如何在Python中进行以下舍入:
四舍五入到最接近的0.05小数位
7.97 -> 7.95
6.72 -> 6.70
31.06 -> 31.05
36.04 -> 36.05
5.25 -> 5.25
希望这样清楚明白。
def round_to(n, precision):
correction = 0.5 if n >= 0 else -0.5
return int( n/precision+correction ) * precision
def round_to_05(n):
return round_to(n, 0.05)
round_to_05(-1)
给出的结果是 -0.95
,这似乎不是正确的结果。 - David Webbround(n / precision) * precision
替换 int()
,则不需要进行 correction
。那么包装函数也不再需要,因为这只是一行简单的代码。但是它确实有一些缺陷,比如 round(0.275/.05)*.05 == 0.30000000000000004
。为了避免这种情况,可以使用 len(str(.05).split('.')[-1])
来获取小数位数,并再次使用 round()
将 0.30000000000000004
更改为 0.30
。 - mo-handef round05(number):
return (round(number * 20) / 20)
更通用的说法是:
def round_to_value(number,roundto):
return (round(number / roundto) * roundto)
唯一的问题是因为您使用浮点数,所以您得到的答案可能不会完全符合您的要求:>>> round_to_value(36.04,0.05)
36.050000000000004
round_to_value(36.04,5)
得到的结果是 35.0
。 - David Webb我们开始吧。
round(VALUE*2.0, 1) / 2.0
敬礼
使用 Lambda 函数:
>>> nearest_half = lambda x: round(x * 2) / 2
>>> nearest_half(5.2)
5.0
>>> nearest_half(5.25)
5.5
>>> nearest_half(5.26)
5.5
>>> nearest_half(5.5)
5.5
>>> nearest_half(5.75)
6.0
这是一个简短的语句
def roundto(number, multiple):
return number+multiple/2 - ((number+multiple/2) % multiple)
要将其四舍五入到您想要的精度:
>>> def foo(x, base=0.05):
... return round(base*round(x/base), 2)
>>> foo(5.75)
5.75
>>> foo(5.775)
5.8
>>> foo(5.77)
5.75
>>> foo(7.97)
7.95
>>> foo(6.72)
6.7
>>> foo(31.06)
31.05
>>> foo(36.04)
36.05
>>> foo(5.25)
5.25
import numpy as np
df['a']=(df["a"]*2).apply(np.ceil)/2
df['a']=(df["a"]*2).apply(np.floor)/2
这是使用numpy进行四舍五入0.5的列处理...
我也遇到过同样的问题,并且由于没有找到“终极”解决方案,所以我提供我的方法。
首先是主要部分(之前已经回答过了):
def round_to_precision(x, precision):
# This correction required due to float errors and aims to avoid cases like:
# 100.00501 / 0.00001 = 10000500.999999998
# It has a downside as well - it may lead to vanishing the difference for case like
# price = 100.5 - (correction - correction/10), precision = 1 => 101 not 100
# 5 decimals below desired precision looks safe enough to ignore
correction = 1e-5 if x > 0 else -1e-5
result = round(x / precision + correction) * precision
return round(result, find_first_meaningful_decimal(precision))
这里唯一有点棘手的地方是find_first_meaningful_decimal,我已经实现了以下代码:
def find_first_meaningful_decimal(x):
candidate = 0
MAX_DECIMAL = 10
EPSILON = 1 / 10 ** MAX_DECIMAL
while round(x, candidate) < EPSILON:
candidate +=1
if candidate > MAX_DECIMAL:
raise Exception('Number is too small: {}'.format(x))
if int(x * 10 ** (candidate + 1)) == 5:
candidate += 1
return candidate
print(round_to_precision(129.950002, 0.0005))
print(round_to_precision(-129.95005, 0.0001))
129.9505
-129.9501
接受答案的扩展。
def round_to(n, precision):
correction = precision if n >= 0 else -precision
return round(int(n/precision+correction)*precision, len(str(precision).split('.')[1]))
test_cases = [101.001, 101.002, 101.003, 101.004, 101.005, 101.006, 101.007, 101.008, 101.009]
[round_to(-x, 0.003) for x in test_cases]
[-101.001, -101.001, -101.001, -101.004, -101.004, -101.004, -101.007, -101.007, -101.007]
20 == 1 / 0.05
。 - Chris Morgan