I thought that in arange the second argument will never be comprised in the list, but see what happens in Python 2.7.3.
我认为在范围内第二个参数永远不会包含在列表中，但是看看Python 2.7.3中会发生什么。
>>> import numpy
>>> numpy.arange(0.2,0.9,0.1)
array([ 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8])
>>> numpy.arange(0.2,0.7,0.1)
array([ 0.2, 0.3, 0.4, 0.5, 0.6])
>>> numpy.arange(0.2,0.6,0.1)
array([ 0.2, 0.3, 0.4, 0.5])
but
但
>>> numpy.arange(0.2,0.8,0.1)
array([ 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8])
Does anybody know the reason for this behaviour ?
有人知道这种行为的原因吗？
2 个解决方案
#1
4
In [169]: val = 0.2
In [170]: for i in range(6): #simulating the loop in arange.
.....: val += 0.1
.....:
In [171]: val
Out[171]: 0.7999999999999999
floating point rounding error. The result being that the last value is also printed.
浮点舍入误差。结果是还打印了最后一个值。
#2
4
This is a problem with floating-point precision.
这是浮点精度的问题。
From the documentation:
从文档：
stop : number
停止：数字
End of interval. The interval does not include this value, except in some cases where step is not an integer and floating point round-off affects the length of out.
间隔结束。间隔不包括此值，除非在某些情况下，step不是整数，浮点舍入会影响out的长度。
To see more about the limitations of floats, look at Python's tutorial.
要了解有关浮点数限制的更多信息，请查看Python的教程。
#1
4
In [169]: val = 0.2
In [170]: for i in range(6): #simulating the loop in arange.
.....: val += 0.1
.....:
In [171]: val
Out[171]: 0.7999999999999999
floating point rounding error. The result being that the last value is also printed.
浮点舍入误差。结果是还打印了最后一个值。
#2
4
This is a problem with floating-point precision.
这是浮点精度的问题。
From the documentation:
从文档：
stop : number
停止：数字
End of interval. The interval does not include this value, except in some cases where step is not an integer and floating point round-off affects the length of out.
间隔结束。间隔不包括此值，除非在某些情况下，step不是整数，浮点舍入会影响out的长度。
To see more about the limitations of floats, look at Python's tutorial.
要了解有关浮点数限制的更多信息，请查看Python的教程。