# tallying occurrences in list

Discussion in 'Python' started by kj, Jun 4, 2010.

1. ### kjGuest

Task: given a list, produce a tally of all the distinct items in
the list (for some suitable notion of "distinct").

Example: if the list is ['a', 'b', 'c', 'a', 'b', 'c', 'a', 'b',
'c', 'a'], then the desired tally would look something like this:

[('a', 4), ('b', 3), ('c', 3)]

I find myself needing this simple operation so often that I wonder:

1. is there a standard name for it?
2. is there already a function to do it somewhere in the Python
standard library?

Granted, as long as the list consists only of items that can be
used as dictionary keys (and Python's equality test for hashkeys
agrees with the desired notion of "distinctness" for the tallying),
then the following does the job passably well:

def tally(c):
t = dict()
for x in c:
t[x] = t.get(x, 0) + 1
return sorted(t.items(), key=lambda x: (-x[1], x[0]))

But, of course, if a standard library solution exists it would be
preferable. Otherwise I either cut-and-paste the above every time
I need it, or I create a module just for it. (I don't like either
of these, though I suppose that the latter is much better than the
former.)

So anyway, I thought I'd ask.

~K

kj, Jun 4, 2010

2. ### Paul RubinGuest

kj <> writes:
> 1. is there a standard name for it?

I don't know of one, or a stdlib for it, but it's pretty trivial.

> def tally(c):
> t = dict()
> for x in c:
> t[x] = t.get(x, 0) + 1
> return sorted(t.items(), key=lambda x: (-x[1], x[0]))

I like to use defaultdict and tuple unpacking for code like that:

from collections import defaultdict
def tally(c):
t = defaultdict(int)
for x in c:
t[x] += 1
return sorted(t.iteritems(), key=lambda (k,v): (-v, k))

Paul Rubin, Jun 4, 2010

3. ### Peter OttenGuest

kj wrote:

>
>
>
>
>
> Task: given a list, produce a tally of all the distinct items in
> the list (for some suitable notion of "distinct").
>
> Example: if the list is ['a', 'b', 'c', 'a', 'b', 'c', 'a', 'b',
> 'c', 'a'], then the desired tally would look something like this:
>
> [('a', 4), ('b', 3), ('c', 3)]
>
> I find myself needing this simple operation so often that I wonder:
>
> 1. is there a standard name for it?
> 2. is there already a function to do it somewhere in the Python
> standard library?
>
> Granted, as long as the list consists only of items that can be
> used as dictionary keys (and Python's equality test for hashkeys
> agrees with the desired notion of "distinctness" for the tallying),
> then the following does the job passably well:
>
> def tally(c):
> t = dict()
> for x in c:
> t[x] = t.get(x, 0) + 1
> return sorted(t.items(), key=lambda x: (-x[1], x[0]))
>
> But, of course, if a standard library solution exists it would be
> preferable. Otherwise I either cut-and-paste the above every time
> I need it, or I create a module just for it. (I don't like either
> of these, though I suppose that the latter is much better than the
> former.)
>
> So anyway, I thought I'd ask.

Python 3.1 has, and 2.7 will have collections.Counter:

>>> from collections import Counter
>>> c = Counter("abcabcabca")
>>> c.most_common()

[('a', 4), ('c', 3), ('b', 3)]

Peter

Peter Otten, Jun 4, 2010
4. ### MagdollGuest

On Jun 4, 11:28 am, Paul Rubin <> wrote:
> kj <> writes:
> > 1. is there a standard name for it?

>
> I don't know of one, or a stdlib for it, but it's pretty trivial.
>
> > def tally(c):
> >     t = dict()
> >     for x in c:
> >         t[x] = t.get(x, 0) + 1
> >     return sorted(t.items(), key=lambda x: (-x[1], x[0]))

>
> I like to use defaultdict and tuple unpacking for code like that:
>
>  from collections import defaultdict
>  def tally(c):
>      t = defaultdict(int)
>      for x in c:
>          t[x] += 1
>      return sorted(t.iteritems(), key=lambda (k,v): (-v, k))

I would also very much like to see this become part of the standard
library. Sure the code is easy to write but I use this incredibly
often and I've always wished I would have a one-line function call
that has the same output as the mysql query:

"SELECT id, count(*) FROM table GROUP BY somefield"

or maybe there is already a short solution to this that I'm not aware
of...

Magdoll, Jun 4, 2010
5. ### MagdollGuest

On Jun 4, 11:33 am, Peter Otten <> wrote:
> kj wrote:
>
> > Task: given a list, produce a tally of all the distinct items in
> > the list (for some suitable notion of "distinct").

>
> > Example: if the list is ['a', 'b', 'c', 'a', 'b', 'c', 'a', 'b',
> > 'c', 'a'], then the desired tally would look something like this:

>
> > [('a', 4), ('b', 3), ('c', 3)]

>
> > I find myself needing this simple operation so often that I wonder:

>
> > 1. is there a standard name for it?
> > 2. is there already a function to do it somewhere in the Python
> >    standard library?

>
> > Granted, as long as the list consists only of items that can be
> > used as dictionary keys (and Python's equality test for hashkeys
> > agrees with the desired notion of "distinctness" for the tallying),
> > then the following does the job passably well:

>
> > def tally(c):
> >     t = dict()
> >     for x in c:
> >         t[x] = t.get(x, 0) + 1
> >     return sorted(t.items(), key=lambda x: (-x[1], x[0]))

>
> > But, of course, if a standard library solution exists it would be
> > preferable.  Otherwise I either cut-and-paste the above every time
> > I need it, or I create a module just for it.  (I don't like either
> > of these, though I suppose that the latter is much better than the
> > former.)

>
> > So anyway, I thought I'd ask.

>
> Python 3.1 has, and 2.7 will have collections.Counter:
>
> >>> from collections import Counter
> >>> c = Counter("abcabcabca")
> >>> c.most_common()

>
> [('a', 4), ('c', 3), ('b', 3)]
>
> Peter

Thanks Peter, I think you just answered my post

Magdoll, Jun 4, 2010
6. ### MRABGuest

kj wrote:
>
>
>
>
> Task: given a list, produce a tally of all the distinct items in
> the list (for some suitable notion of "distinct").
>
> Example: if the list is ['a', 'b', 'c', 'a', 'b', 'c', 'a', 'b',
> 'c', 'a'], then the desired tally would look something like this:
>
> [('a', 4), ('b', 3), ('c', 3)]
>
> I find myself needing this simple operation so often that I wonder:
>
> 1. is there a standard name for it?
> 2. is there already a function to do it somewhere in the Python
> standard library?
>
> Granted, as long as the list consists only of items that can be
> used as dictionary keys (and Python's equality test for hashkeys
> agrees with the desired notion of "distinctness" for the tallying),
> then the following does the job passably well:
>
> def tally(c):
> t = dict()
> for x in c:
> t[x] = t.get(x, 0) + 1
> return sorted(t.items(), key=lambda x: (-x[1], x[0]))
>
> But, of course, if a standard library solution exists it would be
> preferable. Otherwise I either cut-and-paste the above every time
> I need it, or I create a module just for it. (I don't like either
> of these, though I suppose that the latter is much better than the
> former.)
>
> So anyway, I thought I'd ask.
>

In Python 3 there's the 'Counter' class in the 'collections' module.
It'll also be in Python 2.7.

For earlier versions there's this:

http://code.activestate.com/recipes/576611/

MRAB, Jun 4, 2010
7. ### Lie RyanGuest

On 06/05/10 04:38, Magdoll wrote:
> On Jun 4, 11:33 am, Peter Otten <> wrote:
>> kj wrote:
>>
>>> Task: given a list, produce a tally of all the distinct items in
>>> the list (for some suitable notion of "distinct").

>>
>>> Example: if the list is ['a', 'b', 'c', 'a', 'b', 'c', 'a', 'b',
>>> 'c', 'a'], then the desired tally would look something like this:

>>
>>> [('a', 4), ('b', 3), ('c', 3)]

>>
>>> I find myself needing this simple operation so often that I wonder:

>>
>>> 1. is there a standard name for it?
>>> 2. is there already a function to do it somewhere in the Python
>>> standard library?

>>
>>> Granted, as long as the list consists only of items that can be
>>> used as dictionary keys (and Python's equality test for hashkeys
>>> agrees with the desired notion of "distinctness" for the tallying),
>>> then the following does the job passably well:

>>
>>> def tally(c):
>>> t = dict()
>>> for x in c:
>>> t[x] = t.get(x, 0) + 1
>>> return sorted(t.items(), key=lambda x: (-x[1], x[0]))

>>
>>> But, of course, if a standard library solution exists it would be
>>> preferable. Otherwise I either cut-and-paste the above every time
>>> I need it, or I create a module just for it. (I don't like either
>>> of these, though I suppose that the latter is much better than the
>>> former.)

>>
>>> So anyway, I thought I'd ask.

>>
>> Python 3.1 has, and 2.7 will have collections.Counter:
>>
>>>>> from collections import Counter
>>>>> c = Counter("abcabcabca")
>>>>> c.most_common()

>>
>> [('a', 4), ('c', 3), ('b', 3)]
>>
>> Peter

>
>
> Thanks Peter, I think you just answered my post

If you're using previous versions (2.4 and onwards) then:

[(o, len(list(g))) for o, g in itertools.groupby(sorted(myList))]

Lie Ryan, Jun 4, 2010
8. ### kjGuest

Thank you all!

~K

kj, Jun 4, 2010
9. ### Sreenivas Reddy ThatiparthyGuest

On Jun 4, 11:14 am, kj <> wrote:
> Task: given a list, produce a tally of all the distinct items in
> the list (for some suitable notion of "distinct").
>
> Example: if the list is ['a', 'b', 'c', 'a', 'b', 'c', 'a', 'b',
> 'c', 'a'], then the desired tally would look something like this:
>
> [('a', 4), ('b', 3), ('c', 3)]
>
> I find myself needing this simple operation so often that I wonder:
>
> 1. is there a standard name for it?
> 2. is there already a function to do it somewhere in the Python
>    standard library?
>
> Granted, as long as the list consists only of items that can be
> used as dictionary keys (and Python's equality test for hashkeys
> agrees with the desired notion of "distinctness" for the tallying),
> then the following does the job passably well:
>
> def tally(c):
>     t = dict()
>     for x in c:
>         t[x] = t.get(x, 0) + 1
>     return sorted(t.items(), key=lambda x: (-x[1], x[0]))
>
> But, of course, if a standard library solution exists it would be
> preferable.  Otherwise I either cut-and-paste the above every time
> I need it, or I create a module just for it.  (I don't like either
> of these, though I suppose that the latter is much better than the
> former.)
>
> So anyway, I thought I'd ask.
>
> ~K

set([(k,yourList.count(k)) for k in yourList])

Sreenivas Reddy Thatiparthy, Jun 5, 2010
10. ### Paul RubinGuest

Sreenivas Reddy Thatiparthy <> writes: