为什么 Python 中没有 first(iterable) 内置函数?

2022-01-10 00:00:00 python generator iterator

问题描述

我想知道 Python 内置函数中没有 first(iterable) 是否有原因，有点类似于 any(iterable) 和 all(iterable) (它可能藏在某个 stdlib 模块中，但我在 itertools 中看不到它).first 将执行短路生成器评估，从而可以避免不必要的(并且可能是无限数量的)操作；即

I'm wondering if there's a reason that there's no first(iterable) in the Python built-in functions, somewhat similar to any(iterable) and all(iterable) (it may be tucked in a stdlib module somewhere, but I don't see it in itertools). first would perform a short-circuit generator evaluation so that unnecessary (and a potentially infinite number of) operations can be avoided; i.e.

def identity(item):
    return item

def first(iterable, predicate=identity):
    for item in iterable:
        if predicate(item):
            return item
    raise ValueError('No satisfactory value found')

这样你可以表达如下内容:

This way you can express things like:

denominators = (2, 3, 4, 5)
lcd = first(i for i in itertools.count(1)
    if all(i % denominators == 0 for denominator in denominators))

显然你不能在这种情况下执行 list(generator)[0]，因为生成器不会终止.

Clearly you can't do list(generator)[0] in that case, since the generator doesn't terminate.

或者，如果您有一堆正则表达式要匹配(当它们都具有相同的 groupdict 接口时很有用):

Or if you have a bunch of regexes to match against (useful when they all have the same groupdict interface):

match = first(regex.match(big_text) for regex in regexes)

通过避免 list(generator)[0] 和在正匹配时短路，您可以节省大量不必要的处理.

You save a lot of unnecessary processing by avoiding list(generator)[0] and short-circuiting on a positive match.

解决方案

如果你有一个迭代器，你可以调用它的 next 方法.比如:

If you have an iterator, you can just call its next method. Something like:

In [3]: (5*x for x in xrange(2,4)).next()
Out[3]: 10

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