Tro*_*yvs 15 python types iterator exception ipython
我正在使用python 2.7和ipython2.7.在ipython中我尝试过:
class Fib(object):
def __init__(self, max):
super(Fib, self).__init__()
self.max = max
def __iter__(self):
self.a = 0
self.b = 1
return self
def __next__(self):
fib = self.a
if fib > self.max:
raise StopIteration
self.a, self.b = self.b, self.a + self.b
return fib
def main():
fib = Fib(100)
for i in fib:
print i
if __name__ == '__main__':
main()
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那么它报告错误:
类型错误回溯(最近通话最后一个)在()22 23如果名称 == ' 主要 ':---> 24主()25
<ipython-input-21-f10bd2d06666> in main()
18 def main():
19 fib = Fib(100)
---> 20 for i in fib:
21 print i
22
TypeError: iter() returned non-iterator of type 'Fib'
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这段代码实际上来自互联网.语法对我来说似乎没问题,但问题是如何发生的?
谢谢.
Kro*_*ack 35
def __next__(self)
适用于Python 3
对于Python 2,您需要添加方法 next()
此代码将在Python 3和Python 2下运行:
class Fib(object):
def __init__(self, max):
super(Fib, self).__init__()
self.max = max
def __iter__(self):
self.a = 0
self.b = 1
return self
def __next__(self):
fib = self.a
if fib > self.max:
raise StopIteration
self.a, self.b = self.b, self.a + self.b
return fib
def next(self):
return self.__next__()
def main():
fib = Fib(100)
for i in fib:
print(i)
if __name__ == '__main__':
main()
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