Check for (whole only) words in string

问题

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 begun When I was Two I was nearly new"

wwords=['i', 'was', 'three', 'near']

My code goes like:

def popular_words(text: str, words: list) -> dict:
    # your code here

    occurence={}
    text=text.lower()


    for i in words:
        occurence[i]=(text.count(i))

    # incorrectly takes "nearly" as "near"


    print(occurence)
    return(occurence)

popular_words(textt,wwords)

which works almost fine, returning

{'i': 4, 'was': 3, 'three': 0, 'near': 1}

thus counting "near" as a part of the "nearly". It was obviously the authors intention. I, however, cannot find a way to get aroud this other than

"search for words that are not first (index 0) or last (last index) and for these that begin/end with whitespace"

May I ask for a help, please? Building upon this rather childish code, please.

回答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}

回答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]))

来源：https://stackoverflow.com/questions/52302509/check-for-whole-only-words-in-string

标签

python

string

count

find