如何找到具有两位以上小数的浮点数？ （Python）

Question

I need a solution to access all numbers that have more than two decimals. eg

Have:

nums = [0.95, 0.7, 0.0, 0.3234, 0.54563]

Need:

many_decimals = [0.3234, 0.54563]

Thanks a lot:)

Answer 1

You can convert to string, reverse the string, then get index of '.'in the string which is the count of decimal places.

Try something like this:

nums = [0.95, 0.7, 0.0, 0.3234, 0.54563]
print([f for f in nums if str(f)[::-1].find('.') > 2])

Output:

[0.3234, 0.54563]

Solution breakdown:

s = str(0.54563)[::-1] # => '36545.0'
s.find('.')            # => 5

An alternative can use the decimal package and use the exponent, which creates the same output as above.

import decimal
print([f for f in nums if abs(decimal.Decimal(str(f)).as_tuple().exponent) > 2])

Answer 2

10**x*n%1>0 will return True if n has more than x decimals

so you can do many_decimals = [n for n in nums if 10**2*n%1>0]

Explanation

10**x*n%1>0 is ((10**x) * n) % 1 > 0 :

1. 10 to the power of x times n will make decimals shift x spaces to the left
2. modulo 1 of that returns only the decimal part in python, like 213.123 % 1 = 0.123
3. if it's not 0 that means that it had more than x decimals

Answer 3

The fraction of the exact float number's value must end with: 0.0, 0.25, 0.5, 0.75 .

Algorithm:

1. Extract the fraction.
2. Scaled by 4.0.
3. Test fraction against 0.0.

If non-zero, the original value needs more than 2 places for exact decimal representation.