When the decimal places are reserved, there may be a simple solution without rounding

stay Python Run the following code ：

a = 2.358755
print(a)
print(decimal.Decimal.from_float(a))
print('\n Retain 5 The result of decimal places ：')
print(format(a, '.5f'))

Running results ：

2.358755
2.358754999999999935056393951526843011379241943359375

 Retain 5 The result of decimal places ：
2.35875

You can see , The result is not right . actually , Retain 5 The result of decimals is 2.35876
This is due to the principle of the underlying code , Look at number 2 The result of that ,2.358754999999999935056393951526843011379241943359375 , The first 6 The decimal is 4, So we come to 2.35875

Use the following code to solve this problem ：

def format_float(data, n):
    """  Given data , Retain n Decimal place   Example ： format(2.358755, '.5f') , The result is  2.35875, The correct rounding value is  2.35876 """
    digit = len(str(data).split('.')[-1])  #  Calculate the number of decimal places 
    if n == digit - 1:  #  The reserved digits are exactly one less than the total decimal digits , Then fill in the following 1
        data = float(str(data) + '1')
    return format(data, f'.{n}f')

a = 2.358755
print(format_float(a, 5))

Running results ：

2.35876

In fact, it complements the original floating-point number 1 And then we can solve this problem , repair 1 Actual data after ：

print(decimal.Decimal.from_float(2.3587551))

Running results ：

2.358755100000000215487716559437103569507598876953125

You can see the 6 Bit data is now 5, So we can accurately get the rounded value .