Count the occurrence of characters in a string

Question

I'm being asked to count the occurrence of letters and spaces and punctuation characters in a string from user input and and in the printout the letters should appear in order in which they appear in the text, but no letter should appear twice and also lowercase and uppercase should be counted as one. my code so far looks like this.

S = str(input("Enter a string: "))
S = S.lower()
alpha = "abcdefghijklmnopqrstuvwxyz!@#$%^&*()! "

for char in S:
     if char in alpha:
          count = S.count(char)
     print(char,':',count)

output

Enter a string: hello
h : 1
e : 1
l : 2
l : 2
o : 1

i want it to look like this

Enter a string: hello
    h : 1
    e : 1
    l : 2
    o : 1

i was thinking if i turned the string and the the count of occurred chars into a nested list like

Enter a string: hello
[['h', 1], ['e', 1], ['l', 2], ['l', 2], ['o', 1]]

could i remove the list that are the same and leave just one?

John1024 · Accepted Answer · 2015-03-24 05:47:59Z

6

>>> s = 'hello'
>>> s = s.lower()
>>> print('\n'.join('{} : {}'.format(c, s.count(c)) for i, c in enumerate(s) if c in "abcdefghijklmnopqrstuvwxyz!@#$%^&*()! " and c not in s[:i]))
h : 1
e : 1
l : 2
o : 1

edited Mar 24, 2015 at 5:47

answered Mar 24, 2015 at 5:39

John1024

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Comments

Saksham Varma · Accepted Answer · 2015-03-24 05:48:43Z

6

This is a classic situation for using Counter of the collections module:

from collections import Counter

S ='hello'
S = S.lower()
d = Counter(S)
for item, ct in d.items():
    print '%s occured %d times' % (item, ct)

answered Mar 24, 2015 at 5:48

Saksham Varma

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2 Comments

Mike Over a year ago

so Counter is like a dictionary? Counters is a container in which you can store things. I'm not to familiar with them

Saksham Varma Over a year ago

Yes Counter is inherited from dict class. So, it's a special dictionary for maintaining counts.

Paul Rooney · Accepted Answer · 2015-03-24 22:25:16Z

1

Just iterate over a set

S = str(input("Enter a string: ")).lower()
alpha = "abcdefghijklmnopqrstuvwxyz!@#$%^&*()! "

for char in set(S):
    if char in alpha:
        print(char,':',S.count(char))

edited Mar 24, 2015 at 22:25

answered Mar 24, 2015 at 6:29

Paul Rooney

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Comments

user648852 · Accepted Answer · 2015-03-24 21:55:17Z

0

You can do:

S = 'hello'
S = S.lower()
alpha = "abcdefghijklmnopqrstuvwxyz!@#$%^&*()! "

seen=set()

for c in S:
    if c not in seen and c in alpha:
        print '{} : {}'.format(c, S.count(c))
        seen.add(c)

Prints:

h : 1
e : 1
l : 2
o : 1

answered Mar 24, 2015 at 21:55

user648852

Comments

Mike · Accepted Answer · 2015-03-25 03:49:03Z

so i figured out how to do it without using sets and counters.

S = (input("Enter a string: "))
S = S.lower()
alpha = "abcdefghijklmnopqrstuvwxyz!@#$%^&*()! "
L = []
#counts the occurrence of characters in a string.
#then puts them in a list
for char in S:
     if char in alpha:
          count = S.count(char)
     L.append(char)
     L.append(count)
#turns list into a nested list
L2 = []
i = 0
while i <len(L):
        L2.append(L[i:i+2])
        i += 2
#checks that the lists dont appear more than once 
L3 = []
for i in L2:
     if i not in L3:
          L3.append(i)

# print out the letters and count of the letter
for i,k in L3:
     print(i,k)

might be long but it works. would like your opinion on it?

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