Python 3.3.2 round函数并非"四舍五入"

长枪不倒

　　对于一些貌似很简单常见的函数,最好还是去读一下Python文档,否则当你被某个BUG折磨得死去活来时,还不知根源所在.尤其是Python这种不断更新的语言.(python 2.7 的round和3.3.2不一样)

• 3.3.2官方文档对round的定义
  

　　round(number[, ndigits])
　　Return the floating point value number rounded to ndigits digits after the decimal point. If ndigits is omitted, it defaults to zero. Delegates to number.__round__(ndigits).
　　For the built-in types supporting round(), values are rounded to the closest multiple of 10 to the power minus ndigits; if two multiples are equally close, rounding is done toward the even choice (so, for example, both round(0.5) and round(-0.5) are 0, and round(1.5) is 2). The return value is an integer if called with one argument, otherwise of the same type as number.

• 举例查看:
  

>>> round(1.5)
2
>>> round(2.5)
2
>>> round(3.5)
4
>>> round(4.5)
4
>>> round(5.5)
6
>>> round(6.5)
6
　　
　　分析定义中的关键语句:



values are rounded to the closest multiple of 10 to the power minus ndigits

• 假设round(123.45, 1)=A,那么A在数轴上是离123.45最近的点(the closest),且A保留1位小数.
  
• "10 to the power minus ndigits"意思是"10的负ndigits次方",即pow(10, -ndigits).
  
• "multiple of n" 表示n的倍数.倍数只能是整数,这个是关键.(可参考维基百科:http://zh.wikipedia.org/wiki/%E5%9B%A0%E6%95%B8)
  

　　那么,例如round(123.45, 1)就表示数轴上离123.45最近的"0.1的X倍".
　　X=1时,"0.1的X倍"(即0.1)是离123.45最近的值吗?显然不是,因为当X=1000时,"0.1的X倍"是100,比0.1更靠近123.45.
　　按照这个原理进行推导,容易知道当X=1234或1235时,"0.1的X倍"最靠近123.45.那么该取哪个X作为最终结果呢?



if two multiples are equally close, rounding is done toward the even choice
　　"toward the even choice"意思就是说取偶数.即"0.1的1234倍",即123.4.
　　然而,当你测试round(123.45, 1):



>>> round(123.45, 1)
123.5
　　为什么取偶数呢?是BUG吗?非也,官方文档继续解释:



Note
The behavior of round() for floats can be surprising: for example, round(2.675, 2) gives 2.67 instead of the expected 2.68. This is not a bug: it’s a result of the fact that most decimal fractions can’t be represented exactly as a float. See Floating Point Arithmetic: Issues and Limitations for more information
意思就是说计算机需要先将十进制123.45转换为二进制,这个过程会导致二进制的值比123.45略大(比如123.45000001之类的),那么自然就得到123.5这个值了.
这种情形非常普遍:



>>> round(0.05, 1) #should return 0.0
0.1
>>> round(0.15, 1) #should return 0.2
0.1
>>> round(0.25, 1)
0.2
>>> round(0.35, 1) #should return 0.4
0.3
>>> round(0.45, 1) #should return 0.4
0.5