blocks|key|3975248|text|您可以使用lambda。这对我来说很有效：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3975249|yld123cu+=+lambda+u:+(x+*+u+for+x+in+range(1,+4))

for+i+in+yld123cu(2):
++++print(i,+end='+')|code-block|syntax|javascript|3975250|entityMap^0|5|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|$]]

<p>You can use <code>lambda</code>. This works for me:</p>

<pre><code>yld123cu = lambda u: (x * u for x in range(1, 4))

for i in yld123cu(2):
    print(i, end=' ')
</code></pre>


blocks|key|3977637|text|由于生成器表达式不能有自己的作用域，因此首选的方法是按照您的建议定义一个生成器函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3977638|def+yld123u(u):
++++for+i+in+range(1,+4):
++++++++yield+i+*+u|code-block|syntax|javascript|3977639|如果您需要在此过程中更新u，这就是您想要避免上述情况的原因，那么要知道您可以将一个值send到生成器。|offset|length|style|CODE|3977640|def+yld123u(u):
++++for+i+in+range(1,+4):
++++++++received+=+yield+i+*+u+#+a+sent+value+will+be+assigned+to+'received'
++++++++u+=+u+if+received+is+None+else+received|3977641|示例|3977642|如果不使用send，生成器的行为仍然是一样的。|3977643|for+x+in+yld123u(4):
++++print(x)|3977644|输出|3977645|4
8
12+|3977646|您可以使用send更新u的值。|3977647|gen+=+yld123u(4)
v0+=+next(gen)
v1+=+gen.send(0)
v2+=+gen.send(4)
print(v0,+v1,+v2)|3977648|3977649|4+0+12|3977650|输入循环|3977651|如果你需要生成器与用户输入交互，这里是我过去的答案，它建议了一种实现yield-receive+generator+usable+in+a+for-loop的方法。|3977652|class+YieldReceive:
++++stop_iteration+=+object()

++++def+__init__(self,+gen):
++++++++self.gen+=+gen
++++++++self.next+=+next(gen,+self.stop_iteration)

++++def+__iter__(self):
++++++++return+self

++++def+__next__(self):
++++++++if+self.next+is+self.stop_iteration:
++++++++++++raise+StopIteration
++++++++else:
++++++++++++return+self.next

++++def+send(self,+value):
++++++++try:
++++++++++++self.next+=+self.gen.send(value)
++++++++except+StopIteration:
++++++++++++self.next+=+self.stop_iteration|3977653|用法|3977654|it+=+YieldReceive(yld123u(4))
for+x+in+it:
++++print(x)
++++it.send(int(input('Enter+next+u+value:+')))|3977655|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/q/19302530/5079316|1|https://stackoverflow.com/a/50918359/5079316^0|0|0|C|1|16|4|16|4|0|0|0|0|0|0|0|0|5|4|B|1|0|0|0|0|0|Y|18|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1P|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1Q|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1R|8|@$I|1S|J|1T|K|L]|$I|1U|J|1V|K|L]]|9|@$I|1W|J|1X|1|1Y]]|A|$]]|$1|M|3|N|5|D|7|1Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|20|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|21|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|22|8|@]|9|@]|A|$E|F]]|$1|U|3|V|5|6|7|23|8|@]|9|@]|A|$]]|$1|W|3|X|5|D|7|24|8|@]|9|@]|A|$E|F]]|$1|Y|3|Z|5|6|7|25|8|@$I|26|J|27|K|L]|$I|28|J|29|K|L]]|9|@]|A|$]]|$1|10|3|11|5|D|7|2A|8|@]|9|@]|A|$E|F]]|$1|12|3|V|5|6|7|2B|8|@]|9|@]|A|$]]|$1|13|3|14|5|D|7|2C|8|@]|9|@]|A|$E|F]]|$1|15|3|16|5|6|7|2D|8|@]|9|@]|A|$]]|$1|17|3|18|5|6|7|2E|8|@]|9|@$I|2F|J|2G|1|2H]]|A|$]]|$1|19|3|1A|5|D|7|2I|8|@]|9|@]|A|$E|F]]|$1|1B|3|1C|5|6|7|2J|8|@]|9|@]|A|$]]|$1|1D|3|1E|5|D|7|2K|8|@]|9|@]|A|$E|F]]|$1|1F|3|-4|5|6|7|2L|8|@]|9|@]|A|$]]]|1G|$1H|$5|1I|1J|1K|A|$1L|1M]]|1N|$5|1I|1J|1K|A|$1L|1O]]]]

<p>Since a generator expression cannot have its own scope, the preferred approach would be to define a generator function as you suggested.</p>
<pre><code>def yld123u(u):
    for i in range(1, 4):
        yield i * u
</code></pre>
<p>If you need to update <code>u</code> along the way and this is the reason you want to avoid the above, then know that you can <a href="https://stackoverflow.com/q/19302530/5079316"><code>send</code></a> a value to a generator.</p>
<pre><code>def yld123u(u):
    for i in range(1, 4):
        received = yield i * u # a sent value will be assigned to 'received'
        u = u if received is None else received
</code></pre>
<h2>Examples</h2>
<p>The generator still behaves the same if you do not use send.</p>
<pre><code>for x in yld123u(4):
    print(x)
</code></pre>
<h3>Output</h3>
<pre><code>4
8
12 
</code></pre>
<p>You can use <code>send</code> to update the value of <code>u</code>.</p>
<pre><code>gen = yld123u(4)
v0 = next(gen)
v1 = gen.send(0)
v2 = gen.send(4)
print(v0, v1, v2)
</code></pre>
<h3>Output</h3>
<pre><code>4 0 12
</code></pre>
<h2>Input loop</h2>
<p>If you need the generator to interact with user input, here is a past answer of mine which suggest a way to implement a <a href="https://stackoverflow.com/a/50918359/5079316">yield-receive generator usable in a for-loop</a>.</p>
<pre><code>class YieldReceive:
    stop_iteration = object()

    def __init__(self, gen):
        self.gen = gen
        self.next = next(gen, self.stop_iteration)

    def __iter__(self):
        return self

    def __next__(self):
        if self.next is self.stop_iteration:
            raise StopIteration
        else:
            return self.next

    def send(self, value):
        try:
            self.next = self.gen.send(value)
        except StopIteration:
            self.next = self.stop_iteration
</code></pre>
<h3>Usage</h3>
<pre><code>it = YieldReceive(yld123u(4))
for x in it:
    print(x)
    it.send(int(input('Enter next u value: ')))
</code></pre>


<p>so i have this gen object:</p>

<pre><code>yld123cu = (x * u for x in range(1, 4))
</code></pre>

<p>which i can only use like so:</p>

<pre><code>u = 2  # some user given value
for i in yld123cu:
    print(i, end=' ')
</code></pre>

<p>is there any syntax to avoid the extra line with u=.. assignment?
any way to put the u value inside the for statement?</p>

<p>i'd want something similar to below but using gen expression, not function:</p>

<pre><code>def yld123u(u):
    for i in range(1, 4):
        yield i * u

for i in yld123u(2):
    print(i, end=' ')
</code></pre>


Python: passing argument to generator object created by generator expression?

所以我有这个gen对象：yld123cu = (x * u for x in range(1, 4))我只能像这样使用它：u = 2  # some user given valuefor i in yld123cu:    print(i, end=' ')有没有什么语法可以避免用u=..任务？有没有办法把u值放在f...

问Python:向生成器表达式创建的生成器对象传递参数？
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