blocks|key|319303|text|这样如何：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|319304|conditions+=+[a,+b,+c]
if+any(conditions)+and+not+all(conditions):
+++...|code-block|syntax|javascript|319305|其他变体：|319306|if+1+<=+sum(map(bool,+conditions))+<=+2:
+++...|319307|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|$]]

How about:

<pre><code>conditions = [a, b, c]
if any(conditions) and not all(conditions):
 ...
</code></pre>

Other variant:

<pre><code>if 1 &lt;= sum(map(bool, conditions)) &lt;= 2:
 ...
</code></pre>

blocks|key|319361|text|关于：(唯一条件)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|319362|if+(bool(a)+%2B+bool(b)+%2B+bool(c)+==+1):|code-block|syntax|javascript|319363|请注意，如果您也允许两个条件，则可以这样做|319364|if+(bool(a)+%2B+bool(b)+%2B+bool(c)+in+[1,2]):|319365|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 about: (unique condition)

<pre><code>if (bool(a) + bool(b) + bool(c) == 1):
</code></pre>

Notice, if you allow two conditions too you could do that

<pre><code>if (bool(a) + bool(b) + bool(c) in [1,2]):
</code></pre>

blocks|key|319340|text|如果三个条件中只有一个为True，则返回True。这可能就是您在示例代码中想要的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|319341|if+sum(1+for+x+in+(a,b,c)+if+x)+==+1:|code-block|syntax|javascript|319342|entityMap^0|C|4|K|4|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

This returns <code>True</code> if one and only one of the three conditions is <code>True</code>. Probably what you wanted in your example code.

<pre><code>if sum(1 for x in (a,b,c) if x) == 1:
</code></pre>

blocks|key|319436|text|为了清楚起见，您希望根据多少参数为逻辑真(在字符串参数的情况下-不为空)做出决定？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|319437|argsne+=+(1+if+a+else+0)+%2B+(1+if+b+else+0)+%2B+(1+if+c+else+0)|code-block|syntax|javascript|319438|然后你做了一个决定：|319439|if+(+0+<+argsne+<+3+):
+doSth()+|319440|现在逻辑更清晰了。|319441|entityMap^0|0|0|0|0|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]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

To be clear, you want to made your decision based on how much of the parameters are logical TRUE (in case of string arguments - not empty)?

<pre><code>argsne = (1 if a else 0) + (1 if b else 0) + (1 if c else 0)
</code></pre>

Then you made a decision:

<pre><code>if ( 0 &lt; argsne &lt; 3 ):
 doSth() 
</code></pre>

Now the logic is more clear.

blocks|key|102371|text|为什么不直接数一下呢？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|102372|import+sys
a+=+sys.argv
if+len(a)+=+1+:++
++++#+No+arguments+were+given,+the+program+name+count+as+one
elif+len(a)+=+4+:
++++#+Three+arguments+were+given
else+:
++++#+another+amount+of+arguments+was+given|code-block|syntax|javascript|102373|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|$]]

And why not just count them ?

<pre><code>import sys
a = sys.argv
if len(a) = 1 : 
 # No arguments were given, the program name count as one
elif len(a) = 4 :
 # Three arguments were given
else :
 # another amount of arguments was given
</code></pre>

blocks|key|102402|text|如果你不介意有点晦涩，你可以简单地使用0+<+(a+%2B+b+%2B+c)+<+3，如果你有一到两个true语句，它将返回true，如果所有语句都为false或都没有false，则返回false。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|102403|这也简化了您使用函数计算bool的情况，因为您只计算一次变量，这意味着您可以内联编写函数，而不需要临时存储变量。(示例：0+<+(+a(x)+%2B+b(x)+%2B+c(x)+)+<+3。)|102404|entityMap^0|J|J|1M|4|0|1O|U|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|R|8|@]|D|@]|E|$]]]|I|$]]

If you don't mind being a bit cryptic you can simly roll with <code>0 &lt; (a + b + c) &lt; 3</code> which will return <code>true</code> if you have between one and two true statements and false if all are false or none is false.

This also simplifies if you use functions to evaluate the bools as you only evaluate the variables once and which means you can write the functions inline and do not need to temporarily store the variables. (Example: <code>0 &lt; ( a(x) + b(x) + c(x) ) &lt; 3</code>.)

blocks|key|319414|text|问题指出您需要所有三个参数(a、b和c)，或者不需要(a、b或c)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|319415|这提供了：|319416|(a、b和c)或不是(a、b或c)|319417|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

The question states that you need either all three arguments (a and b and c) or none of them (not (a or b or c))

This gives:

(a and b and c) or not (a or b or c)

blocks|key|319395|text|如果你使用条件的迭代器，它的访问速度可能会很慢。但是，您不需要多次访问每个元素，也不必总是读取所有元素。这是一个适用于无限生成器的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|319396|#!/usr/bin/env+python3
from+random+import+randint
from+itertools+import+tee

def+generate_random():
++++while+True:
++++++++yield+bool(randint(0,1))

def+any_but_not_all2(s):+#+elegant
++++t1,+t2+=+tee(s)
++++return+False+in+t1+and+True+in+t2+#+could+also+use+"not+all(...)+and+any(...)"

def+any_but_not_all(s):+#+simple
++++hadFalse+=+False
++++hadTrue+=+False
++++for+i+in+s:
++++++++if+i:
++++++++++++hadTrue+=+True
++++++++else:
++++++++++++hadFalse+=+True
++++++++if+hadTrue+and+hadFalse:
++++++++++++return+True
++++return+False


r1,+r2+=+tee(generate_random())
assert+any_but_not_all(r1)
assert+any_but_not_all2(r2)

assert+not+any_but_not_all([True,+True])
assert+not+any_but_not_all2([True,+True])

assert+not+any_but_not_all([])
assert+not+any_but_not_all2([])

assert+any_but_not_all([True,+False])
assert+any_but_not_all2([True,+False])|code-block|syntax|javascript|319397|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 work with an iterator of conditions, it could be slow to access. But you don't need to access each element more than once, and you don't always need to read all of it. Here's a solution that will work with infinite generators:

<pre><code>#!/usr/bin/env python3
from random import randint
from itertools import tee

def generate_random():
 while True:
 yield bool(randint(0,1))

def any_but_not_all2(s): # elegant
 t1, t2 = tee(s)
 return False in t1 and True in t2 # could also use "not all(...) and any(...)"

def any_but_not_all(s): # simple
 hadFalse = False
 hadTrue = False
 for i in s:
 if i:
 hadTrue = True
 else:
 hadFalse = True
 if hadTrue and hadFalse:
 return True
 return False


r1, r2 = tee(generate_random())
assert any_but_not_all(r1)
assert any_but_not_all2(r2)

assert not any_but_not_all([True, True])
assert not any_but_not_all2([True, True])

assert not any_but_not_all([])
assert not any_but_not_all2([])

assert any_but_not_all([True, False])
assert any_but_not_all2([True, False])
</code></pre>

blocks|key|102423|text|当每个给定的bool都是True，或者当每个给定的bool都是False时...|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|102424|它们都是平等的！|102425|因此，我们只需要找到两个元素，它们的计算结果是不同的bool|102426|要知道至少有一个True和至少一个False。|102427|我的简短解决方案是：|102428|not+bool(a)==bool(b)==bool(c)|code-block|syntax|javascript|102429|我相信它是短路的，因为AFAIK+a==b==c等于a==b+and+b==c。|102430|我的通用解决方案是：|102431|def+_any_but_not_all(first,+iterable):+#doing+dirty+work
++++bool_first=bool(first)
++++for+x+in+iterable:
++++++++if+bool(x)+is+not+bool_first:
++++++++++++return+True
++++return+False

def+any_but_not_all(arg,+*args):+#takes+any+amount+of+args+convertable+to+bool
++++return+_any_but_not_all(arg,+args)

def+v_any_but_not_all(iterable):+#takes+iterable+or+iterator
++++iterator=iter(iterable)
++++return+_any_but_not_all(next(iterator),+iterator)|102432|我还写了一些处理多个迭代器的代码，但我从这里删除了它，因为我认为它没有意义。然而，它仍然可以在here上使用。|102433|entityMap|0|LINK|mutability|MUTABLE|url|http://rextester.com/TNPOV26244^0|6|4|C|4|P|4|V|5|0|0|Q|4|0|8|4|H|5|0|0|0|H|7|Q|D|0|0|0|1B|4|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1H|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|1I|8|@$9|1J|A|1K|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|1L|8|@$9|1M|A|1N|B|C]|$9|1O|A|1P|B|C]]|D|@]|E|$]]|$1|L|3|M|5|6|7|1Q|8|@]|D|@]|E|$]]|$1|N|3|O|5|P|7|1R|8|@]|D|@]|E|$Q|R]]|$1|S|3|T|5|6|7|1S|8|@$9|1T|A|1U|B|C]|$9|1V|A|1W|B|C]]|D|@]|E|$]]|$1|U|3|V|5|6|7|1X|8|@]|D|@]|E|$]]|$1|W|3|X|5|P|7|1Y|8|@]|D|@]|E|$Q|R]]|$1|Y|3|Z|5|6|7|1Z|8|@]|D|@$9|20|A|21|1|22]]|E|$]]|$1|10|3|-4|5|6|7|23|8|@]|D|@]|E|$]]]|11|$12|$5|13|14|15|E|$16|17]]]]

When every given <code>bool</code> is <code>True</code>, or when every given <code>bool</code> is <code>False</code>... 
they all are equal to each other! 

So, we just need to find two elements which evaluates to different <code>bool</code>s 
to know that there is at least one <code>True</code> and at least one <code>False</code>.

<h1>My short solution:</h1>

<pre><code>not bool(a)==bool(b)==bool(c)
</code></pre>

I belive it short-circuits, cause AFAIK <code>a==b==c</code> equals <code>a==b and b==c</code>.

<h1>My generalized solution:</h1>

<pre><code>def _any_but_not_all(first, iterable): #doing dirty work
 bool_first=bool(first)
 for x in iterable:
 if bool(x) is not bool_first:
 return True
 return False

def any_but_not_all(arg, *args): #takes any amount of args convertable to bool
 return _any_but_not_all(arg, args)

def v_any_but_not_all(iterable): #takes iterable or iterator
 iterator=iter(iterable)
 return _any_but_not_all(next(iterator), iterator)
</code></pre>

I wrote also some code dealing with multiple iterables, but I deleted it from here because I think it's pointless. It's however still available <a href="http://rextester.com/TNPOV26244" rel="nofollow">here</a>.

I have a python script that can receive either zero or three command line arguments. (Either it runs on default behavior or needs all three values specified.)

What's the ideal syntax for something like:

<pre><code>if a and (not b or not c) or b and (not a or not c) or c and (not b or not a):
</code></pre>

?

Python syntax for "if a or b or c but not all of them"

我有一个python脚本，它可以接收零个或三个命令行参数。(它要么以默认行为运行，要么需要指定所有三个值。)对于这样的东西，理想的语法是什么：if a and (not b or not c) or b and (not a or not c) or c and (not b or not a):

问"if a、b或c，但不是全部“的Python语法
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回答 9

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