blocks|key|5575973|text|我在这里使用re.findall，然后只计算找到的匹配项的数量：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5575974|inp+=+"blah+code+blah+cole+blah+core+blah"
print(len(re.findall(r'\bco[a-z]e\b',+inp)))++++#+prints+3|code-block|syntax|javascript|5575975|这里使用的正则表达式模式是\bco[a-z]e\b，它可以匹配任何单词code，其中第三个字母d实际上可以是任何字母。|5575976|entityMap^0|6|A|0|0|D|C|Z|4|1B|1|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|N|$]]

I would use <code>re.findall</code> here, and then just take the count of how many matches were found:

<pre><code>inp = "blah code blah cole blah core blah"
print(len(re.findall(r'\bco[a-z]e\b', inp))) # prints 3
</code></pre>

The regex pattern used here is <code>\bco[a-z]e\b</code>, which matches any word <code>code</code>, where the third letter <code>d</code> can actually be any letter.

blocks|key|4107777|text|您混淆了索引和字符的内容：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4107778|str.find(x)+%2B+1+==+‘o’|code-block|syntax|javascript|4107779|这绝不可能是真的:您的find调用将返回字符串中第一个这样的字符的位置。将该索引加1。你如何想象这个整数和可以等于字符'o'？整数和字符不能相等。|offset|length|style|CODE|4107780|你需要备份并重新设计你的逻辑。当您这样做时，请参阅debugging+help的这个可爱的参考资料。您需要学习如何在自己的代码中跟踪问题。|4107781|entityMap|0|LINK|mutability|MUTABLE|url|https://ericlippert.com/2014/03/05/how-to-debug-small-programs^0|0|0|B|4|0|P|E|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]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|11|8|@]|9|@$I|12|J|13|1|14]]|A|$]]|$1|O|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

You've confused the index and the contents of the characters:

<pre><code>str.find(x) + 1 == ‘o’
</code></pre>

This can never be true: your <code>find</code> call will return the position of the first such character in the string. You add 1 to that index. How do you imagine that this integer sum can be equal to the character 'o'? An integer and a character cannot be equal.

You need to back up and redesign your logic. When you do that, see this lovely reference for <a href="https://ericlippert.com/2014/03/05/how-to-debug-small-programs" rel="nofollow noreferrer">debugging help</a>. You need to learn how to trace problems in your own code.

blocks|key|4107868|text|正如Tim提到的，如果你需要一种更简单的方法(不是最有效的方法，而是一种方法)，我会在这类任务中使用正则表达式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4107869|import+string


text+=+'elephant+code+joke+cole+coke+man'
total+=+0
for+letter+in+string.ascii_lowercase:
++++total+%2B=+text.count(f'co{letter}e')
print(f'{total}')|code-block|syntax|javascript|4107870|请注意，在您的解决方案尝试中，存在以下情况：|4107871|for+x+in+str:|4107872|除了在这种情况下应该避免名称冲突str之外，对于1000个字符的文本，您将使用这种方法进行1000次迭代，减少到26次迭代(字母表)。|offset|length|style|CODE|4107873|entityMap^0|0|0|0|0|G|3|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

As Tim mentioned I would go for regular expressions in such tasks however, if you need a simpler way(not the most efficient but one way to do it):

<pre><code>import string


text = 'elephant code joke cole coke man'
total = 0
for letter in string.ascii_lowercase:
 total += text.count(f'co{letter}e')
print(f'{total}')
</code></pre>

Note that in your solution attempt, there is this:

<pre><code>for x in str:
</code></pre>

Apart from the fact that you should avoid name collisions <code>str</code> in that case, for a text of size 1000 characters, you'll be doing 1000 iterations, reduced to 26 iterations(letters of the alphabet) using this approach.

blocks|key|1790595|text|使用正则表达式，它是快速，方便和简短的代码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1790596|#CODE
import+re

#findall()+Returns+a+list+containing+all+matches
str="cole+code+cope"
result+=+re.findall("co.e",+str)+
print(len(result))
#END|code-block|syntax|javascript|1790597|希望这能满足你的要求。|1790598|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|$]]

Use regular expression for this, It's fast, convenient and short code.

<pre><code>#CODE
import re

#findall() Returns a list containing all matches
str="cole code cope"
result = re.findall("co.e", str) 
print(len(result))
#END
</code></pre>

Hope this will fulfill your requirement.

I was practicing from coding bat and the question was to find the number of times the word ‘code’ appears in a string, but the letter ‘d’ in ‘code’ can be substituted for any letter. So “code” has count 1, ‘cope’ also 1. I wrote some code but it doesn't work. I don't understand why it doesn't work. The function returns 0 for every input. Here is a link to what I tried <a href="https://codingbat.com/prob/p186048" rel="nofollow noreferrer">https://codingbat.com/prob/p186048</a>

<pre><code>def count_code(str):
 count = 0
 str = str.lower()
 for x in str:
 if x == ‘c’ and str.find(x) + 1 == “o” and str.find(x) + 3 == “e”:
 count += 1
 return count 
</code></pre>

Counting number of times words appear In strings

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在练习编写bat代码，问题是找出单词“code”在字符串中出现的次数，但“code”中的字母“d”可以替换任何字母。所以“code”的计数是1，“cope”也是1。我写了一些代码，但它不能工作。我不明白为什么它不能工作。该函数为每一个输入返回0。这是我尝试过的的链接def count_code(str):     ...

问统计单词在字符串中出现的次数
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