I’m trying to find the most frequent letter in a list of words. I’m struggling with the algorithm because I need to count the letter frequency in a word only once skipping duplicates, so I need help finding a way to count the frequency of the letters in the entire list with only one occurrence per word, ignoring the second occurrence.

For example if i have:

words = ["tree", "bone", "indigo", "developer"]

The frequency will be:

letters={a:0, b:1, c:0, d:2, e:3, f:0, g:1, h:0, i:1, j:0, k:0, l:1, m:0, n:2, o:3, p:1, q:0, r:2, s:0, t:1, u:0, v:1, w:0, x:0, y:0, z:0}

As you can see from the letters dictionary: ‘e’ is 3 and not 5 because if ‘e’ repeats more than once in the same word it should be ignored.

This is the algorithm that I came up with, it’s implemented in Python:

for word in words:
    count=0;

    for letter in word:
        if(letter.isalpha()):
            if((letters[letter.lower()] > 0  && count == 0) ||
               (letters[letter.lower()] == 0 && count == 0)):

                    letters[letter.lower()]+=1
                    count=1

            elif(letters[letter.lower()]==0 && count==1):   
                letters[letter.lower()]+=1

But it still requires work and I can’t think about anything else, I’d be glad to anyone who will help me to think about a working solution.

A variation on @Primusa answer without using update:

from collections import Counter

words = ["tree", "bone", "indigo", "developer"]
counts = Counter(c for word in words for c in set(word.lower()) if c.isalpha())

Output

Counter({'e': 3, 'o': 3, 'r': 2, 'd': 2, 'n': 2, 'p': 1, 'i': 1, 'b': 1, 'v': 1, 'g': 1, 'l': 1, 't': 1})

Basically convert each word to a set and then iterate over each set.

Create a counter object and then update it with sets for each word:

from collections import Counter

wordlist = ["tree","bone","indigo","developer"]

c = Counter()
for word in wordlist:
    c.update(set(word.lower()))

print(c)

Output:

Counter({'e': 3, 'o': 3, 'r': 2, 'n': 2, 'd': 2, 't': 1, 'b': 1, 'i': 1, 'g': 1, 'v': 1, 'p': 1, 'l': 1})

Note that although letters that weren’t present in wordlist aren’t present in in the Counter, this is fine because a Counter behaves like a defaultdict(int), so accessing a value not present automatically returns a default value of 0.

One without Counter

words=["tree","bone","indigo","developer"]
d={}
for word in words:         # iterate over words
    for i in set(word):    # to remove the duplication of characters within word
        d[i]=d.get(i,0)+1

Output

{'b': 1,
 'd': 2,
 'e': 3,
 'g': 1,
 'i': 1,
 'l': 1,
 'n': 2,
 'o': 3,
 'p': 1,
 'r': 2,
 't': 1,
 'v': 1}

Comparing speed of the solutions presented so far:

def f1(words):
    c = Counter()
    for word in words:
        c.update(set(word.lower()))
    return c

def f2(words):
    return Counter(
        c
        for word in words
        for c in set(word.lower()))

def f3(words):
    d = {}
    for word in words:
        for i in set(word.lower()):
            d[i] = d.get(i, 0) + 1
    return d

My timing function (using different sizes for the list of words):

word_list = [
    'tree', 'bone', 'indigo', 'developer', 'python',
    'language', 'timeit', 'xerox', 'printer', 'offset',
]

for exp in range(5):
    words = word_list * 10**exp

    result_list = []
    for i in range(1, 4):
        t = timeit.timeit(
            'f(words)',
            'from __main__ import words,  f{} as f'.format(i),
            number=100)
        result_list.append((i, t))

    print('{:10,d} words | {}'.format(
        len(words),
        ' | '.join(
            'f{} {:8.4f} sec'.format(i, t) for i, t in result_list)))

The results:

        10 words | f1   0.0028 sec | f2   0.0012 sec | f3   0.0011 sec
       100 words | f1   0.0245 sec | f2   0.0082 sec | f3   0.0113 sec
     1,000 words | f1   0.2450 sec | f2   0.0812 sec | f3   0.1134 sec
    10,000 words | f1   2.4601 sec | f2   0.8113 sec | f3   1.1335 sec
   100,000 words | f1  24.4195 sec | f2   8.1828 sec | f3  11.2167 sec

The Counter with list comprehension (here as f2()) seems to be the fastest. Using counter.update() seems to be a slow point (here as f1()).

The other solutions are good, but they specifically don’t include the letters with zero frequency. Here’s an approach which does, but is approximately 2-3 times slower than the others.

import string
counts = {c: len([w for w in words if c in w.lower()]) for c in string.ascii_lowercase}

which produces a dict like this:

{'a': 4, 'b': 2, 'c': 2, 'd': 4, 'e': 7, 'f': 2, 'g': 2, 'h': 3, 'i': 7, 'j': 0, 'k': 0, 'l': 4, 'm': 5, 'n': 4, 'o': 4, 'p': 1, 'q': 0, 'r': 5, 's': 3, 't': 3, 'u': 2, 'v': 0, 'w': 3, 'x': 0, 'y': 2, 'z': 1}

Here’s my update of Ralf’s timings:

        10 words | f1   0.0004 sec | f2   0.0004 sec | f3   0.0003 sec | f4   0.0010 sec
       100 words | f1   0.0019 sec | f2   0.0014 sec | f3   0.0013 sec | f4   0.0034 sec
     1,000 words | f1   0.0180 sec | f2   0.0118 sec | f3   0.0140 sec | f4   0.0298 sec
    10,000 words | f1   0.1960 sec | f2   0.1278 sec | f3   0.1542 sec | f4   0.2648 sec
   100,000 words | f1   2.0859 sec | f2   1.3971 sec | f3   1.6815 sec | f4   3.5196 sec

based on the following code and the word list from https://github.com/dwyl/english-words/

import string
import timeit
import random
from collections import Counter

def f1(words):
    c = Counter()
    for word in words:
        c.update(set(word.lower()))
    return c

def f2(words):
    return Counter(
        c
        for word in words
        for c in set(word.lower()))

def f3(words):
    d = {}
    for word in words:
        for i in set(word.lower()):
            d[i] = d.get(i, 0) + 1
    return d


def f4(words):
    d = {c: len([w for w in words if c in w.lower()]) for c in string.ascii_lowercase} 
    return d


with open('words.txt') as word_file:
    valid_words = set(word_file.read().split())

for exp in range(5):

    result_list = []
    for i in range(1, 5):
        t = timeit.timeit(
            'f(words)',
            'from __main__ import f{} as f, valid_words, exp; import random; words = random.sample(valid_words, 10**exp)'.format(i),
            number=100)
        result_list.append((i, t))

    print('{:10,d} words | {}'.format(
        len(words),
        ' | '.join(
            'f{} {:8.4f} sec'.format(i, t) for i, t in result_list)))

print(f4(random.sample(valid_words, 10000)))
print(f4(random.sample(valid_words, 1000)))
print(f4(random.sample(valid_words, 100)))
print(f4(random.sample(valid_words, 10)))

Try using a dictionary comprehension:

import string
print({k:max(i.count(k) for i in words) for k in string.ascii_lowercase})

A bit too late to the party, but here you go:

freq = {k: sum(k in word for word in words) for k in set(''.join(words))}

which returns:

{'i': 1, 'v': 1, 'p': 1, 'b': 1, 'e': 3, 'g': 1, 't': 1, 'n': 2, 'd': 2, 'o': 3, 'l': 1, 'r': 2}

from collections import Counter  
import string  

words=["tree","bone","indigo","developer"]  
y=Counter(string.ascii_lowercase)  
new_dict=dict(y) 

for k in new_dict:  
    new_dict[k]=0  
trial = 0  
while len(words) > trial:  
    for let in set(words[trial]):    
        if let in new_dict:  
            new_dict[str(let)]=new_dict[str(let)]+1  

    trial = trial +1  
print(new_dict)

import collections
import itertools
import string


def main():
    words = ["tree", "bone", "indigo", "developer"]
    no_repeated_letters = (set(word) for word in words)
    letter_stream = itertools.chain.from_iterable(no_repeated_letters)
    counter = collections.Counter(letter_stream)
    # set zeros for unseen letters, to match poster's answer.
    for letter in string.ascii_lowercase:
        if letter not in counter:
            counter[letter] = 0
    # print result.
    for key in sorted(counter):
        print(key, counter[key])


if __name__ == '__main__':
    main()