出了什么问题:

if word in mystring: 
   print('success')

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} '

你可以在"word"前后加一个空格。

x = raw_input("Type your word: ")
if " word " in x:
    print("Yes")
elif " word " not in x:
    print("Nope")

这样它会查找“word”前后的空格。

>>> Type your word: Swordsmith
>>> Nope
>>> Type your word:  word 
>>> Yes

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")

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.')