Check For (whole Only) Words In String

Training on Checkio. The task is called Popular words. The task is to search for words from a list (of strings) in a given string. For example: textt='When I was One I had just beg

Solution 1:

you'd be better off splitting your sentence, then count the words, not the substrings:

textt="When I was One I had just begun When I was Two I was nearly new"
wwords=['i', 'was', 'three', 'near']
text_words = textt.lower().split()
result = {w:text_words.count(w) for w in wwords}

print(result)

prints:

{'three': 0, 'i': 4, 'near': 0, 'was': 3}

if the text has punctuation now, you're better off with regular expressions to split the string according to non-alphanum:

import re

textt="When I was One, I had just begun.I was Two when I was nearly new"

wwords=['i', 'was', 'three', 'near']
text_words = re.split("\W+",textt.lower())
result = {w:text_words.count(w) for w in wwords}

result:

{'was': 3, 'near': 0, 'three': 0, 'i': 4}

(another alternative is to use findall on word characters: text_words = re.findall(r"\w+",textt.lower()))

Now if your list of "important" words is big, maybe it's better to count all the words, and filter afterwards, using the classical collections.Counter:

text_words = collections.Counter(re.split("\W+",textt.lower()))
result = {w:text_words.get(w) for w in wwords}

Solution 2:

Your simple solution would be this one:

from collections import Counter

textt="When I was One I had just begun When I was Two I was nearly new".lower()
wwords=['i', 'was', 'three', 'near']

txt = textt.split()

keys = Counter(txt)

for i in wwords:
    print(i + ' : ' + str(keys[i]))