TL;DR:
round(x)
将对其进行四舍五入并将其更改为整数.
您没有给任何变量赋值round(h)
.当你调用round(h)
时,它返回整数,但不做其他任何操作;您必须将该行更改为:
h = round(h)
将新值指定给h
.
正如@plowman在 comments 中所说的,Python的round()
并不像人们通常所期望的那样工作,这是因为数字作为变量存储的方式通常不是你在屏幕上看到的方式.有lots of answers种解释了这种行为.
避免这个问题的一种方法是使用this answer表示的小数.
为了让这个答案在不使用额外库的情况下正常工作,可以方便地使用自定义舍入函数.我提出了以下解决方案,就我测试而言,它避免了所有存储问题.它基于使用字符串表示法,用repr()
(而不是str()
!)获得.它看起来很黑,但这是我找到的解决所有案件的唯一方法.它同时适用于Python2和Python3.
def proper_round(num, dec=0):
num = str(num)[:str(num).index('.')+dec+2]
if num[-1]>='5':
return float(num[:-2-(not dec)]+str(int(num[-2-(not dec)])+1))
return float(num[:-1])
测验:
>>> print(proper_round(1.0005,3))
1.001
>>> print(proper_round(2.0005,3))
2.001
>>> print(proper_round(3.0005,3))
3.001
>>> print(proper_round(4.0005,3))
4.001
>>> print(proper_round(5.0005,3))
5.001
>>> print(proper_round(1.005,2))
1.01
>>> print(proper_round(2.005,2))
2.01
>>> print(proper_round(3.005,2))
3.01
>>> print(proper_round(4.005,2))
4.01
>>> print(proper_round(5.005,2))
5.01
>>> print(proper_round(1.05,1))
1.1
>>> print(proper_round(2.05,1))
2.1
>>> print(proper_round(3.05,1))
3.1
>>> print(proper_round(4.05,1))
4.1
>>> print(proper_round(5.05,1))
5.1
>>> print(proper_round(1.5))
2.0
>>> print(proper_round(2.5))
3.0
>>> print(proper_round(3.5))
4.0
>>> print(proper_round(4.5))
5.0
>>> print(proper_round(5.5))
6.0
>>>
>>> print(proper_round(1.000499999999,3))
1.0
>>> print(proper_round(2.000499999999,3))
2.0
>>> print(proper_round(3.000499999999,3))
3.0
>>> print(proper_round(4.000499999999,3))
4.0
>>> print(proper_round(5.000499999999,3))
5.0
>>> print(proper_round(1.00499999999,2))
1.0
>>> print(proper_round(2.00499999999,2))
2.0
>>> print(proper_round(3.00499999999,2))
3.0
>>> print(proper_round(4.00499999999,2))
4.0
>>> print(proper_round(5.00499999999,2))
5.0
>>> print(proper_round(1.0499999999,1))
1.0
>>> print(proper_round(2.0499999999,1))
2.0
>>> print(proper_round(3.0499999999,1))
3.0
>>> print(proper_round(4.0499999999,1))
4.0
>>> print(proper_round(5.0499999999,1))
5.0
>>> print(proper_round(1.499999999))
1.0
>>> print(proper_round(2.499999999))
2.0
>>> print(proper_round(3.499999999))
3.0
>>> print(proper_round(4.499999999))
4.0
>>> print(proper_round(5.499999999))
5.0
最后,正确的答案是:
# Having proper_round defined as previously stated
h = int(proper_round(h))
测验:
>>> proper_round(6.39764125, 2)
6.31 # should be 6.4
>>> proper_round(6.9764125, 1)
6.1 # should be 7
这里的问题是,第dec
位小数可以是9,如果第dec+1
位>=5 9将变成0,1应被带到第dec-1
位.
如果我们考虑到这一点,我们会得到:
def proper_round(num, dec=0):
num = str(num)[:str(num).index('.')+dec+2]
if num[-1]>='5':
a = num[:-2-(not dec)] # integer part
b = int(num[-2-(not dec)])+1 # decimal part
return float(a)+b**(-dec+1) if a and b == 10 else float(a+str(b))
return float(num[:-1])
在上面描述的情况下,b = 10
和之前的版本只是连接a
和b
,这将导致连接10
,其中尾随0将消失.这个版本将b
转换为基于dec
的右小数位,作为正确的进位.