“'int'对象不可迭代"在一段时间内

问题描述

def rect_extend(x):
  m, n = 1
  while 1 < x:
    m = m + 1
    n = n + 1
  return m, n

此简单函数返回:

iPython中的

错误.我不知道为什么这样做，while函数不起作用-条件似乎是true.

error in iPython. I don't know why it does this, while function doesn't work - condition seems to be true.

(而的条件是有意简化的；原始代码没有此条件)

(while's condition was simplified on purpose; original code doesn't have it)

推荐答案

我认为你想要

m = 1
n = 1

或

m = n = 1

而不是m, n = 1.

此(序列解压缩)[http://docs.python.org/3/tutorial/datastructures.html#tuples-and-sequences]:

This (sequence unpacking)[http://docs.python.org/3/tutorial/datastructures.html#tuples-and-sequences]:

x, y = z

与您认为的有所不同.

这实际上意味着:

x = z[0]    # The first item in z
y = z[1]    # The second element of z

例如，您可以执行以下操作:

For instance, you could do this:

x, y, z = (1, 2, 4)

然后:

>>> x
1
>>> y
2
>>> z
4

在您的情况下，这是行不通的，因为1是整数，没有元素，因此会出错.

In your case, you this doesn't work, because 1 is an integer, it doesn't have elements, hence the error.

与元组(和splat运算符-*)结合使用的序列拆包的有用功能:

Useful features of sequence unpacking combined with tuples (and the splat operator - *):

此:

a, b = b, a

交换a和b的值.

解压range，对常量有用:

>>> RED, GREEN, BLUE = range(3)
>>> RED
0
>>> GREEN
1
>>> BLUE
2

splat运算符:

>>> first, *middle, last = 1, 2, 3, 4, 5, 6, 7, 8, 9
>>> first
1
>>> middle
[2, 3, 4, 5, 6, 7, 8]
>>> last
9

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