blocks|key|4308321|text|你的循环是错误的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4308322|这应该是可行的：|4308323|++for+s+in+str:
+++++if+ch+==+s:
+++++++...|code-block|syntax|javascript|4308324|这样就不会使用index变量，您可以删除它。如果要使用index，请将for更改为：|offset|length|style|CODE|4308325|for+index+in+range(len(str)):
+++...+(rest+is+OK+but+...)
+++...+(do+not+increase+index+in+loop+body)|4308326|您还可以通过%2B=运算符递增变量，如下所示：|4308327|cnt+%2B=+1|4308328|因此，完成的代码将如下所示：|4308329|def+countLetters(str,+ch):
+++count+=+0
+++for+s+in+str:
+++++if+ch+==+s:
+++++++count+%2B=+1
+++print+count|4308330|entityMap^0|0|0|0|7|5|R|5|Z|3|0|0|6|2|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|13|8|@$K|14|L|15|M|N]|$K|16|L|17|M|N]|$K|18|L|19|M|N]]|9|@]|A|$]]|$1|O|3|P|5|F|7|1A|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|1B|8|@$K|1C|L|1D|M|N]]|9|@]|A|$]]|$1|S|3|T|5|F|7|1E|8|@]|9|@]|A|$G|H]]|$1|U|3|V|5|6|7|1F|8|@]|9|@]|A|$]]|$1|W|3|X|5|F|7|1G|8|@]|9|@]|A|$G|H]]|$1|Y|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|Z|$]]

Your loop is wrong.

This should work:

<pre><code> for s in str:
 if ch == s:
 ...
</code></pre>

this way <code>index</code> variable will not be used and you can remove it. If you want to use <code>index</code> then change <code>for</code> into:

<pre><code>for index in range(len(str)):
 ... (rest is OK but ...)
 ... (do not increase index in loop body)
</code></pre>

You can also increment variable by <code>+=</code> operator like:

<pre><code>cnt += 1
</code></pre>

So finished code will look like:

<pre><code>def countLetters(str, ch):
 count = 0
 for s in str:
 if ch == s:
 count += 1
 print count
</code></pre>

blocks|key|1669384|text|完全未经测试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1669385|def+count_letters(s,+c):
++++return+sum(1+for+x+in+s+if+x+==+c)|code-block|syntax|javascript|1669386|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|$]]

Completely untested:

<pre><code>def count_letters(s, c):
 return sum(1 for x in s if x == c)
</code></pre>

blocks|key|2706103|text|你正在覆盖你的'ch‘变量：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2706104|def+countLetters(str,+ch):
#++++++++++++++++++++++%5E+the+character+you+are+looking+for
++++count=0
++++index=0
++++for+ch+in+str:
#++++++++%5E+the+string+character+you+are+trying+to+check
++++++++if+ch==str[index]:++#+???
++++++++++++count=count%2B1
++++++++index=index%2B1
++++print+count|code-block|syntax|javascript|2706105|(而且，返回值通常比仅仅打印它更有用)。|2706106|内置方法是str.count：|2706107|"aaabb".count("a")++->+3|2706108|如何重写代码：|2706109|def+countLetters(search_in,+search_for):
++++count+=+0
++++for+s+in+search_in:++++#+iterate+by+string+char,+not+by+index
++++++++if+s==search_for:
++++++++++++count+%2B=+1
++++return+count|2706110|和一个快速的巨蟒替换：|2706111|def+countLetters(search_in,+search_for):
++++return+sum(1+for+s+in+search_in+if+s==search_for)|2706112|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]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|12|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

You are over-writing your 'ch' variable:

<pre><code>def countLetters(str, ch):
# ^ the character you are looking for
 count=0
 index=0
 for ch in str:
# ^ the string character you are trying to check
 if ch==str[index]: # ???
 count=count+1
 index=index+1
 print count
</code></pre>

(also, it is usually more useful to return the value than to just print it).

The built-in method is str.count:

<pre><code>"aaabb".count("a") -&gt; 3
</code></pre>

How you could rewrite your code:

<pre><code>def countLetters(search_in, search_for):
 count = 0
 for s in search_in: # iterate by string char, not by index
 if s==search_for:
 count += 1
 return count
</code></pre>

and a quick pythonic replacement:

<pre><code>def countLetters(search_in, search_for):
 return sum(1 for s in search_in if s==search_for)
</code></pre>

blocks|key|639226|text|从逻辑上考虑运行代码时会发生什么:由于循环中的测试在第一次迭代中成功，因此每次都会成功！您只需检查Python中的迭代是否有效。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|639227|正确的表述是|639228|def+count(s,+input):
++++count+=+0
++++for+c+in+s:
++++++++if+c+==+input:
++++++++++++count+%2B=+1|code-block|syntax|javascript|639229|或者，等同地，|639230|def+count(input):
++++return+sum(c+==+input+for+c+in+s)|639231|但是你也可以这样做：|639232|s.count(c)|639233|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Think logically about what happens when you run your code: since the test in the loop succeeds on the first iteration, it is guaranteed to succeed every time! You are simply checking that iteration in Python works.

The correct formulation is

<pre><code>def count(s, input):
 count = 0
 for c in s:
 if c == input:
 count += 1
</code></pre>

Or, equivalently,

<pre><code>def count(input):
 return sum(c == input for c in s)
</code></pre>

But you could just as well do:

<pre><code>s.count(c)
</code></pre>

The question is how to write a program that measures how many times a character appears in a string in a generalizable way in python.

The code that I wrote:

<pre><code>def countLetters(str, ch):
 count=0
 index=0
 for ch in str:
 if ch==str[index]:
 count=count+1
 index=index+1
 print count
</code></pre>

when I use this function, it measures the length of the string instead of how many times the character occurs in the string. What did I do wrong? What is the right way to write this code?

Exercise 7.9 in "How to Think Like a Computer Scientist (python)" measuring occurrences of a character in a string

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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问题是如何用python编写一个程序来测量一个字符在字符串中出现的次数。我写的代码：def countLetters(str, ch):   count=0   index=0   for ch in str:     if ch==str[index]:       count=count+1     index=i...

问“如何像计算机科学家一样思考(python)”中的练习7.9测量字符在字符串中的出现次数
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问“如何像计算机科学家一样思考(python)”中的练习7.9测量字符在字符串中的出现次数EN