枚举（）-在Python中生成

Question

我想知道当我将生成器函数的结果传递给python的enumerate（）时会发生什么。示例：

def veryBigHello():
    i = 0
    while i < 10000000:
        i += 1
        yield "hello"

numbered = enumerate(veryBigHello())
for i, word in numbered:
    print i, word

枚举是缓慢地迭代，还是将所有内容都放入第一个枚举？我99.999%确信它是懒惰的，所以我能把它和生成器函数一模一样地对待吗，或者我需要注意什么？

Answer 1

由于可以调用此函数而不必获取内存异常，因此它确实很懒

def veryBigHello():
    i = 0
    while i < 1000000000000000000000000000:
        yield "hello"

numbered = enumerate(veryBigHello())
for i, word in numbered:
    print i, word

Answer 2

比之前的建议更容易说出来：

$ python
Python 2.5.5 (r255:77872, Mar 15 2010, 00:43:13)
[GCC 4.3.4 20090804 (release) 1] on cygwin
Type "help", "copyright", "credits" or "license" for more information.
>>> abc = (letter for letter in 'abc')
>>> abc
<generator object at 0x7ff29d8c>
>>> numbered = enumerate(abc)
>>> numbered
<enumerate object at 0x7ff29e2c>

如果枚举不执行延迟计算，它将返回[(0,'a'), (1,'b'), (2,'c')]或一些（几乎）等效值。

当然，enumerate只是一个奇特的生成器：

def myenumerate(iterable):
   count = 0
   for _ in iterable:
      yield (count, _)
      count += 1

for i, val in myenumerate((letter for letter in 'abc')):
    print i, val

Answer 3

太懒了。很容易证明事实如此：

>>> def abc():
...     letters = ['a','b','c']
...     for letter in letters:
...         print letter
...         yield letter
...
>>> numbered = enumerate(abc())
>>> for i, word in numbered:
...     print i, word
...
a
0 a
b
1 b
c
2 c