Python中的序列查找功能 [英] Sequence find function in Python

问题描述

如何在满足特定条件的序列中找到对象?

How do I find an object in a sequence satisfying a particular criterion?

列表理解和过滤遍历整个列表.唯一的替代方法是手工制作吗?

List comprehension and filter go through the entire list. Is the only alternative a handmade loop?

mylist = [10, 2, 20, 5, 50]
find(mylist, lambda x:x>10) # Returns 20

推荐答案

这是我使用的模式:

mylist = [10, 2, 20, 5, 50]
found = next(i for i in mylist if predicate(i))

或者，在Python 2.4/2.5中，next()不是内置的:

Or, in Python 2.4/2.5, next() is a not a builtin:

found = (i for i in mylist if predicate(i)).next()

请注意，如果未找到任何元素，则next()会引发StopIteration.在大多数情况下，这可能很好.您要求输入第一个元素，不存在这样的元素，因此程序可能无法继续.

Do note that next() raises StopIteration if no element was found. In most cases, that's probably good. You asked for the first element, no such element exists, and so the program probably cannot continue.

如果另一方面，您知道知道该怎么做，则可以将默认值提供给next():

If, on the other hand, you do know what to do in that case, you can supply a default to next():

conf_files = ['~/.foorc', '/etc/foorc']
conf_file = next((f for f in conf_files if os.path.exists(f)),
                 '/usr/lib/share/foo.defaults')

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