英文:
Unpack a list with ternary operator
问题
以下是翻译好的部分：
I stumbled upon a weird behaviour when I want to unpack a list only if condition is true.
我在尝试仅在条件为真时解包列表时遇到了奇怪的行为。
How can I use the unpacking () based on a condition?
如何根据条件使用解包 ()？
Example:
示例：
def foo1(x1, x2):
print(x1, x2)
def foo2(x):
print(x)
l = [7,8]
foo1(*l if True else l) # this works
foo2(*l if False else l) # this does not work
foo1(l if not True else *l) # this does not work
英文:
I stumbled upon a weird behaviour when I want to unpack a list only if condition is true.
How can I use the unpacking (*) based on a condition?
Example:
def foo1(x1, x2):
print(x1, x2)
def foo2(x):
print(x)
l = [7,8]
foo1(*l if True else l) # this works
foo2(*l if False else l) # this does not work
foo1(l if not True else *l) # this does not work
答案1
得分: 5
这不是解析为在左侧选择*l和右侧的l。 l if True else l 被评估，然后结果被无条件地解包。
没有办法有一个函数调用表达式有条件地解包或不解包参数。如果你在那里放一个*，就会有东西被解包。你可以做的是将l包装在另一个列表中，然后解包：
foo(*(l if condition else [l]))
英文:
That's not parsed as choosing between *l on the left and l on the right. l if True else l is evaluated, and then the result is unpacked, unconditionally.
There's no way to have a function call expression conditionally either unpack or not unpack an argument. Something's getting unpacked if you put a * there. What you can do is wrap l in another list and unpack that:
foo(*(l if condition else [l]))
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