Pythonic way to count sequences
Denis McMahon
denismfmcmahon at gmail.com
Thu Apr 25 19:29:28 EDT 2013
On Wed, 24 Apr 2013 22:05:52 -0700, CM wrote:
> I have to count the number of various two-digit sequences in a list such
> as this:
>
> mylist = [(2,4), (2,4), (3,4), (4,5), (2,1)] # (Here the (2,4) sequence
> appears 2 times.)
>
> and tally up the results, assigning each to a variable. The inelegant
> first pass at this was something like...
>
> # Create names and set them all to 0 alpha = 0 beta = 0 delta = 0 gamma
> = 0 # etc...
>
> # loop over all the tuple sequences and increment appropriately for
> sequence_tuple in list_of_tuples:
> if sequence_tuple == (1,2):
> alpha += 1
> if sequence_tuple == (2,4):
> beta += 1
> if sequence_tuple == (2,5):
> delta +=1
> # etc... But I actually have more than 10 sequence types.
>
> # Finally, I need a list created like this:
> result_list = [alpha, beta, delta, gamma] #etc...in that order
>
> I can sense there is very likely an elegant/Pythonic way to do this, and
> probably with a dict, or possibly with some Python structure I don't
> typically use. Suggestions sought. Thanks.
mylist = [ (3,3), (1,2), "fred", ("peter",1,7), 1, 19, 37, 28.312,
("monkey"), "fred", "fred", (1,2) ]
bits = {}
for thing in mylist:
if thing in bits:
bits[thing] += 1
else:
bits[thing] = 1
for thing in bits:
print thing, " occurs ", bits[thing], " times"
outputs:
(1, 2) occurs 2 times
1 occurs 1 times
('peter', 1, 7) occurs 1 times
(3, 3) occurs 1 times
28.312 occurs 1 times
fred occurs 3 times
19 occurs 1 times
monkey occurs 1 times
37 occurs 1 times
if you want to check that thing is a 2 int tuple then use something like:
for thing in mylist:
if isinstance( thing, tuple ) and len( thing ) == 2 and isinstance
( thing[0], ( int, long ) ) and isinstance( thing[1], ( int, long) ):
if thing in bits:
bits[thing] += 1
else:
bits[thing] = 1
--
Denis McMahon, denismfmcmahon at gmail.com
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