将数字舍入到最接近的偶数 [英] Rounding a number to nearest even
问题描述
内置函数round（）将始终向上舍入，即1.5是
舍入为2.0，2.5舍入为3.0。
如果我想要舍入到最近的偶数，那就是
my_round（1.5）= 2＃正如预期的那样
my_round（2.5）= 2 #Not 3，这是一个奇怪的数字
我对舍入x.5形式的数字很感兴趣。取决于
x是奇数还是偶数。关于如何实现它的任何想法？
The built-in function round( ) will always "round up", that is 1.5 is
rounded to 2.0 and 2.5 is rounded to 3.0.
If I want to round to the nearest even, that is
my_round(1.5) = 2 # As expected
my_round(2.5) = 2 # Not 3, which is an odd num
I''m interested in rounding numbers of the form "x.5" depending upon
whether x is odd or even. Any idea about how to implement it ?
推荐答案
On 11 avr，12：14，bdsatish< bdsat ... @ gmail.comwrote：
On 11 avr, 12:14, bdsatish <bdsat...@gmail.comwrote:
内置函数round（）将始终向上舍入，即1.5是
四舍五入为2.0，2.5为四舍五入到3.0。
如果我想要舍入到最近的偶数，那就是
my_round（1.5）= 2＃正如预期的那样/>
my_round（2.5）= 2＃不是3，这是一个奇怪的数字
我对舍入x.5形式的数字感兴趣取决于
x是奇数还是偶数。有关如何实现它的任何想法？
The built-in function round( ) will always "round up", that is 1.5 is
rounded to 2.0 and 2.5 is rounded to 3.0.
If I want to round to the nearest even, that is
my_round(1.5) = 2 # As expected
my_round(2.5) = 2 # Not 3, which is an odd num
I''m interested in rounding numbers of the form "x.5" depending upon
whether x is odd or even. Any idea about how to implement it ?
当你说舍入到最近的偶数时，你的意思是new_round（3）< 3？
是这样，你可以尝试：
在[37]中：def new_round（x）：
....：返回回合（x / 2。 ）* 2
....：
在[38]中：new_round（1.5）
Out [38]： 2.0
在[39]：new_round（2.5）
Out [39]：2.0
在[ 40]：new_round（3）
Out [40]：4.0
When you say "round to the nearest even", you mean new_round(3) <3?
Is so, you can try:
In [37]: def new_round(x):
....: return round(x/2.)*2
....:
In [38]: new_round(1.5)
Out[38]: 2.0
In [39]: new_round(2.5)
Out[39]: 2.0
In [40]: new_round(3)
Out[40]: 4.0
4月11日下午3:27，colas.fran .. 。@ gmail.com写道：
On Apr 11, 3:27 pm, colas.fran...@gmail.com wrote:
11 avr，12：14，bdsatish< bdsat ... @ gmail.comwrote：
On 11 avr, 12:14, bdsatish <bdsat...@gmail.comwrote:
内置函数round（）将始终向上舍入，即1.5是
舍入为2.0，2.5舍入为3.0。
The built-in function round( ) will always "round up", that is 1.5 is
rounded to 2.0 and 2.5 is rounded to 3.0.
如果我想舍入到最近的偶数，那就是
If I want to round to the nearest even, that is
my_round（1.5）= 2＃正如预期的那样
my_round（2.5）= 2＃Not 3，这是一个奇数num
my_round(1.5) = 2 # As expected
my_round(2.5) = 2 # Not 3, which is an odd num
我对四舍五入x.5形式的数字很感兴趣。取决于
x是奇数还是偶数。有关如何实现它的任何想法？
I''m interested in rounding numbers of the form "x.5" depending upon
whether x is odd or even. Any idea about how to implement it ?
当你说舍入到最近的偶数时，你的意思是new_round（3）< 3？
When you say "round to the nearest even", you mean new_round(3) <3?
完全没有。条款最接近偶数只有当数字的形式为x.5时，才会出现图片。或者它与内置round（）相同。
new_round（3.0）必须是3.0本身。这是我想要的数学定义
：
如果''n''是一个整数，
new_round（n + 0.5）= n如果n / 2是整数
new_round（n + 0.5）=（n + 1）if（n + 1）/ 2是整数
在所有其他情况下，new_round（）的行为与round（）相似。这是
我期望的结果：
new_round（3.2）= 3
new_round（3.6）= 4
new_round（3.5）= 4
new_round（2.5）= 2
new_round（-0.5）= 0.0
new_round（ -1.5）= -2.0
new_round（-1.3）= -1.0
new_round（-1.8）= -2
new_round（-2.5 ）= -2.0
内置功能无法满足我对圆形（-2.5）或
轮（2.5）的需求
No. not at all. The clause "nearest even" comes into picture only when
a number is of form "x.5" or else it''s same as builtin round( ).
new_round(3.0) must be 3.0 itself. Here is the mathematical definition
of what I want:
If ''n'' is an integer,
new_round(n+0.5) = n if n/2 is integer
new_round(n+0.5) = (n+1) if (n+1)/2 is integer
In all other cases, new_round() behave similarly as round( ). Here are
the results I expect:
new_round(3.2) = 3
new_round(3.6) = 4
new_round(3.5) = 4
new_round(2.5) = 2
new_round(-0.5) = 0.0
new_round(-1.5) = -2.0
new_round(-1.3) = -1.0
new_round(-1.8) = -2
new_round(-2.5) = -2.0
The built-in function doesnt meet my needs for round(-2.5) or
round(2.5)
你不能这么做。
#untested
new_round（n）：
回答=回合（n）
＃现在回答奇数
如果回答％2：
返回答案 - 1
else：
返回答案
couldn''t you just do.
#untested
new_round(n):
answer = round(n)
# is answer now odd
if answer % 2:
return answer - 1
else:
return answer
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