我这样做: def walk(samples):
for d in range(samples):
yield d
def walk200():
for d in walk(200):
yield d
但实际上,我想这是什么,使代码更短: def walk200():
reyield walk(200)
我该怎么办reyield?
正如问题所述,将一个值传递给第一个iterator.next()调用是否有意义,还是会被忽略?作为一个例子: function* foo(x) {
var y = 2 * (yield(x + 1));
var z = yield(y/3);
return (x + y + z);
}
var it = foo(5);
//
我在合并和拆分部分偶然发现了来自pymotw.com的这段代码。 from itertools import *
def make_iterables_to_chain():
yield [1, 2, 3]
yield ['a', 'b', 'c']
for i in chain.from_iterable(make_iterables_to_chain()):
any一旦在任何迭代器中发现单个真值(无论是迭代器,生成器,列表等),就会发生短路 对于普通的发电机组,这是真的: Python 3.6.2 (v3.6.2:5fd33b5, Jul 8 2017, 04:14:34) [MSC v.1900 32 bit (Intel)] on win32
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