blocks|key|2057822|text|是像这样吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2057823|for+i+in+range(nIterations):+y+=+f(y)|code-block|syntax|javascript|2057824|只有一个命令的for循环可以写成一行。|2057825|编辑|offset|length|style|BOLD|2057826|或者稍微干净一点：|2057827|for+_+in+xrange(nIterations):+y+=+f(y)|2057828|既然你不想有一个可以被分成两个单独的语句的东西(我想)，下面是另一个：|2057829|reduce(lambda+y,+_:+f(y),+xrange(nIterations),+initValue)|2057830|尽管如此，我还是建议只使用您的原始代码，它更直观，可读性更好。还要注意Guido+van+Rossum对循环和repeat的看法。顺便说一句，(在Python2.x中)对于大型nIterations，xrange比range更有效，因为它返回的是实际的迭代器，而不是分配的列表。|CODE|2057831|entityMap|0|LINK|mutability|MUTABLE|url|http://www.artima.com/weblogs/viewpost.jsp?thread=211200^0|0|0|0|0|2|0|0|0|0|0|A|9|1J|6|2S|6|2Z|5|Z|G|0|0^^$0|@$1|2|3|4|5|6|7|17|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|18|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|19|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|1A|8|@$K|1B|L|1C|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1D|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|1E|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|1F|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|1G|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|1H|8|@$K|1I|L|1J|M|N]|$K|1K|L|1L|M|Y]|$K|1M|L|1N|M|Y]|$K|1O|L|1P|M|Y]]|9|@$K|1Q|L|1R|1|1S]]|A|$]]|$1|Z|3|-4|5|6|7|1T|8|@]|9|@]|A|$]]]|10|$11|$5|12|13|14|A|$15|16]]]]

like this?

<pre><code>for i in range(nIterations): y = f(y)
</code></pre>

A for loop with one command can be written as a single line.

EDIT

Or maybe slightly cleaner:

<pre><code>for _ in xrange(nIterations): y = f(y)
</code></pre>

Since you don't want to have a something that can be split into two separate statements (i think), here's another one:

<pre><code>reduce(lambda y, _: f(y), xrange(nIterations), initValue)
</code></pre>

Still, I would recommend to just use your original code, which is much more intuitive and readable. Also note what <a href="http://www.artima.com/weblogs/viewpost.jsp?thread=211200" rel="nofollow noreferrer">Guido van Rossum</a> has to say on loops versus <code>repeat</code>.
Note by the way that (in python 2.x) <code>xrange</code> is more efficient than <code>range</code> for large nIterations as it returns an actual iterator and not an allocated list.

blocks|key|1893833|text|你是说像这样吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1893834|for+i+in+range(nIterations):+y+=+f(y)|code-block|syntax|javascript|1893835|虽然这看起来很不错，但我认为(正如你的帖子下面的评论所做的那样)，这个不会提高可读性，最好是留成2行。|offset|length|style|BOLD|1893836|entityMap^0|0|0|Z|2|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]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

So like this you mean? 

<pre><code>for i in range(nIterations): y = f(y)
</code></pre>

While this might seem nice and pretty, I'd argue (as has been done in the comments below your post) that this doesn't improve readability, and is best off left as 2 lines.

blocks|key|2057888|text|只需将所有内容放在一行中，就像这样：for+i+in+range(nIterations):+y+=+f(y)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2057889|将代码放在一行还是多行的决定已经争论了很多年--没有性能提升--只是展示一下你喜欢它的方式，以及你的如何才能读得最好。|2057890|entityMap^0|I|11|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|N|8|@]|D|@]|E|$]]]|I|$]]

Just stick it all on one line like this: <code>for i in range(nIterations): y = f(y)</code>

The decision to have code on one line or multiple has been an argument for years - there is no performance increase - just lay it out how you like it and how you can read it best.

blocks|key|590985|text|您的问题缺少上下文，但可以使用map函数或列表理解(这两个一行程序)重写|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|590986|entityMap^0|F|3|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

Your question lacks context, but this could be rewritten using <code>map</code> function or list comprehension (both one-liners)

blocks|key|1893874|text|不是只有一行，而是一旦定义了函数的幂运算：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1893875|def+f_pow(f,+n):
++if+n+==+1:
++++return+f
++else:
++++return+lambda+x:+f_pow(f,+n-1)(f(x))|code-block|syntax|javascript|1893876|你可以这样写：|1893877|f_pow(f,+nIterations)(y)|1893878|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|$]]

Not exactly one line, but once you define the power operation for functions:

<pre><code>def f_pow(f, n):
 if n == 1:
 return f
 else:
 return lambda x: f_pow(f, n-1)(f(x))
</code></pre>

you can write this:

<pre><code>f_pow(f, nIterations)(y)
</code></pre>

blocks|key|1893884|text|reduce(lambda+y,_:+f(y),xrange(niterations),y)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1893885|entityMap^0|0|1A|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

<code>reduce(lambda y,_: f(y),xrange(niterations),y)</code>

blocks|key|591093|text|好的，这可能是reduce函数的一种非常奇怪的，难以理解的用法，所以对于真正的代码，我会坚持使用你所拥有的。但为了好玩，这里是这样的：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|591094|reduce(lambda+a,+b:+f(a),+range(nIterations),+y)|code-block|syntax|javascript|591095|entityMap^0|7|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Ok this is probably a very weird an incomprehensible use of the <code>reduce</code> function, so for real code I'd stick with what you have. But just for the fun of it, here goes:

<pre><code>reduce(lambda a, b: f(a), range(nIterations), y)
</code></pre>

blocks|key|591208|text|如果您将y设置为可变的，那么您可以使用列表理解。但这不是我会在真正的代码中使用的东西，除非真的是必要的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|BOLD|entityRanges|data|591209|def+f(y):
++++y[0]+%2B=+5

y+=+[0]
[f(y)+for+_+in+xrange(10)]
print+y[0]+#+=>+50|code-block|syntax|javascript|591210|entityMap^0|4|1|19|3|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]|$9|Q|A|R|B|D]]|E|@]|F|$]]|$1|G|3|H|5|I|7|S|8|@]|E|@]|F|$J|K]]|$1|L|3|-4|5|6|7|T|8|@]|E|@]|F|$]]]|M|$]]

If you make <code>y</code> mutable, then you can use list comprehension. But this isn't something I'd use in real code, unless really necessary.

<pre><code>def f(y):
 y[0] += 5

y = [0]
[f(y) for _ in xrange(10)]
print y[0] # =&gt; 50
</code></pre>

blocks|key|2058117|text|虽然我建议您保留原始代码片段，因为它更清晰，但您可以使用reduce函数通过一行代码完成此操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2058118|reduce(lambda+a,b:+f(a),+xrange(nIterations),+y)|code-block|syntax|javascript|2058119|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|$]]

While I'd suggest you keep the original code snippet as it is much clearer, you can accomplish this with a single line of code using the reduce function:

<pre><code>reduce(lambda a,b: f(a), xrange(nIterations), y)
</code></pre>

blocks|key|591227|text|您可以使用分号;创建这样的代码片段。如果您需要在循环中执行多条指令，下面是一个示例：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|591228|for+i+in+xrange(nIterations):+x=f(i);+y=f(x);+z=f(y)|code-block|syntax|javascript|591229|entityMap^0|7|1|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

You can create such snippets using semicolons ; if you need to execute more than one instruction inside the loop, here is an example:

<pre><code>for i in xrange(nIterations): x=f(i); y=f(x); z=f(y)
</code></pre>

blocks|key|2058179|text|y=[f(y)+for+i+in+range(niteration)]|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2058180|希望这能有所帮助;)|2058181|entityMap^0|0|Z|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|N|8|@]|D|@]|E|$]]]|I|$]]

<code>y=[f(y) for i in range(niteration)]</code>

hope that helps ;)

I have some code of the form:

<pre><code>for i in range(nIterations):
 y = f(y)
</code></pre>

Where <code>f</code> is a function defined elsewhere. hopefully the idea of that code is that after it's run <code>y</code> will have had <code>f</code> applied to it <code>nIterations</code> times.

Is there a way in python to write this in a single line?

Iteration in a single line

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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我有一些表单的代码：for i in range(nIterations):    y = f(y)其中f是在其他地方定义的函数。希望这段代码的想法是，在运行之后，y将在nIterations次对其应用f。在python中有没有一种方法可以把它写成一行？

问单行迭代
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