSomething3)] for regex, fun in tbl: match = regexp.match(line) if match: fun(line) break Thank you for the idea. This is a bit more difficult when functions need to work with a common context, but in that case I could store the context in an object and use the object's methods. Thanks, Ernesto -- http://mail.python.org/mailman/listinfo/python-list
Cascading ifs
Hi experts, How would you do this without the more and more indenting cascade of ifs?: match = my_regex.search(line) if match: doSomething(line) else: match = my_regex2.search(line) if match: doSomething2(line) else: match = my_regex3.search(line) if match: doSomething3(line) etc. Thanks in advance and regards, Ernesto -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a commas-in-between idiom?
I've collected a bunch of list pydioms and other notes here: http://effbot.org/zone/python-list.htm Thank you for the suggestion. Ernesto -- http://mail.python.org/mailman/listinfo/python-list
Is there a commas-in-between idiom?
Hi experts, it's very common that I have a list and I want to print it with commas in between. How do I do this in an easy manner, whithout having the annoying comma in the end? code list = [1,2,3,4,5,6] # the easy way for element in list: print element, ',', print # this is what I really want. is there some way better? if (len(list) 0): print list[0], for element in list[1:]: print ',', element, /code Thx, Ernesto -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a commas-in-between idiom?
Ernesto García García wrote: Hi experts, it's very common that I have a list and I want to print it with commas in between. How do I do this in an easy manner, whithout having the annoying comma in the end? code list = [1,2,3,4,5,6] # the easy way for element in list: print element, ',', print # this is what I really want. is there some way better? if (len(list) 0): print list[0], for element in list[1:]: print ',', element, /code Thx, Ernesto mylist = [1,2,3,4,5,6] print ','.join(map(str, mylist)) Great solution! Thank all of you, Ernesto -- http://mail.python.org/mailman/listinfo/python-list
Re: Is there a commas-in-between idiom?
Tim Peters wrote: More idiomatic as if len(list) 0: and even more so as plain if list: print list[0], for element in list[1:]: print ',', element, Do you really want a space before and after each inter-element comma? No, but it was only an example. I usually go for string concatenation. An often-overlooked alternative to playing with ,.join() is: print str(list)[1:-1] That's funny! Not that I like it more that the join solution, but funny nevertheless. Thank you, Ernesto -- http://mail.python.org/mailman/listinfo/python-list
[Newbie] List from a generator function
Hi all, I'm sure there is a better way to do this: [random.choice(possible_notes) for x in range(length)] Regards, Ernesto -- http://mail.python.org/mailman/listinfo/python-list
Re: [Newbie] List from a generator function
I'm sure there is a better way to do this: [random.choice(possible_notes) for x in range(length)] There is at least a better way to ask the question. The subject has nothing to do with the body of your post. Or am I missing something? Sorry, I should have explained better. I just want to build a fix length list made up of elements generated by a function, in this case random.choice(). I don't like my list comprehension, because I'm using that dumb variable x and the range() list. Ernesto -- http://mail.python.org/mailman/listinfo/python-list
Re: building lists of dictionaries
parinfo = [{'value':0., 'fixed':0, 'limited':[0,0], 'limits':[0.,0.]}]*6
With this, you are creating a list with 6 references to the same list.
Note that the left operand of '*' is evaluated only once before
multiplying it six times.
Regards,
Tito
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Re: [Newbie] List from a generator function
Thank you guys. So the answer is to keep with the original form, perhaps with xrange. Ernesto -- http://mail.python.org/mailman/listinfo/python-list
Re: building lists of dictionaries
parinfo = [{'value':0., 'fixed':0, 'limited':[0,0],
'limits':[0.,0.]}.copy() for i in xrange(0,6)]
However, this will still reference internal lists that have
been referenced multiple times, such that
parinfo[5]['limited']
[0, 0]
parinfo[4]['limited'][0] = 2
parinfo[5]['limited']
[2, 0]
Interesting. Cut-and-paste to my python prompt and I get
parinfo[5]['limited']
[0, 0]
Tried both Python 2.4.1 and 2.5 beta, Linux, GCC 4.0.2
Of course. The expression within the list comprehension is evaluated for
each iteration, so that the objects are recreated each time. The copy()
for the dictionary is also not needed:
parinfo = [{'value':0., 'fixed':0, 'limited':[0,0],
'limits':[0.,0.]} for i in xrange(0,6)]
parinfo
[{'limited': [0, 0], 'fixed': 0, 'limits': [0.0, 0.0], 'value': 0.0},
{'limited': [0, 0], 'fixed': 0, 'limits': [0.0, 0.0], 'value': 0.0},
{'limited': [0, 0], 'fixed': 0, 'limits': [0.0, 0.0], 'value': 0.0},
{'limited': [0, 0], 'fixed': 0, 'limits': [0.0, 0.0], 'value': 0.0},
{'limited': [0, 0], 'fixed': 0, 'limits': [0.0, 0.0], 'value': 0.0},
{'limited': [0, 0], 'fixed': 0, 'limits': [0.0, 0.0], 'value': 0.0}]
parinfo[0]['limited']
[0, 0]
parinfo[0]['limited'][0]=1
parinfo
[{'limited': [1, 0], 'fixed': 0, 'limits': [0.0, 0.0], 'value': 0.0},
{'limited': [0, 0], 'fixed': 0, 'limits': [0.0, 0.0], 'value': 0.0},
{'limited': [0, 0], 'fixed': 0, 'limits': [0.0, 0.0], 'value': 0.0},
{'limited': [0, 0], 'fixed': 0, 'limits': [0.0, 0.0], 'value': 0.0},
{'limited': [0, 0], 'fixed': 0, 'limits': [0.0, 0.0], 'value': 0.0},
{'limited': [0, 0], 'fixed': 0, 'limits': [0.0, 0.0], 'value': 0.0}]
Ernesto
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[Newbie] Referring to a global variable inside a function
Hi experts,
I've built a class for parsing a user-defined list of files and matching
lines with a user-defined list of regular expressions. It looks like this:
code
import re
import glob
class LineMatcher:
Parses a list of text files, matching their lines with the given
regular expressions and executing associated actions.
def __init__(self):
# List of file names to parse for matching.
self.file_names = []
# Association of reg expressions to actions to execute when match.
self.regexp_action = {}
def go(self):
for file_name in self.file_names:
file = open(file_name)
for line in file:
for regexp, action in self.regexp_action.items():
match = regexp.match(line)
if match:
action(line, match.groupdict())
def add_files(self, file_pattern):
self.file_names.extend(glob.glob(file_pattern))
def add_action(self, regexp_string, action):
self.regexp_action[re.compile(regexp_string)] = action
/code
But then, when I try to use my class using actions with memory it will
fail:
code
import LineMatcher
global count
count = 0
def line_action(line, match_dictionary):
count = count + 1
line_matcher = LineMatcher.LineMatcher()
line_matcher.add_files('*')
line_matcher.add_action(r'(?Pline.*)', line_action)
line_matcher.go()
/code
The error is:
console
Traceback (most recent call last):
File Test.py, line 12, in ?
line_matcher.go()
File LineMatcher.py, line 21, in go
action(line, match.groupdict())
File Test.py, line 7, in line_action
count = count + 1
UnboundLocalError: local variable 'count' referenced before assignment
/console
How would you do this?
Regards,
Tito
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Re: [Newbie] Referring to a global variable inside a function
How would you do this? def line_action(line, match_dictionary): global count # make it a module-global variable, not a function-local count = count + 1 /F OK, I had put it on the global block. Thanks, Ernesto -- http://mail.python.org/mailman/listinfo/python-list
