CODEWITHSUNDEEP
pythonstring
Elektron Atom
Elektron Atom

Reputation: 51

Counting letters with python

What I am trying to do is count how many times each letter appear in a string. The I want to store the details in a dictionary.
The following is my attempt to do so:

def scan(string):
    list_string = []
    for letter in string:
         list_string.append(letter)

    list_string.sort()
    scanned = {}

    for k in range(0,len(list_string)):
        count = 0
        for kk in range(k,len(list_string)):
            if list_string[k] == list_string[kk]:
                count += 1

        scanned.update({list_string[k]:count})

    return scanned

However every key is having the value 1 even though there are times when a letter appears more than one time.

It works fine when I try this

print("Letter '{}': \t {}.format(list_string[k],count))

Can anyone help?

Upvotes: 2

Views: 141

Answers (3)

willeM_ Van Onsem
willeM_ Van Onsem

Reputation: 476503

Easy fix

You can simply count with the Counter:

from collection import Counter

ctr = Counter(list_string)

Now ctr is a Counter: a special sort of dictionary. It maps the elements that were in the list onto numbers (the amount of time it is in the list). An element that has not been seen, is mapped on 0.

The error

That's because you overwrite it each time you recount that value.

Say the list is:

['a','b','a','a']

Now you first set the k-cursor (^) to the first element, and the kk-cursor (v) to the same:

#v
['a','b','a','a']
#^

Now we count the number of as and obtain 3. Next we advance, the k-counter and count the number of bs (1). Now the cursor is moved again, to the second a:

#        v
['a','b','a','a']
#        ^

Now we count the number of 'a's again, but now we obtain 2: because the kk-cursor ignores the first one. Finally we move the cursor to the last 'a', and count the number of 'a's, and guess what? We obtain 1. So we rewrite it to 1.

#            v
['a','b','a','a']
#            ^

Regardless of the number of times an element occurs in the list, we will always count it a last time, and then the count will obviously always be 1.

Upvotes: 7

cs95
cs95

Reputation: 402253

letter_count = { x : list(mystring).count(x) for x in set(mystring)}

As rightly pointed out, a pretty pythonic one liner with bad asymptotic complexity if your elements are hashable, in which case using count(...) will become a linear operation.

Upvotes: 1

Hugh Bothwell
Hugh Bothwell

Reputation: 56624

from collections import Counter

letter_count = Counter(mystring)

Upvotes: 7

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