Python 3.x：测试生成器是否剩余元素 [英] Python 3.x: Test if generator has elements remaining

问题描述

当我在for循环中使用生成器时，它似乎知道，当没有更多的元素产生时。现在，我必须使用没有for循环的生成器，并手动使用 next （）来获取下一个元素。我的问题是，我怎么知道，如果没有更多元素？

When I use a generator in a for loop, it seems to "know", when there are no more elements yielded. Now, I have to use a generator WITHOUT a for loop, and use next() by hand, to get the next element. My problem is, how do I know, if there are no more elements?

我只知道： next （）引发异常（StopIteration） ），如果没有什么可以留下的，但对于这样一个简单的问题，这也不是一个例外吗？是不是像 has_next （）那样的方法？

I know only: next() raises an exception (StopIteration), if there is nothing left, BUT isn't an exception a little bit too "heavy" for such a simple problem? Isn't there a method like has_next() or so?

以下几行应该清楚，我的意思是：

The following lines should make clear, what I mean:

#!/usr/bin/python3

# define a list of some objects
bar = ['abc', 123, None, True, 456.789]

# our primitive generator
def foo(bar):
    for b in bar:
        yield b

# iterate, using the generator above
print('--- TEST A (for loop) ---')
for baz in foo(bar):
    print(baz)
print()

# assign a new iterator to a variable
foobar = foo(bar)

print('--- TEST B (try-except) ---')
while True:
    try:
        print(foobar.__next__())
    except StopIteration:
        break
print()

# assign a new iterator to a variable
foobar = foo(bar)

# display generator members
print('--- GENERATOR MEMBERS ---')
print(', '.join(dir(foobar)))

输出如下：

--- TEST A (for loop) ---
abc
123
None
True
456.789

--- TEST B (try-except) ---
abc
123
None
True
456.789

--- GENERATOR MEMBERS ---
__class__, __delattr__, __doc__, __eq__, __format__, __ge__, __getattribute__, __gt__, __hash__, __init__, __iter__, __le__, __lt__, __name__, __ne__, __new__, __next__, __reduce__, __reduce_ex__, __repr__, __setattr__, __sizeof__, __str__, __subclasshook__, close, gi_code, gi_frame, gi_running, send, throw

感谢大家，祝你有个美好的一天！ ：）

Thanks to everybody, and have a nice day! :)

推荐答案

你写的两个语句都是以完全相同的方式查找生成器的结尾。 for循环只调用.next（），直到引发StopIteration异常，然后终止。

The two statements you wrote deal with finding the end of the generator in exactly the same way. The for-loop simply calls .next() until the StopIteration exception is raised and then it terminates.

http://docs.python.org/tutorial/classes.html#iterators

因此我不喜欢我认为等待StopIteration异常是处理问题的'重'方式，它是设计使用的生成器的方式。

As such I don't think waiting for the StopIteration exception is a 'heavy' way to deal with the problem, it's the way that generators are designed to be used.

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