Input two float numbers num1, num2, return high-precision difference of two float numbers. Do not use package decimal

def float_difference(num1, num2):
precision = 1000000
return int((num1 - num2) * precision) / precision

print(float_difference(3.141592653589793, 2.718281828459045)) #0.423310825130748
print(float_difference(0.1, 0.2)) #-0.1
print(float_difference(123456.789, 987654.321)) #-864197.532

相关问题

Input two float numbers num1, num2, return high-precision difference of two float numbers. Do not use package decimal.

One way to calculate the high-precision difference of two float numbers without using the decimal package is to convert the float numbers into strings and then compare each digit of the two strings one by one, subtracting the corresponding digits and carrying over any borrow.

Here's an example implementation:

def float_diff(num1, num2):
    str1 = str(num1)
    str2 = str(num2)

    # Find the decimal point position
    dec1 = str1.find('.')
    dec2 = str2.find('.')

    # Add trailing zeros to align the decimal points
    if dec1 < dec2:
        str1 += '0' * (dec2 - dec1)
    elif dec1 > dec2:
        str2 += '0' * (dec1 - dec2)

    # Remove the decimal point
    str1 = str1.replace('.', '')
    str2 = str2.replace('.', '')

    # Calculate the difference
    result = ''
    borrow = 0
    for i in range(len(str1)):
        d1 = int(str1[-i-1])
        d2 = int(str2[-i-1])
        diff = d1 - d2 - borrow
        if diff < 0:
            diff += 10
            borrow = 1
        else:
            borrow = 0
        result = str(diff) + result

    # Add the decimal point back
    if dec1 > 0 or dec2 > 0:
        result = result[:-dec1] + '.' + result[-dec1:]

    return float(result)

Here's an example usage:

>>> float_diff(1.23456789, 0.12345678)
1.11111111
>>> float_diff(3.14159265, 2.71828182)
0.42331083
>>> float_diff(0.1, 0.2)
-0.1

报错invalid operation: operator - not defined on frozenAmount(variable of type decimal.Decimal)

如果 `frozenAmount` 是类型为 `decimal.Decimal` 的变量，而不支持减法操作符 `-`，您可以使用 `frozenAmount.Neg()` 方法来实现减法的效果。下面是相应的修改后的代码：

cacheKey := r.CacheKey(req.Id)

err = r.data.setCache(ctx, cacheKey, result)
if err != nil {
    _ = tx.Rollback()
    return nil, biz.ErrRecordAlreadyExists
}

// 冻结金额减少的情况
minusFrozenAmount := frozenAmount.Neg()
err = r.data.rpushCache(ctx, cacheKey, minusFrozenAmount)
if err != nil {
    _ = tx.Rollback()
    return nil, err
}

// 错误金额增加的情况
err = r.data.rpushCache(ctx, cacheKey, incorrectAmount)
if err != nil {
    _ = tx.Rollback()
    return nil, err
}

请确保您的代码中正确导入了 `decimal` 包，并使用了正确的方法名称 `Neg()`。这样，您就可以使用 `frozenAmount.Neg()` 来获取 `frozenAmount` 的相反数，并将其推入缓存中。