Returning number of items in dictionary-CodePudding

I want to modify this function to return the number of items in the dictionary that begin with each letter. For example:

list= ['I', 'say', 'what', 'I', 'mean', 'and', 'I', 'mean', 'what', 'I', 'say']
print(letterCount(list))
{'I': 4, 's': 2, 'w': 2, 'm': 2, 'a': 1}

So far I can return the first letter and words, but not the number of letters that are different.

def letterCounter(wrdLst):
    output = {}
    for word in wrdLst:
        letter = word[0]
        if letter in output:
            output[letter].append(word)
        else:
            output[letter] = [word]
    return output

list = ["apple", "pear", "brown", "red"]
print(letterCounter(list))

this would return the letters instead of number: output:

{'I': ['I'], 's': ['say'], 'w': ['what'], 'm': ['mean'], 'a': ['and']}

CodePudding user response：

Don't append the word, increment a counter.

def letterCounter(wrdLst):
    output = {}
    for word in wrdLst:
        letter = word[0]
        if letter in output:
            output[letter]  = 1
        else:
            output[letter] = 1
    return output

You can also use the standard collections module.

from collections import Counter

def letterCounter(wrdLst):
    return Counter(word[0] for word in wrdLst)

CodePudding user response：

If you are cool using numpy, you can just use its unique function to do what you are trying to do. Below inside of the np.unique call, I do a list comprehension to pull the first letter off of each word and make sure it is lowercase. Then the unique function returns the unique starting letters and with return_counts enabled gives their number of occurrences.

import numpy as np

words = ['I', 'say', 'what', 'I', 'mean', 'and', 'I', 'mean', 'what', 'I', 'say']
unique, counts = np.unique([word[0].lower() for word in words], return_counts=True)

for (val, count) in zip(unique, counts):
    print (val, count)

CodePudding user response：

Another Counter way:

from collections import Counter

def letterCount(words):
    return Counter(next(zip(*words)))

words = ['I', 'say', 'what', 'I', 'mean', 'and', 'I', 'mean', 'what', 'I', 'say']
print(letterCount(words))

CodePudding user response：

Using a defaultdict will make the code more concise as follows:

from collections import defaultdict

def letterCounter(wrdLst):
    d = defaultdict(int)
    for w in wrdLst:
        d[w[0]]  = 1
    return dict(d) # return a regular dictionary
    
print(letterCounter(['I', 'say', 'what', 'I', 'mean', 'and', 'I', 'mean', 'what', 'I', 'say']))

Output:

{'I': 4, 's': 2, 'w': 2, 'm': 2, 'a': 1}

CodePudding user response：

You can also make the initial letters to a string and then use str.count() to make the dict:

def letter_count(wrdLst):
   s = ''
   for i in wrdLst:
      s  = i[0]
   d = {}
   for i in s:
       d[i] = s.count(i)
   return d