blocks|key|54092|text|您要查找的语法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|54093|lambda+x:+True+if+x+%25+2+==+0+else+False|code-block|syntax|javascript|54094|但是您不能在lambda中使用print或raise。|offset|length|style|CODE|54095|entityMap^0|0|0|F|5|L|5|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|$]]

The syntax you're looking for:

<pre><code>lambda x: True if x % 2 == 0 else False
</code></pre>

But you can't use <code>print</code> or <code>raise</code> in a lambda.

blocks|key|2539605|text|为什么不直接定义一个函数呢？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2539606|def+f(x):
++++if+x+==+2:
++++++++print(x)
++++else:
++++++++raise+ValueError|code-block|syntax|javascript|2539607|在这种情况下，确实没有理由使用lambda。|2539608|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|$]]

why don't you just define a function?

<pre><code>def f(x):
 if x == 2:
 print(x)
 else:
 raise ValueError
</code></pre>

there really is no justification to use lambda in this case.

blocks|key|2749698|text|可能是我到目前为止写过的最糟糕的python代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2749699|f+=+lambda+x:+sys.stdout.write(["2\n",][2*(x==2)-2])|code-block|syntax|javascript|2749700|如果打印x+==+2，|2749701|如果x+!=+2，你提高。|2749702|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]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Probably the worst python line I've written so far: 

<pre><code>f = lambda x: sys.stdout.write(["2\n",][2*(x==2)-2])
</code></pre>

If x == 2 you print, 

if x != 2 you raise.

blocks|key|2749686|text|如果你真的想这样做，你可以很容易地在lambda中引发一个异常。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2749687|def+Raise(exception):
++++raise+exception
x+=+lambda+y:+1+if+y+<+2+else+Raise(ValueError("invalid+value"))|code-block|syntax|javascript|2749688|这是个好主意吗？一般来说，我的直觉是将错误报告保留在lambdas之外；让它的值为None，并在调用者中引发错误。我不认为这本质上是邪恶的--我认为"y+if+x+else+z“语法本身更糟糕--只是要确保您没有试图在lambda主体中填充太多内容。|2749689|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|$]]

You can easily raise an exception in a lambda, if that's what you really want to do.

<pre><code>def Raise(exception):
 raise exception
x = lambda y: 1 if y &lt; 2 else Raise(ValueError("invalid value"))
</code></pre>

Is this a good idea? My instinct in general is to leave the error reporting out of lambdas; let it have a value of None and raise the error in the caller. I don't think this is inherently evil, though--I consider the "y if x else z" syntax itself worse--just make sure you're not trying to stuff too much into a lambda body.

blocks|key|53925|text|Python中的Lambdas对允许使用的内容有相当大的限制。具体来说，在它们的主体中不能有任何关键字(除了and、not、or等运算符)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|53926|所以，您不可能在示例中使用…(因为您不能使用raise)，但是如果您愿意承认这个lambda您可以使用：|53927|f+=+lambda+x:+x+==+2+and+x+or+None|code-block|syntax|javascript|53928|entityMap^0|1I|3|1M|3|1Q|2|0|M|5|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|N|$]]

Lambdas in Python are fairly restrictive with regard to what you're allowed to use. Specifically, you can't have any keywords (except for operators like <code>and</code>, <code>not</code>, <code>or</code>, etc) in their body.

So, there's no way you could use a lambda for your example (because you can't use <code>raise</code>), but if you're willing to concede on that… You could use:

<pre><code>f = lambda x: x == 2 and x or None
</code></pre>

blocks|key|54129|text|注意:您可以在您的lambda定义中使用几个else...if语句：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|54130|f+=+lambda+x:+1+if+x>0+else+0+if+x+==0+else+-1|code-block|syntax|javascript|54131|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|$]]

note you can use several else...if statements in your lambda definition:

<pre class="lang-py prettyprint-override"><code>f = lambda x: 1 if x&gt;0 else 0 if x ==0 else -1
</code></pre>

blocks|key|2749780|text|如果您还想打印，可以导入将来的模块|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2749781|from+__future__+import+print_function

f+=+lambda+x:+print(x)+if+x%252+==+0+else+False|code-block|syntax|javascript|2749782|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|$]]

If you still want to print you can import future module

<pre><code>from __future__ import print_function

f = lambda x: print(x) if x%2 == 0 else False
</code></pre>

blocks|key|2749791|text|您还可以使用逻辑运算符将条件|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2749792|func+=+lambda+element:+(expression+and+DoSomething)+or+DoSomethingIfExpressionIsFalse|code-block|syntax|javascript|2749793|您可以查看有关逻辑运算符here的更多信息|offset|length|2749794|entityMap|0|LINK|mutability|MUTABLE|url|https://www.tutorialspoint.com/python/logical_operators_example.htm^0|0|0|C|4|0|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]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

You can also use Logical Operators to have something like a Conditional

<pre><code>func = lambda element: (expression and DoSomething) or DoSomethingIfExpressionIsFalse
</code></pre>

You can see more about Logical Operators <a href="https://www.tutorialspoint.com/python/logical_operators_example.htm" rel="nofollow noreferrer">here</a>

blocks|key|2539774|text|这段代码应该会对您有所帮助：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2539775|x+=+lambda+age:+'Older'+if+age+>+30+else+'Younger'

print(x(40))|code-block|syntax|javascript|2539776|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|$]]

This snippet should help you:

<pre class="lang-html prettyprint-override"><code>x = lambda age: 'Older' if age &gt; 30 else 'Younger'

print(x(40))
</code></pre>

blocks|key|2539762|text|你真正需要的是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2539763|def+fun():
++++raise+Exception()
f+=+lambda+x:print+x+if+x==2+else+fun()|code-block|syntax|javascript|2539764|现在以您需要的方式调用该函数|2539765|f(2)
f(3)|2539766|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|$]]

what you need exactly is

<pre><code>def fun():
 raise Exception()
f = lambda x:print x if x==2 else fun()
</code></pre>

now call the function the way you need

<pre><code>f(2)
f(3)
</code></pre>

blocks|key|54064|text|在lambda中执行if的一种简单方法是使用列表理解。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|54065|您不能在lambda中引发异常，但在Python3.x中，这是一种接近示例的方法：|54066|f+=+lambda+x:+print(x)+if+x==2+else+print("exception")|code-block|syntax|javascript|54067|另一个例子：|54068|如果M返回1，否则返回0|54069|f+=+lambda+x:+1+if+x=="M"+else+0|54070|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|V|8|@]|9|@]|A|$G|H]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

An easy way to perform an if in lambda is by using list comprehension. 

You can't raise an exception in lambda, but this is a way in Python 3.x to do something close to your example:

<pre><code>f = lambda x: print(x) if x==2 else print("exception")
</code></pre>

Another example:

return 1 if M otherwise 0

<pre><code>f = lambda x: 1 if x=="M" else 0
</code></pre>

blocks|key|2749870|text|给定场景的解决方案是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2749871|f+=+lambda+x+:+x+if+x+==+2+else+print("number+is+not+2")
f(30)++#+number+is+not+2
f(2)+++#2|code-block|syntax|javascript|2749872|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|$]]

the solution for the given scenerio is:
<pre class="lang-py prettyprint-override"><code>f = lambda x : x if x == 2 else print(&quot;number is not 2&quot;)
f(30) # number is not 2
f(2) #2
</code></pre>

blocks|key|2539756|text|下面的示例代码适用于我。不确定它是否与这个问题直接相关，但希望它能在其他情况下有所帮助。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2539757|a+=+''.join(map(lambda+x:+str(x*2)+if+x%252==0+else+"",+range(10)))|code-block|syntax|javascript|2539758|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|$]]

Following sample code works for me. Not sure if it directly relates to this question, but hope it helps in some other cases.

<pre><code>a = ''.join(map(lambda x: str(x*2) if x%2==0 else "", range(10)))
</code></pre>

blocks|key|2749836|text|希望这篇文章能有所帮助。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2749837|您可以通过以下方式解决此问题|2749838|f+=+lambda+x:++x==2+++

if+f(3):
++print("do+logic")
else:
++print("another+logic")|code-block|syntax|javascript|2749839|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Hope this will help a little
you can resolve this problem in the following way
<pre><code>f = lambda x: x==2 

if f(3):
 print(&quot;do logic&quot;)
else:
 print(&quot;another logic&quot;)
</code></pre>

blocks|key|2539797|text|我想这就是你要找的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2539798|>>>+f+=+lambda+x+:+print(x)+if+x==2+else+print("ERROR")
>>>+f(23)
ERROR
>>>+f(2)
2
>>>+|code-block|syntax|javascript|2539799|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|$]]

I think this is what you were looking for
<pre><code>&gt;&gt;&gt; f = lambda x : print(x) if x==2 else print(&quot;ERROR&quot;)
&gt;&gt;&gt; f(23)
ERROR
&gt;&gt;&gt; f(2)
2
&gt;&gt;&gt; 
</code></pre>

In Python 2.6, I want to do:
<pre><code>f = lambda x: if x==2 print x else raise Exception()
f(2) #should print &quot;2&quot;
f(3) #should throw an exception
</code></pre>
This clearly isn't the syntax. Is it possible to perform an <code>if</code> in <code>lambda</code> and if so how to do it?

Is there a way to perform "if" in python's lambda?

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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在Python2.6中，我想做的是：f = lambda x: if x==2 print x else raise Exception()f(2) #should print "2"f(3) #should throw an exception这显然不是语法。可以在lambda中执行if吗?如果可以，如何执行？

问有没有办法在python的lambda中执行"if“？
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问有没有办法在python的lambda中执行"if“？EN

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问有没有办法在python的lambda中执行"if“？
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