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what i'm finding hard is that you can't nest "i for i in [list]" inside Boolean logic checks, and not wanting to list all 26 characters

Snippyro
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    [`collections.Counter("ababac")`](https://docs.python.org/3/library/collections.html#collections.Counter)? – jonrsharpe Jan 04 '19 at 12:17
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    Possible duplicate of [Counting Letter Frequency in a String (Python)](https://stackoverflow.com/questions/40985203/counting-letter-frequency-in-a-string-python) – pmcarpan Jan 04 '19 at 12:24

3 Answers3

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You can use "aabc".count('a') to give you the amount of times a appears in aabc, for example. In this example, it gives 2.

Building on from this, you could implement some simple code to check how many times every letter of the alphabet occurs like so:

import string

s = 'ababac'

for letter in string.ascii_lowercase:
    print("%s appears %s times" % (letter, s.count(letter)))

Which produces the output:

a appears 3 times
b appears 2 times
c appears 1 times
d appears 0 times
e appears 0 times
f appears 0 times
g appears 0 times
...

NB: you don't have to list all 26 characters in the alphabet because string.ascii_lowercase has that for you. Note there is also string.ascii_uppercase and so on. More information: https://docs.python.org/3/library/string.html

For more information about str.count(sub[, start[, end]]): https://docs.python.org/3/library/stdtypes.html#str.count

Also collections.Counter can create a dictionary for all the letters in the string. Using this, you could rewrite the above code like so:

import string
from collections import Counter

counts=Counter('ababac')
# counts = {'a': 3, 'b': 2, 'c': 1}

for letter in string.ascii_lowercase:
    if letter in counts:
        print("%s appears %s times" % (letter, counts[letter]))
    else:
        print("%s doesn't appear" % letter)

Output:

a appears 3 times
b appears 2 times
c appears 1 times
d doesn't appear
e doesn't appear
f doesn't appear
g doesn't appear
...

Edit NB:

print("%s appears %s times" % (letter, s.count(letter)))

can also be written as:

print("{} appears {} times".format(letter, s.count(letter)))

using the newer str.format interface. More information: https://docs.python.org/3/library/stdtypes.html#str.format.

  • thanks! this solves my problem, but I don't get how: `("%s appears %s times" % (letter, s.count(letter)))` works. where can I check about this specific string usage? – Snippyro Jan 04 '19 at 12:46
  • It's called string formatting. Some more information: https://docs.python.org/3/library/stdtypes.html#printf-style-string-formatting. There are actually a newer and better ways to do this too nowadays, more information about another way: https://docs.python.org/3/library/stdtypes.html#str.format –  Jan 04 '19 at 12:53
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using defualtdict you can do this way .

from collections import defaultdict
def fun():
    return 0

dic=defaultdict(fun)

string ='ababa'

for i in string:
    dic[i]+=1

for i in dic:
    print(i,dic[i])

output:

 a 3 
 b 2
sahasrara62
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  • rather than doing `defaultdict(fun)` you could use a lambda method like so `defaultdict(lambda: 0)`, which means you don't need to define `fun` –  Jan 04 '19 at 12:36
  • @AbhinavBhandari i know, my main motive is to let the OP to understand the code and solve it in future and also no use of any library – sahasrara62 Jan 04 '19 at 13:55
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You can get the each available characters count in a given string using below code

def count_dict(mystring):
    d = {}
# count occurances of character
    for w in mystring: 
        d[w] = mystring.count(w)
# print the result
    for k in sorted(d):
        print (k + ': ' + str(d[k]))

mystring='qwertyqweryyyy'
count_dict(mystring)

You can find the live demo here

Suraj Kumar
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    Suraj, you're already linking to the demo in your answer, there is no need to attempt to edit the link into prashant's answer as well: https://stackoverflow.com/review/suggested-edits/21843571 – Nick is tired Jan 04 '19 at 12:40