CODEWITHSUNDEEP
pythoncounting
lfhfsfhjsf
lfhfsfhjsf

Reputation: 63

Correct output for function that counts occurrences of each digit in a string

I want the output of the code to be something like this if the user enters a string of numbers like let's say... 122033

Enter string of numbers: 122033
0 occurs 1 time
1 occurs 1 time
2 occurs 2 times
3 occurs 2 times


def count_digits(s):
    res = [0]*10
    for x in s:
        res[int(x)] += 1
    while 0 in res:
        res.remove(0)
    return res

def main():
    s=input("Enter string of numbers: ")

    print(count_digits(s))
main()

This is the program that I have so far. At it's current state, if a user enters something like 122033 the output is: [1,1,2,2]

Note: I cannot use collections for this.

Upvotes: 0

Views: 3194

Answers (3)

user2555451
user2555451

Reputation:

Without collections.Counter, here is a pretty short and efficient solution:

>>> def count_digits(inp):
...     for a,b in sorted((c, inp.count(c)) for c in set(inp)):
...         print("{} occurs {} times".format(a, b))
...
>>> mystr = input("Enter string of numbers: ")
Enter string of numbers: 122033
>>> count_digits(mystr)
0 occurs 1 times
1 occurs 1 times
2 occurs 2 times
3 occurs 2 times
>>>

As Peter DeGlopper notes in the comment below, this solution will work for any character set, not just digits. If however you want it to only work with digits, all you need to do is make a slight modification to the for-loop line:

for a,b in sorted((c, inp.count(c)) for c in set(inp) if c.isdigit()):

Adding if c.isdigit() to the end of that will make it only capture digits.

Upvotes: 2

Burhan Khalid
Burhan Khalid

Reputation: 174624

An approach that does not use counters:

d = {}

for i in somestring:
   if i not in d:
     d[i] = 1
   else:
     d[i] += 1
for k,v in d.iteritems():
   print('{0} occurs {1} times'.format(k,v))

Upvotes: 0

Peter DeGlopper
Peter DeGlopper

Reputation: 37319

You're pretty close to a working solution, but removing all the 0-count entries changes the indices of your list. You already need to write some custom pretty printing code, so just leave the 0s in and skip elements where the count is 0. Maybe something like this:

def count_digits(s):
    res = [0]*10
    for x in s:
        res[int(x)] += 1
    return res

def print_counts(counts):
    for (index, count) in enumerate(counts):
        if count == 1:
            print("%d occurs %d time" % (index, count))
        elif count > 1:
            print("%d occurs %d times" % (index, count))

def main():
    s=input("Enter string of numbers: ")

    print_counts(count_digits(s))

Upvotes: 2

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