使用next()时停止迭代错误

Question

我无法澄清我next()在python(3)中使用的自我.

我有一个数据:

chr pos ms01e_PI    ms01e_PG_al ms02g_PI    ms02g_PG_al ms03g_PI    ms03g_PG_al ms04h_PI    ms04h_PG_al
2   15881989    4   C|C 6   A|C 7   C|C 7   C|C
2   15882091    4   A|T 6   A|T 7   T|A 7   A|A
2   15882148    4   T|T 6   T|T 7   T|T 7   T|G

我读它就像:

工作良好

c = csv.DictReader(io.StringIO(data), dialect=csv.excel_tab)
print(c)
print(list(c))

工作良好

c = csv.DictReader(io.StringIO(data), dialect=csv.excel_tab)
print(c)
keys = next(c)
print('keys:', keys)

但是,现在有一个问题.

c = csv.DictReader(io.StringIO(data), dialect=csv.excel_tab)
print(c)
print(list(c))
keys = next(c)
print('keys:', keys)

错误信息:

Traceback (most recent call last):
2   15882601    4   C|C 9   C|C 6   C|C 5   T|C

  File "/home/everestial007/Test03.py", line 24, in <module>
keys = next(c)
  File "/home/everestial007/anaconda3/lib/python3.5/csv.py", line 110, in __next__

    row = next(self.reader)

StopIteration

为什么print(keys)后来print(list(c))给出StopIteration？我阅读了文档,但我不清楚这个特定的例子.

Answer 1

错误不在于print声明.这是keys = next(c)线.考虑一个更简单的例子来重现您的问题.

a = (i ** 2 for i in range(5))  

a           # `a` is a generator object
<generator object <genexpr> at 0x150e286c0>

list(a)     # calling `list` on `a` will exhaust the generator
[0, 1, 4, 9, 16]

next(a)     # calling `next` on an exhausted generator object raises `StopIteration`
---------------------------------------------------------------------------
StopIteration                             Traceback (most recent call last)
<ipython-input-2076-3f6e2eea332d> in <module>()
----> 1 next(a)

StopIteration: 

发生的事情是c迭代器对象(非常类似于上面的生成器 a),并且意味着迭代一次直到它耗尽.调用list此对象将耗尽它,以便可以将元素收集到列表中.

一旦对象耗尽,它将不再产生任何元素.此时,StopIteration如果您尝试迭代它,即使在耗尽之后,生成器机制也会设计为引发a .诸如for循环之类的构造侦听此错误,然后静默地吞下它,然后,next一旦引发它就会返回原始异常.