On Feb 5, 4:05 pm, marduk <mar...@python.net> wrote:
>
> Although that implementation also scans the string 10 times (s.count()),
> which may not be as efficient (although it is happening in C, so perhaps
> not).
>
> A better solution involves only scanning the string once.
agreed. i was specifically showing how to reverse the loop.
using the much-better-suited Counter class:
from collections import Counter
s=input("Enter a string, eg(4856w23874): ")
checkS=['0','1','2','3','4','5','6','7','8','9']
cnt = Counter()
for char in s:
cnt[char] += 1
for char, tally in sorted(cnt.items()):
if char in checkS and tally > 0:
if tally == 1:
print(char,"occurs 1 time.")
else:
print(char, "occurs", tally,"times.")
>>>
Enter a string, eg(4856w23874): 192398209asdfbc12903348955
0 occurs 2 times.
1 occurs 2 times.
2 occurs 3 times.
3 occurs 3 times.
4 occurs 1 time.
5 occurs 2 times.
8 occurs 2 times.
9 occurs 5 times.
>>>
HTH,
Don
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