python:将/ float / string格式化为特定的小数/分数?

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我在python中找到了很多关于舍入的帖子,但据我所知,到目前为止还没有解决我的问题。

提问于
用户回答回答于

可以使用十进制模块对于高度可定制的四舍五入:

import decimal 

def rnd(f,n,r=decimal.ROUND_HALF_DOWN):
    pat='1.'+'0'*n
    return str(decimal.Decimal(str(f)).quantize(decimal.Decimal(pat),r))

现在来测试:

for n in (500.99,100.39,5.019):
    for i in (0,1,2):
        print('n={:10}, i={:2}, Down: {:>10}, Closest: {:>10}'.format(n,i, rnd(n,i,decimal.ROUND_FLOOR),rnd(n,i)))   
n=    500.99, i= 0, Down:        500, Closest:        501
n=    500.99, i= 1, Down:      500.9, Closest:      501.0
n=    500.99, i= 2, Down:     500.99, Closest:     500.99
n=    100.39, i= 0, Down:        100, Closest:        100
n=    100.39, i= 1, Down:      100.3, Closest:      100.4
n=    100.39, i= 2, Down:     100.39, Closest:     100.39
n=     5.019, i= 0, Down:          5, Closest:          5
n=     5.019, i= 1, Down:        5.0, Closest:        5.0
n=     5.019, i= 2, Down:       5.01, Closest:       5.02

我认为你只需要用fmod:

import math 

def rnd(n,f,prec,direction='down'):
    if direction=='down':
        if prec==0:
            return str(int(n))
        return str(round(n-math.fmod(n,f),prec))
    else:
        if prec==0:
            return str(int(round(n)))
        n+=f
        return str(round(n-math.fmod(n,f),prec))

所有例子如下:

>>> rnd(500.99,0.1,0)
'500'
>>> rnd(500.99,0.1,0,'')
'501'
>>> rnd(100.39,0.2,1,'')
'100.4'
>>> rnd(100.39,0.2,1)
'100.2'
>>> rnd(100.59,0.2,1,'')
'100.6'
>>> rnd(100.59,0.2,1)
'100.4'
>>> rnd(5.019,0.01,2)
'5.01'
>>> rnd(5.019,0.01,2,'')
'5.02'
def rnd(n,f,prec):
    n1=int(n) if prec==0 else round(n-math.fmod(n,f),prec)
    n2=float(int(round(n))) if prec==0 else round((n+f)-math.fmod((n+f),f),prec)
    return str(min((n1,n2), key=lambda e: math.fabs(e-n)))
用户回答回答于

像这样的怎么样?

def round_accordingly(num):

    if num>=500:
        return str(round(num))
    elif num>=100:
        return str(round(num, 1))
    elif num>=5:
        return str(round(num, 2))


print(round_accordingly(550.73424))
print(round_accordingly(200.234245))
print(round_accordingly(7.564743))

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