regex expression to include optional decimals

Asked Jan 31 '20 at 19:24

Active Jan 31 '20 at 19:24

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I'm trying to write a python script using regex that will take a string such as this one:

string = "C:66:66:0:0%:1E0 C:18:18:0:0%:1E0 C:26:26:0:0%:1E0 C:95:95:0:0%:1E0 Y:106:104:2:1.89%:2.4882E-1

and break it up into chunks such as this

C:66:66:0:0%:1E0

The code I have right now is:

res = re.findall("\w+:\d+:\d+:\d+:\d+%:\w+", string)

but the output leaves out everything following the "Y" due to the decimals

['C:66:66:0:0%:1E0', 'C:18:18:0:0%:1E0', 'C:26:26:0:0%:1E0', 'C:95:95:0:0%:1E0']

Does someone know how the regex can be modified so that it includes optional decimals in the string? Thank you!

asked Jan 31 '20 at 19:24

bleep

What is the expected output? – Wiktor Stribiżew Jan 31 '20 at 19:28
I need it to include other letters as well, such as the "Y:106:104:2:1.89%:2.4882E-1" part of the string – bleep Jan 31 '20 at 19:29
Then why not use `s.split()`? What are the pattern requirements (rules)? – Wiktor Stribiżew Jan 31 '20 at 19:29
I was definitely making this more complicated than it needs to be. Thank you! – bleep Jan 31 '20 at 19:32

0 Answers0