問題描述

我想知道 Python 內(nèi)置函數(shù)中沒有 first(iterable) 是否有原因，有點類似于 any(iterable) 和 all(iterable) (它可能藏在某個 stdlib 模塊中，但我在 itertools 中看不到它).first 將執(zhí)行短路生成器評估，從而可以避免不必要的(并且可能是無限數(shù)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')

這樣你可以表達如下內(nèi)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))

顯然你不能在這種情況下執(zhí)行 list(generator)[0]，因為生成器不會終止.

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

或者，如果您有一堆正則表達式要匹配(當(dāng)它們都具有相同的 groupdict 接口時很有用):

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)

通過避免 list(generator)[0] 和在正匹配時短路，您可以節(jié)省大量不必要的處理.

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

推薦答案

如果你有一個迭代器，你可以調(diào)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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