改写python round()函数，解决四舍五入问题 round(1.365,2)=1.36

Posted 2021-01-22 dinghanhua

 round()函数四舍五入存在一个问题，遇到5不一定进一。如下图所示：

print(round(1.365,2)) #1.36 没进一
print(\'%.2f\'%1.365)
print(round(1.3651,2)) #1.37 对的
print(\'%.2f\'%1.3651)
print(round(1.465,2)) #1.47 对的
print(\'%.2f\'%1.465)

 

 

没想到什么好办法，先改写了一下 

def round_rewrite(data,i=0):
    \'\'\'
    四舍五入,解决round(7.35)=7.3的问题
    :param data:
    :param i: 保留的位数,默认0
    :return:
    \'\'\'
    if not isinstance(i,int): #i不是整数报错
        raise TypeError(\'the second param must be int\')
    else:
        mi = 10**i
        datax = data*mi
        f = datax - int(datax)
        if f >=0.5:
            res = (int(datax)+1)/mi
        elif f <=-0.5:
            res = (int(datax)-1)/mi
        else:
            res = int(datax)/mi

        if i <= 0:
            res = int(res)
    return res

 

data = 1.365
print(round(data,2))
print(round_rewrite(data,2))