拆分字符串，剥离单词和标点符号怎么样?

w in [ws.strip(',.?!') for ws in p.split()]

如有需要，请注意小写或大写:

w.lower() in [ws.strip(',.?!') for ws in p.lower().split()]

也许是这样:

def wcheck(word, phrase):
    # Attention about punctuation and about split characters
    punctuation = ',.?!'
    return word.lower() in [words.strip(punctuation) for words in phrase.lower().split()]

示例:

print(wcheck('CAr', 'I own a caR.'))

我没有检查性能……

出了什么问题:

if word in mystring: 
   print('success')

Find返回一个整数，表示搜索项所在位置的索引。如果没有找到，则返回-1。

haystack = 'asdf'

haystack.find('a') # result: 0
haystack.find('s') # result: 1
haystack.find('g') # result: -1

if haystack.find(needle) >= 0:
  print('Needle found.')
else:
  print('Needle not found.')

如果你想知道一个完整的单词是否在一个以空格分隔的单词列表中，只需使用:

def contains_word(s, w):
    return (' ' + w + ' ') in (' ' + s + ' ')

contains_word('the quick brown fox', 'brown')  # True
contains_word('the quick brown fox', 'row')    # False

这种优雅的方法也是最快的。与Hugh Bothwell和daSong的方法相比:

>python -m timeit -s "def contains_word(s, w): return (' ' + w + ' ') in (' ' + s + ' ')" "contains_word('the quick brown fox', 'brown')"
1000000 loops, best of 3: 0.351 usec per loop

>python -m timeit -s "import re" -s "def contains_word(s, w): return re.compile(r'\b({0})\b'.format(w), flags=re.IGNORECASE).search(s)" "contains_word('the quick brown fox', 'brown')"
100000 loops, best of 3: 2.38 usec per loop

>python -m timeit -s "def contains_word(s, w): return s.startswith(w + ' ') or s.endswith(' ' + w) or s.find(' ' + w + ' ') != -1" "contains_word('the quick brown fox', 'brown')"
1000000 loops, best of 3: 1.13 usec per loop

编辑:Python 3.6+略有不同，同样快:

def contains_word(s, w):
    return f' {w} ' in f' {s} '

if 'seek' in 'those who seek shall find':
    print('Success!')

但请记住，这匹配的是一个字符序列，而不一定是一个完整的单词——例如，'swordsmith'中的'word'是True。如果你只想匹配整个单词，你应该使用正则表达式:

import re

def findWholeWord(w):
    return re.compile(r'\b({0})\b'.format(w), flags=re.IGNORECASE).search

findWholeWord('seek')('those who seek shall find')    # -> <match object>
findWholeWord('word')('swordsmith')                   # -> None

高级的方法来检查确切的单词，我们需要在一个长字符串中找到:

import re
text = "This text was of edited by Rock"
#try this string also
#text = "This text was officially edited by Rock" 
for m in re.finditer(r"\bof\b", text):
    if m.group(0):
        print("Present")
    else:
        print("Absent")