enumerate() - 在 Python 中生成一个生成器
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enumerate()-ing a generator in Python
提问by Adam
I'd like to know what happens when I pass the result of a generator function to python's enumerate(). Example:
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。例子：
def veryBigHello():
i = 0
while i < 10000000:
i += 1
yield "hello"
numbered = enumerate(veryBigHello())
for i, word in numbered:
print i, word
Is the enumeration iterated lazily, or does it slurp everything into the first? I'm 99.999% sure it's lazy, so can I treat it exactly the same as the generator function, or do I need to watch out for anything?
枚举是懒惰地迭代，还是将所有内容都放入第一个？我 99.999% 确定它是懒惰的，所以我可以将它与生成器函数完全相同，还是需要注意什么？
采纳答案by Dave Webb
It's lazy. It's fairly easy to prove that's the case:
它很懒。很容易证明情况确实如此：
>>> 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
回答by Nikolaus Gradwohl
Since you can call this function without getting out of memory exceptions it definitly is lazy
由于您可以调用此函数而不会出现内存不足异常，因此它绝对是懒惰的
def veryBigHello():
i = 0
while i < 1000000000000000000000000000:
yield "hello"
numbered = enumerate(veryBigHello())
for i, word in numbered:
print i, word
回答by Wayne Werner
It's even easier to tell than either of the previous suggest:
比之前的任何一个建议都更容易判断：
$ 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>
If enumerate didn't perform lazy evaluation it would return [(0,'a'), (1,'b'), (2,'c')]or some (nearly) equivalent.
如果 enumerate 没有执行惰性求值，它将返回[(0,'a'), (1,'b'), (2,'c')]或一些（几乎）等价的。
Of course, enumerate is really just a fancy generator:
当然， enumerate 实际上只是一个花哨的生成器：
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
