Python解包中的默认值 [英] Default value in Python unpacking

问题描述

如果与变量列表相比，要解包的值的数量太少，有没有办法获得默认值?

例如:

a, b, c = read_json(request)

如果 read_json 返回三个或更多变量的数组，则此方法有效.如果它只返回两个，我会在分配 c 时遇到异常.那么，如果不能正确解包，有没有办法将 c 设置为默认值?类似的东西:

a, b, (c=2) = read_json(request)

这与定义带有默认参数的函数时所做的类似.

谢谢！

解决方案

你可以试试 * 解包一些后处理:

a, b, *c = read_json(request)c = c[0] 如果 c 否则 2

这将正常分配 a 和 b.如果 c 被赋值，它将是一个带有一个元素的 list.如果只解压了两个值，它将是一个空的 list.如果存在，则第二条语句将其第一个元素分配给 c，否则为 2 的默认值.

<预><代码>>>>a, b, *c = 1, 2, 3>>>c = c[0] 如果 c 否则 2>>>一种1>>>乙2>>>C3>>>a, b, *c = 1, 2>>>c = c[0] 如果 c 否则 2>>>一种1>>>乙2>>>C2

Is there a way to have a default value if the number of values to unpack is too little compared to the variable list?

For example:

a, b, c = read_json(request)

This works if read_json returns an array of three or more variable. If it only returns two, I get an exception while assigning c. So, is there a way to set c to a default value if it can't be unpacked properly? Something like:

a, b, (c=2) = read_json(request)

Which is similar to what you do when defining a function with default arguments.

Thank you!

解决方案

You could try * unpacking with some post-processing:

a, b, *c = read_json(request)
c = c[0] if c else 2

This will assign a and b as normal. If c is assigned something, it will be a list with one element. If only two values were unpacked, it will be an empty list. The second statement assigns to c its first element if there is one, or the default value of 2 otherwise.

>>> a, b, *c = 1, 2, 3
>>> c = c[0] if c else 2
>>> a
1
>>> b
2
>>> c
3
>>> a, b, *c = 1, 2
>>> c = c[0] if c else 2
>>> a
1
>>> b
2
>>> c
2

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