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?
列表理解和过滤器会遍历整个列表.唯一的选择是手工循环吗?
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() 不是内置函数:
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，如果没有找到元素，next() 会引发 StopIteration.在大多数情况下，这可能很好.您要求第一个元素，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.
另一方面，如果您确实知道在这种情况下该怎么做，则可以为 next() 提供默认值:
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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