在 python 中处理 list.index(might-not-exist) 的最佳方法? [英] Best way to handle list.index(might-not-exist) in python?

问题描述

我的代码看起来像这样:

I have code which looks something like this:

thing_index = thing_list.index(thing)
otherfunction(thing_list, thing_index)

好的，这样简化了，但你明白了.现在 thing 可能实际上不在列表中，在这种情况下，我想将 -1 作为 thing_index 传递.在其他语言中，这是您期望 index() 在找不到元素时返回的内容.事实上，它会抛出一个 ValueError.

ok so that's simplified but you get the idea. Now thing might not actually be in the list, in which case I want to pass -1 as thing_index. In other languages this is what you'd expect index() to return if it couldn't find the element. In fact it throws a ValueError.

我可以这样做:

try:
    thing_index = thing_list.index(thing)
except ValueError:
    thing_index = -1
otherfunction(thing_list, thing_index)

但这感觉很脏，而且我不知道是否会因其他原因引发 ValueError.我基于生成器函数想出了以下解决方案，但看起来有点复杂:

But this feels dirty, plus I don't know if ValueError could be raised for some other reason. I came up with the following solution based on generator functions, but it seems a little complex:

thing_index = ( [(i for i in xrange(len(thing_list)) if thing_list[i]==thing)] or [-1] )[0]

有没有更简洁的方法来实现同样的目标?假设列表未排序.

Is there a cleaner way to achieve the same thing? Let's assume the list isn't sorted.

推荐答案

使用 try-except 子句并没有什么肮脏"之处.这是pythonic的方式.ValueError 只会由 .index 方法引发，因为它是你在那里唯一的代码！

There is nothing "dirty" about using try-except clause. This is the pythonic way. ValueError will be raised by the .index method only, because it's the only code you have there!

回复评论:
在 Python 中，请求宽恕比获得许可更容易 哲学已经确立，no index 不会为任何其他问题引发此类错误.不是我能想到的.

To answer the comment:
In Python, easier to ask forgiveness than to get permission philosophy is well established, and no index will not raise this type of error for any other issues. Not that I can think of any.

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