blocks|key|966213|text|问题出在哪里：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|966214|if+word+in+mystring:+
+++print('success')|code-block|syntax|javascript|966215|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

What is wrong with:
<pre><code>if word in mystring: 
 print('success')
</code></pre>

blocks|key|966306|text|if+'seek'+in+'those+who+seek+shall+find':
++++print('Success!')|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|966307|但请记住，这与字符序列匹配，而不一定是整个单词-例如，'word'+in+'swordsmith'为True。如果你只想匹配整个单词，你应该使用正则表达式：|unstyled|offset|length|style|CODE|966308|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|966309|entityMap^0|0|R|M|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|P|8|@$G|Q|H|R|I|J]]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|M|3|-4|5|F|7|T|8|@]|9|@]|A|$]]]|N|$]]

<pre><code>if 'seek' in 'those who seek shall find':
 print('Success!')
</code></pre>

but keep in mind that this matches a sequence of characters, not necessarily a whole word - for example, <code>'word' in 'swordsmith'</code> is True. If you only want to match whole words, you ought to use regular expressions:

<pre><code>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') # -&gt; &lt;match object&gt;
findWholeWord('word')('swordsmith') # -&gt; None
</code></pre>

blocks|key|966259|text|find返回一个整数，表示找到搜索项的位置的索引。如果没有找到，则返回-1。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|966260|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.')|code-block|syntax|javascript|966261|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

find returns an integer representing the index of where the search item was found. If it isn't found, it returns -1.
<pre><code>haystack = 'asdf'

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

if haystack.find(needle) &gt;= 0:
 print('Needle found.')
else:
 print('Needle not found.')
</code></pre>

blocks|key|751028|text|您可以将字符串拆分为单词并检查结果列表。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|751029|if+word+in+string.split():
++++print("success")|code-block|syntax|javascript|751030|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can split string to the words and check the result list.
<pre><code>if word in string.split():
 print(&quot;success&quot;)
</code></pre>

blocks|key|966410|text|这个小函数比较给定文本中的所有搜索词。如果在文本中找到所有搜索词，则返回搜索长度，否则返回False。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|966411|还支持unicode字符串搜索。|966412|def+find_words(text,+search):
++++"""Find+exact+words"""
++++dText+++=+text.split()
++++dSearch+=+search.split()

++++found_word+=+0

++++for+text_word+in+dText:
++++++++for+search_word+in+dSearch:
++++++++++++if+search_word+==+text_word:
++++++++++++++++found_word+%2B=+1

++++if+found_word+==+len(dSearch):
++++++++return+lenSearch
++++else:
++++++++return+False|code-block|syntax|javascript|966413|用法：|966414|find_words('çelik+güray+ankara',+'güray+ankara')|966415|entityMap^0|19|5|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|R|$]]

This small function compares all search words in given text. If all search words are found in text, returns length of search, or <code>False</code> otherwise.

Also supports unicode string search.

<pre><code>def find_words(text, search):
 """Find exact words"""
 dText = text.split()
 dSearch = search.split()

 found_word = 0

 for text_word in dText:
 for search_word in dSearch:
 if search_word == text_word:
 found_word += 1

 if found_word == len(dSearch):
 return lenSearch
 else:
 return False
</code></pre>

usage:

<pre><code>find_words('çelik güray ankara', 'güray ankara')
</code></pre>

blocks|key|750817|text|如果匹配一个字符序列还不够，您需要匹配整个单词，这里有一个简单的函数可以完成这项工作。它主要是在需要的地方添加空格，并在字符串中进行搜索：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|750818|def+smart_find(haystack,+needle):
++++if+haystack.startswith(needle%2B"+"):
++++++++return+True
++++if+haystack.endswith("+"%2Bneedle):
++++++++return+True
++++if+haystack.find("+"%2Bneedle%2B"+")+!=+-1:
++++++++return+True
++++return+False|code-block|syntax|javascript|750819|这假设逗号和其他标点符号已经被去掉。|750820|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

If matching a sequence of characters is not sufficient and you need to match whole words, here is a simple function that gets the job done. It basically appends spaces where necessary and searches for that in the string:

<pre><code>def smart_find(haystack, needle):
 if haystack.startswith(needle+" "):
 return True
 if haystack.endswith(" "+needle):
 return True
 if haystack.find(" "+needle+" ") != -1:
 return True
 return False
</code></pre>

This assumes that commas and other punctuations have already been stripped out.

blocks|key|966577|text|由于您要求的是单词而不是字符串，因此我想提出一个对前缀/后缀不敏感且忽略大小写的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|966578|#!/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'(%5E%7C[%5E\w]){}([%5E\w]%7C$)'.format(word)
++++pattern+=+re.compile(pattern,+re.IGNORECASE)
++++matches+=+re.search(pattern,+text)
++++return+bool(matches)


if+__name__+==+'__main__':
++++import+doctest
++++doctest.testmod()|code-block|syntax|javascript|966579|如果您的单词可能包含正则表达式特殊字符(例如%2B)，那么您需要re.escape(word)|offset|length|style|CODE|966580|entityMap^0|0|0|M|1|U|F|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

As you are asking for a word and not for a string, I would like to present a solution which is not sensitive to prefixes / suffixes and ignores case:

<pre><code>#!/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
 --------
 &gt;&gt;&gt; is_word_in_text("Python", "python is awesome.")
 True

 &gt;&gt;&gt; is_word_in_text("Python", "camelCase is pythonic.")
 False

 &gt;&gt;&gt; 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()
</code></pre>

If your words might contain regex special chars (such as <code>+</code>), then you need <code>re.escape(word)</code>

blocks|key|966509|text|检查我们需要在长字符串中找到的确切单词的高级方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|966510|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")|code-block|syntax|javascript|966511|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Advanced way to check the exact word, that we need to find in a long string:
<pre><code>import re
text = &quot;This text was of edited by Rock&quot;
#try this string also
#text = &quot;This text was officially edited by Rock&quot; 
for m in re.finditer(r&quot;\bof\b&quot;, text):
 if m.group(0):
 print(&quot;Present&quot;)
 else:
 print(&quot;Absent&quot;)
</code></pre>

blocks|key|751094|text|拆分字符串和剥离单词标点符号怎么样？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|751095|w+in+[ws.strip(',.?!')+for+ws+in+p.split()]|code-block|syntax|javascript|751096|如果需要，请注意小写/大写：|751097|w.lower()+in+[ws.strip(',.?!')+for+ws+in+p.lower().split()]|751098|也许是这样的：|751099|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()]|751100|示例：|751101|print(wcheck('CAr',+'I+own+a+caR.'))|751102|我没有检查性能...|751103|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|14|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

What about to split the string and strip words punctuation?
<pre><code>w in [ws.strip(',.?!') for ws in p.split()]
</code></pre>
If need, do attention to lower/upper case:
<pre><code>w.lower() in [ws.strip(',.?!') for ws in p.lower().split()]
</code></pre>
Maybe that way:
<pre><code>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()]
</code></pre>
Sample:
<pre><code>print(wcheck('CAr', 'I own a caR.'))
</code></pre>
I didn't check performance...

blocks|key|966460|text|你可以在"word“前后加一个空格。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|966461|x+=+raw_input("Type+your+word:+")
if+"+word+"+in+x:
++++print("Yes")
elif+"+word+"+not+in+x:
++++print("Nope")|code-block|syntax|javascript|966462|这样它就会查找"word“前后的空格。|966463|>>>+Type+your+word:+Swordsmith
>>>+Nope
>>>+Type+your+word:++word+
>>>+Yes|966464|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

You could just add a space before and after &quot;word&quot;.
<pre><code>x = raw_input(&quot;Type your word: &quot;)
if &quot; word &quot; in x:
 print(&quot;Yes&quot;)
elif &quot; word &quot; not in x:
 print(&quot;Nope&quot;)
</code></pre>
This way it looks for the space before and after &quot;word&quot;.
<pre><code>&gt;&gt;&gt; Type your word: Swordsmith
&gt;&gt;&gt; Nope
&gt;&gt;&gt; Type your word: word 
&gt;&gt;&gt; Yes
</code></pre>

blocks|key|751129|text|我相信这个答案更接近于最初的问题：Find+substring+in+string+but+only+if+whole+words?|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|751130|它使用了一个简单的正则表达式：|751131|import+re

if+re.search(r"\b"+%2B+re.escape(word)+%2B+r"\b",+string):
++print('success')|code-block|syntax|javascript|751132|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/4154961/find-substring-in-string-but-only-if-whole-words^0|H|1D|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|C|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

I believe this answer is closer to what was initially asked: <a href="https://stackoverflow.com/questions/4154961/find-substring-in-string-but-only-if-whole-words">Find substring in string but only if whole words?</a>
It is using a simple regex:
<pre><code>import re

if re.search(r&quot;\b&quot; + re.escape(word) + r&quot;\b&quot;, string):
 print('success')
</code></pre>

I'm working with Python v2, and I'm trying to find out if you can tell if a word is in a string.
I have found some information about identifying if the word is in the string - using .find, but is there a way to do an IF statement. I would like to have something like the following:
<pre><code>if string.find(word):
 print(&quot;success&quot;)
</code></pre>
Thanks for any help.

Python - Check If Word Is In A String

我正在使用Python v2，我想知道您是否能分辨出一个单词是否在字符串中。我找到了一些关于识别单词是否在字符串中的信息-使用.find，但有没有方法来执行if语句。我想要像下面这样的东西：if string.find(word):    print("success")谢谢你的帮助。

问Python -检查Word是否在字符串中
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回答 11

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问Python -检查Word是否在字符串中EN

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问Python -检查Word是否在字符串中
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