Separate Float Into Digits

I have a float: pi = 3.141 And want to separate the digits of the float and put them into a list as intergers like this: #[3, 1, 4, 1] I know to to separate them and put them int

Solution 1:

You need to loop through the string representation of the number and check if the number is a digit. If yes, then add it to the list. This can be done by using a list comprehension.

>>>pi = 3.141>>>[int(i) for i instr(pi) if i.isdigit()]
[3, 1, 4, 1]

Another way using Regex (Not - preffered)

>>>map(int,re.findall('\d',str(pi)))
[3, 1, 4, 1]

A final way - Brute force

>>>pi = 3.141>>>x = list(str(pi))>>>x.remove('.')>>>map(int,x)
[3, 1, 4, 1]

Few references from the docs

• str.isdigit
• re.findall
• map

The timeit results

python -m timeit "pi = 3.141;[int(i) for i in str(pi) if i.isdigit()]"100000 loops, best of 3: 2.56 usec per loop
python -m timeit "s = 3.141; list(map(int, str(s).replace('.','')))" # Avinash's Method
100000 loops, best of 3: 2.54 usec per loop
python -m timeit "import re;pi = 3.141; map(int,re.findall('\d',str(pi)))"100000 loops, best of 3: 5.72 usec per loop
python -m timeit "pi = 3.141; x = list(str(pi));x.remove('.'); map(int,x);"100000 loops, best of 3: 2.48 usec per loop

As you can see the brute force method is the fastest. The Regex answer as known is the slowest.

Solution 2:

You could also use string.replace along with map, list functions.

>>>s = 3.141>>>list(map(int, str(s).replace('.','')))
[3, 1, 4, 1]