find out: In how many word a character is present in a given Sentence: PYTHON

Question

Before explaining my requirement, let me explain what I don't want:

s= 'Hello satya, have you achieved your goal'
s.count('l') #o/p: 3
#I don't want 3 instead  I need 2. How? explained below.

Expected Output:

{'H':3, 'e':3, 'l':2, 'o':4, 's':1, 'a':3....etc}

That is : take all unique letters/characters from that sentence. Then find "in how many words that character is present" , so count that no of occurence and store in dictionary.

For example character 'l' is present 3 times in that sentence, but present in 2 WORDS(in hello it is repeated but I want it to be counted as 1).

Please suggest. If i missed something to make my problem more understandable, please highlight.

Here is what I tried:

def char_word(s):
    s_d = {}
    chars = list(s.replace(' ', ''))
    print(chars)
    for char in chars:
        c_count = 0
        for word in s.split():
            if char in word:
                c_count += 1
            s_d[char] = c_count
    print(s_d)
    return s_d

Answer 1

You can do

sum(['l' in i for i in s.split()])

s.split() splits your sentence into words (it splits by ). Then the for-loop checks if the character l is in that particular word, yielding True or False if it is or is not. The sum function counts all the True s

Answer 2

First, split the string into individual words and create the output dictionary. I recommend using the collections.Counter c.ass, which is just a dict subclass designed to count occurrences:

import collections
counter = collections.Counter()
words = s.split()

Now loop through all of the words and create a set from each word. Since set can't have duplicate values, it's guaranteed to have exactly one copy of each characters:

for word in words:
    characters = set(word)

And finally, loop through the now unique characters of the word, incrementing the counter's value for that character:

    for character in characters:
        counter[character] += 1

Also, you can shorten this quite a lot, even down to a one-liner:

>>> counter = collections.Counter(c for word in s.split() for c in set(word))

To get the count of a particular character, just access it through counter[c] :

>>> counter['l']
2
>>> counter['x']
0
>>> counter['H']
1

Answer 3

You can do it, too, within groupby from itertools module and dict comprenesion like this way:

from itertools import groupby

a = 'Hello satya, have you achieved your goal'
a = a.replace(',','').split()
sub = [j.lower() for k in a for j in set(k)]
final = {k:len(list(v)) for k,v in groupby(sorted(sub), lambda x: x)}
print(final)

Outut:

{'h': 3, 'u': 2, 's': 1, 'i': 1, 'c': 1, 'a': 4, 'd': 1, 'e': 3, 't': 1, 'v': 2, 'l': 2, 'o': 4, 'r': 1, 'g': 1, 'y': 3}