Re: [Tutor] Counting help
I'll do this again, just because I like sending email.
Very similar to Alan G.'s -- but with the do first, ask forgiveness later
a = [Joe Smith, Joe Smith, Jack Smith, Sam Love, Joe Smith]
b = {}
for x in a:
try: b[x] += 1
except KeyError: b[x] = 1
Access count like this, of course,
b[Joe Smith]
3
Jacob Schmidt
- Original Message -
From: Scott Oertel [EMAIL PROTECTED]
To: Tutor@python.org
Sent: Tuesday, August 23, 2005 1:01 PM
Subject: [Tutor] Counting help
I have extracted a list of names, i.e.
Joe Smith
Joe Smith
Jack Smith
Sam Love
Joe Smith
I need to be able to count the occurances of these names and I really
don't have any idea where to begin.
Any ideas? excuse me this is my first post to this list, I hope I
included enough information.
-Scott Oertel
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Re: [Tutor] Counting help
Kent Johnson wrote:
Scott Oertel wrote:
The next problem I have though is creating the dict,
i have a loop, but i can't figure out how to compile the dict, it is
returning this: ('Joey Gale', ('Scott Joe', 'This is lame' )))
listofnames = []
while (cnt number[1][0]):
if (date[2] == today[2]):
test = regex.findall(M.fetch(int(number[1][0]) - cnt,
'(BODY[HEADER.FIELDS (FROM)])')[1][0][1].rstrip())
cnt += 1
if (nameofsender != []):
print nameofsender[0]
listofnames = nameofsender[0], listofnames
I think you want
listofnames.append(nameofsender[0])
which will add nameofsender[0] to the list. What you have -
listofnames = nameofsender[0], listofnames
is making a tuple (a pair) out of the new name and the old list, and assigning it to listofnames. Kind of like CONS in LISP - but Python lists are more like arrays than like LISP lists.
Kent
else:
no_name += 1
else: break
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Thank you everyone, this is exactly it, I'm going to take that link
from byron and read up on dicts/lists.
-Scott Oertel
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[Tutor] Counting help
I have extracted a list of names, i.e. Joe Smith Joe Smith Jack Smith Sam Love Joe Smith I need to be able to count the occurances of these names and I really don't have any idea where to begin. Any ideas? excuse me this is my first post to this list, I hope I included enough information. -Scott Oertel ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Counting help
On 8/23/05, Scott Oertel [EMAIL PROTECTED] wrote:
I have extracted a list of names, i.e.
Joe Smith
Joe Smith
Jack Smith
Sam Love
Joe Smith
I need to be able to count the occurances of these names and I really
don't have any idea where to begin.
Any ideas? excuse me this is my first post to this list, I hope I
included enough information.
-Scott Oertel
Ideally, you would put your names into a list or dictionary to make
working with them easier. If all you're trying to do is count them
(and your list of names is long), you might consider a dictionary
which you would use like so:
#This is just the first thing I considered.
l = ['a list of names']
d = {}
for name in namelist:
if d.has_key(name):
x = d.get(name)
d[name] = x + 1
else:
d[name] = 1
Luis
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Re: [Tutor] Counting help
Luis N wrote:
Ideally, you would put your names into a list or dictionary to make
working with them easier. If all you're trying to do is count them
(and your list of names is long), you might consider a dictionary
which you would use like so:
#This is just the first thing I considered.
l = ['a list of names']
d = {}
for name in namelist:
if d.has_key(name):
x = d.get(name)
d[name] = x + 1
else:
d[name] = 1
100% agreed. I have used this approach before and it works great...
Byron
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Re: [Tutor] Counting help
Byron wrote:
Luis N wrote:
Ideally, you would put your names into a list or dictionary to make
working with them easier. If all you're trying to do is count them
(and your list of names is long), you might consider a dictionary
which you would use like so:
#This is just the first thing I considered.
l = ['a list of names']
d = {}
for name in namelist:
if d.has_key(name):
x = d.get(name)
d[name] = x + 1
else:
d[name] = 1
100% agreed. I have used this approach before and it works great...
Byron
Thanks for the snipplet, it's perfect for what I'm doing, I wasn't
aware of the has_key() or get(), this is very usefull.
-Scott Oertel
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Re: [Tutor] Counting help
Scott Oertel wrote:
Byron wrote:
Luis N wrote:
Ideally, you would put your names into a list or dictionary to make
working with them easier. If all you're trying to do is count them
(and your list of names is long), you might consider a dictionary
which you would use like so:
#This is just the first thing I considered.
l = ['a list of names']
d = {}
for name in namelist:
if d.has_key(name):
x = d.get(name)
d[name] = x + 1
else:
d[name] = 1
100% agreed. I have used this approach before and it works great...
Byron
Thanks for the snipplet, it's perfect for what I'm doing, I wasn't
aware of the has_key() or get(), this is very usefull.
-Scott Oertel
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The next problem I have though is creating the dict,
i have a loop, but i can't figure out how to compile the dict, it is
returning this: ('Joey Gale', ('Scott Joe', ('This is lame' )))
listofnames = []
while (cnt number[1][0]):
if (date[2] == today[2]):
test = regex.findall(M.fetch(int(number[1][0]) - cnt,
'(BODY[HEADER.FIELDS (FROM)])')[1][0][1].rstrip())
cnt += 1
if (nameofsender != []):
print nameofsender[0]
listofnames = nameofsender[0], listofnames
else:
no_name += 1
else: break
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Re: [Tutor] Counting help
I have extracted a list of names, i.e.
Joe Smith
Joe Smith
Jack Smith
Sam Love
Joe Smith
I need to be able to count the occurances of these names and I
really don't have any idea where to begin.
The classic way to do this kind of thing is with a dictionary:
names = [
Joe Smith,
Joe Smith,
Jack Smith,
Sam Love,
Joe Smith]
counts = {}
for name in names:
if name in counts:
counts[name] += 1
else: counts[name] = 1
print counts
You could also use a list comprehension combined with the list
count() method but I doubt if its much faster.
HTH,
Alan G
Author of the Learn to Program web tutor
http://www.freenetpages.co.uk/hp/alan.gauld
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Re: [Tutor] Counting help
Scott Oertel wrote:
The next problem I have though is creating the dict,
i have a loop, but i can't figure out how to compile the dict, it is
returning this: ('Joey Gale', ('Scott Joe', 'This is lame' )))
listofnames = []
while (cnt number[1][0]):
if (date[2] == today[2]):
test = regex.findall(M.fetch(int(number[1][0]) - cnt,
'(BODY[HEADER.FIELDS (FROM)])')[1][0][1].rstrip())
cnt += 1
if (nameofsender != []):
print nameofsender[0]
listofnames = nameofsender[0], listofnames
I think you want
listofnames.append(nameofsender[0])
which will add nameofsender[0] to the list. What you have -
listofnames = nameofsender[0], listofnames
is making a tuple (a pair) out of the new name and the old list, and assigning
it to listofnames. Kind of like CONS in LISP - but Python lists are more like
arrays than like LISP lists.
Kent
else:
no_name += 1
else: break
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Re: [Tutor] Counting help
Luis N wrote:
Ideally, you would put your names into a list or dictionary to make
working with them easier. If all you're trying to do is count them
(and your list of names is long), you might consider a dictionary
which you would use like so:
#This is just the first thing I considered.
l = ['a list of names']
d = {}
for name in namelist:
if d.has_key(name):
x = d.get(name)
d[name] = x + 1
else:
d[name] = 1
dict.get() allows a default argument that will be used if the key doesn't
exist, so this can be shortened to
for name in namelist:
d[name] = d.get(name, 0) + 1
Kent
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Re: [Tutor] Counting help
Hi Scott, The site ( http://www.greenteapress.com ) has a wonderful tutorial on it for Python that quickly teaches one (within a few minutes) how to work with dictionaries. I would highly recommend that you check it out. It's well worth it... Byron ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
[Tutor] Counting help
listofnames = nameofsender[0], listofnames does not add a name to a list. Rather it creates a tuple of the new name and the list and then binds the tuple to the list name. That's why you wind up with the lisp style list. To add a name to the head of the list use listofnames.insert(0, nameofsender[0]) If you are using a version of Python that supports sets, using sets would be much simpler since the duplicates get discarded automatically. import sets # python2.3 setofnames = sets.Set() while. setofnames.add(nameofsender[0]) len(setofnames) # count of distinct names -- Lloyd Kvam Venix Corp ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
