Split Data by Ignoring Semicolon Inside Square Bracket in Python-CodePudding

I want to split the data by Semi Colon, But i need to ignore the semi colon inside the Square Bracket.


val = "L=1nH;Tol=0.3nH;Idc=1000mA;Isat=NA;Iacpp=NA;Fr=10GHz;ESR=0.10ohm;Q=[100MegHz 7; 500MegHz 23;1.8GHz 47;2.4GHz 57];dL@Isat=NA;dT@Idc=NA"
val.split(";")

CodePudding user response：

Using re.findall we can try:

val = "L=1nH;Tol=0.3nH;Idc=1000mA;Isat=NA;Iacpp=NA;Fr=10GHz;ESR=0.10ohm;Q=[100MegHz 7; 500MegHz 23;1.8GHz 47;2.4GHz 57];dL@Isat=NA;dT@Idc=NA"
parts = re.findall(r'\w =\[.*?\]|\w =[^;] ', val)
print(parts)

The regex pattern used here says to match:

\w =\[.*?\]  a key = a term in [...]
|            OR
\w =[^;]     a key = some other term

The regex alternation places the key=[...] variant first, so that we avoid trying to match to semicolons internal to the bracketed value.

CodePudding user response：

You could split on capturing the data from an opening till closing square bracket in a capture group to keep it in the output, or split on ; using an alternation | in the pattern.

The pattern matches:

( Capture group 1 (keep after split)
- \[[^][]*] Match from [...]
) Close group 1
| Or
; Match a semicolon

Example

import re

val = "L=1nH;Tol=0.3nH;Idc=1000mA;Isat=NA;Iacpp=NA;Fr=10GHz;ESR=0.10ohm;Q=[100MegHz 7; 500MegHz 23;1.8GHz 47;2.4GHz 57];dL@Isat=NA;dT@Idc=NA"
pattern = r"(\[[^][]*])|;"
result = [s for s in  re.split(pattern, val) if s]
print(result)

Output

['L=1nH', 'Tol=0.3nH', 'Idc=1000mA', 'Isat=NA', 'Iacpp=NA', 'Fr=10GHz', 'ESR=0.10ohm', 'Q=', '[100MegHz 7; 500MegHz 23;1.8GHz 47;2.4GHz 57]', 'dL@Isat=NA', 'dT@Idc=NA']