python - Letter Count Dict

Write a Python function called LetterCount() which takes a string as an argument and returns a dictionary of letter counts.

The line:

print LetterCount("Abracadabra, Monsignor")

Should produce the output:

{'a': 5, 'c': 1, 'b': 2, 'd': 1, 'g': 1, 'i': 1, 'm': 1, 'o': 2, 'n': 2, 's': 1, 'r': 3}

I tried:

import collections
c = collections.Counter('Abracadabra, Monsignor')
print c
print list(c.elements())

the answer I am getting looks like this

{'a': 4, 'r': 3, 'b': 2, 'o': 2, 'n': 2, 'A': 1, 'c: 1, 'd': 1, 'g': 1, ' ':1, 'i':1, 'M':1 ',':1's': 1, }
['A', 'a','a','a','a','c','b','b','d','g', and so on 

okay now with this code import collections c = collections.Counter('Abracadabra, Monsignor'.lower())

print c am getting this {'a': 5, 'r': 3, 'b': 2, '开发者_运维问答o': 2, 'n': 2, 'c: 1, 'd': 1, 'g': 1, ' ':1, 'i':1, ',':1's': 1, }

but answer should be this {'a': 5, 'c': 1, 'b': 2, 'd': 1, 'g': 1, 'i': 1, 'm': 1, 'o': 2, 'n': 2, 's': 1, 'r': 3}

You are close. Note that in the task description, the case of the letters is not taken into account. They want {'a': 5}, where you have {'a': 4, 'A': 1}.

So you have to convert the string to lower case first (I'm sure you will find out how).

Use dictionary for the letter count:

s = "string is an immutable object"
d = {}
for i in s:
    d[i] = d.get(i,0)+1

print d

Output:

{'a': 2, ' ': 4, 'c': 1, 'b': 2, 'e': 2, 'g': 1, 'i': 3, 'j': 1, 'm': 2, 'l': 1, 'o': 1, 'n': 2, 's': 2, 'r': 1, 'u': 1, 't': 3}

Maybe this to count just letters, exclude spaces and overlook its case and sort them alphabetically?:

    t = "The cat is out of the bag."
    word_count = {}

    for i in t.casefold():
        if i.isalnum(): 
            word_count[i] = word_count.get(i,0)+1

    for letter, count in sorted(word_count.items()):
        print(letter, count)

Output: a 2 b 1 c 1 e 2 f 1 g 1 h 2 i 1 o 2 s 1 t 4 u 1