give labels according to string

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Elysi Cochin
Elysi Cochin am 20 Nov. 2019
Kommentiert: Star Strider am 20 Nov. 2019
I have a cell array of strings with the following pattern
String Label
'Abc\a1\L\XYZ1R08' 1
'Abc\a1\R\XYZ1R09' 1
'Abc\a1\R\XYZ1R10' 1
'Abc\b2\L\XYZ2L01' 2
'Abc\b2\R\XYZ2L02' 2
'Abc\b2\R\XYZ2L03' 2
'Abc\c3\L\XYZ2L04' 3
'Abc\c3\R\XYZ2L05' 3
'Abc\d4\L\XYZ2L06' 4
'Abc\d4\R\XYZ2L07' 4
i wanted to give a new variable, "Label', values according to a1,b2,c3,d4, etc
  3 Kommentare
Rik
Rik am 20 Nov. 2019
Is that label guaranteed to be between the first two slashes?
Elysi Cochin
Elysi Cochin am 20 Nov. 2019
Bearbeitet: Elysi Cochin am 20 Nov. 2019
yes sir, based on the first two slashes
find the different elements, and give same value for same elements
(a1, b2, c3, d4 - the names can change. This is just an example)
in my example i have 4 different values a1, b2, c3, d4 (it can be greater than 4 also)
So where all i have a1, i want to give 1 , for next value (b2) next value 2 and so on
% its a cell array
my_string = {
'Abc\a1\L\XYZ1R08';
'Abc\a1\R\XYZ1R09';
'Abc\a1\R\XYZ1R10';
'Abc\b2\L\XYZ2L01';
'Abc\b2\R\XYZ2L02';
'Abc\b2\R\XYZ2L03';
'Abc\c3\L\XYZ2L04';
'Abc\c3\R\XYZ2L05';
'Abc\d4\L\XYZ2L06';
'Abc\d4\R\XYZ2L07'};

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Star Strider
Star Strider am 20 Nov. 2019
Try this:
String = {'Abc\a1\L\XYZ1R08'
'Abc\a1\R\XYZ1R09'
'Abc\a1\R\XYZ1R10'
'Abc\b2\L\XYZ2L01'
'Abc\b2\R\XYZ2L02'
'Abc\b2\R\XYZ2L03'
'Abc\c3\L\XYZ2L04'
'Abc\c3\R\XYZ2L05'
'Abc\d4\L\XYZ2L06'
'Abc\d4\R\XYZ2L07'};
grpstr = regexp(String, '(?<=\\)\w{2}', 'once','match'); % Get Two Letters After First ‘\’
Group = findgroups(grpstr);
Result = {String, Group} % Cell Array
T1 = table(string(String), Group) % Table
producing:
T1 =
10×2 table
Var1 Group
__________________ _____
"Abc\a1\L\XYZ1R08" 1
"Abc\a1\R\XYZ1R09" 1
"Abc\a1\R\XYZ1R10" 1
"Abc\b2\L\XYZ2L01" 2
"Abc\b2\R\XYZ2L02" 2
"Abc\b2\R\XYZ2L03" 2
"Abc\c3\L\XYZ2L04" 3
"Abc\c3\R\XYZ2L05" 3
"Abc\d4\L\XYZ2L06" 4
"Abc\d4\R\XYZ2L07" 4
  5 Kommentare
Elysi Cochin
Elysi Cochin am 20 Nov. 2019
Star Strider
Star Strider am 20 Nov. 2019
As always, my pleasure! (And our pleasure!)

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