I have a list of decimals, I want to filter it using python, and get only decimals with one digit on the right like 2.1 and 1.8, not 2.15:
my_list = [2.1, 2.15, 1.8, 1.995, 1.31, 2.2]
needed output:
[2.1, 1.8, 2.2]
CodePudding user response：
You can use round method in both numpy and pure Python:
numpy:
import numpy as np
my_list = [2.1, 2.15, 1.8, 1.995, 1.31, 2.2]
x = np.array(my_list)
y = x[np.around(x, 1) == x]
>>> y
array([2.1, 1.8, 2.2])
>>> y.tolist()
[2.1, 1.8, 2.2]
Python:
my_list = [2.1, 2.15, 1.8, 1.995, 1.31, 2.2]
>>> [n for n in my_list if round(n, 1) == n]
[2.1, 1.8, 2.2]
Note that it doesn't work in general for numbers that differs by small amounts like epsilon = np.finfo(float).eps. For example 2.2 - epsilon is returned to be decimal in both numpy and Python.
CodePudding user response：
Multiply each element by 100 and check if the last digit is zero:
(my_number * 100) % 10 == 0
CodePudding user response：
condition = lambda x: (num*10)%1==0 and (nums%1)!=1 # Has only 1 trailing digit
res = [num for num in my_list if condition(x)]
Explanation
This program writes a condition that returns true if a number is:
- Not an integer
- Is an integer if shifted over by one
This means that it returns True if it has only 1 decimal place, no more, no less.
The second part simply just gets all the numbers in that list that meet that condition, i.e. having 1 decimal place.