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")

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")

使用regex是一种解决方案，但对于这种情况来说太复杂了。

您可以简单地将文本分割成单词列表。使用split(separator, num)方法。它返回字符串中所有单词的列表，使用分隔符作为分隔符。如果separator未指定，则对所有空格进行分割(您可以选择将分割的数量限制为num)。

list_of_words = mystring.split()
if word in list_of_words:
    print('success')

这将不工作的字符串与逗号等。例如:

mystring = "One,two and three"
# will split into ["One,two", "and", "three"]

如果你也想拆分所有的逗号等，使用分隔符参数如下:

# whitespace_chars = " \t\n\r\f" - space, tab, newline, return, formfeed
list_of_words = mystring.split( \t\n\r\f,.;!?'\"()")
if word in list_of_words:
    print('success')

由于您要求的是一个词，而不是一个字符串，我想提出一个解决方案，这是不敏感的前缀/后缀和忽略大小写:

#!/usr/bin/env python

import re


def is_word_in_text(word, text):
    """
    Check if a word is in a text.

    Parameters
    ----------
    word : str
    text : str

    Returns
    -------
    bool : True if word is in text, otherwise False.

    Examples
    --------
    >>> is_word_in_text("Python", "python is awesome.")
    True

    >>> is_word_in_text("Python", "camelCase is pythonic.")
    False

    >>> is_word_in_text("Python", "At the end is Python")
    True
    """
    pattern = r'(^|[^\w]){}([^\w]|$)'.format(word)
    pattern = re.compile(pattern, re.IGNORECASE)
    matches = re.search(pattern, text)
    return bool(matches)


if __name__ == '__main__':
    import doctest
    doctest.testmod()

如果你的单词可能包含正则表达式特殊字符(例如+)，那么你需要re.escape(word)

出了什么问题:

if word in mystring: 
   print('success')

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……