Effacer les filtres
Effacer les filtres

Covert a string to an array of individual characters.

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Dimitrios Anagnostou
Dimitrios Anagnostou le 28 Mar 2023
Modifié(e) : Dyuman Joshi le 28 Mar 2023
I apologize if my question is trivial.
I have the following datasets.The letters in each dataset represent student names enrolled in each course :
Course1=['A','B','C','E','F','G','I','P','Q'];
Course2=['B','E','F','H','K','Q','R','S','T','U','V','Z'];
Course3=['C','E','G','H','J','K','O','Q','Z'];
So for example
Course1orCourse2 = union(Course1,Course2)
Course1orCourse2 = 'ABCEFGHIKPQRSTUVZ'
Course1andCourse2 = intersect(Course1,Course2)
Course1andCourse2 = 'BEFQ'
How can I get an output in the form ['A', 'B', 'C', 'E', 'F', 'G', 'H', 'I', 'K', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'Z'] and ['B', 'E', 'F', 'G']? Thank you very much?
  1 commentaire
Stephen23
Stephen23 le 28 Mar 2023
Modifié(e) : Stephen23 le 28 Mar 2023
Note that square brackets are a concatentation operator (not a list operator like in some other langauges), so your examples are just complicated ways of writing character vectors. For example, this code:
Course1 = ['A','B','C','E','F','G','I','P','Q'];
is just a longer way of writing this equivalent character vector:
Course1 = 'ABCEFGIPQ';
The same applies to your requested outputs, e.g. this:
['A', 'B', 'C', 'E', 'F', 'G', 'H', 'I', 'K', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'Z']
is exactly equivalent to this simple character vector:
'ABCEFGHIKPQRSTUVZ'
And that is already what UNION and INTERSECT are giving you: character vectors which consist of lots of individual characters. Understanding what square brackets actually do, and what character vectors are, is very important when using MATLAB:

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Dyuman Joshi
Dyuman Joshi le 28 Mar 2023
Modifié(e) : Dyuman Joshi le 28 Mar 2023
If you need every alphabet presented individually, you will have to define them as strings.
Either manually define each input with double inverted commas
Course1=["A","B","C","E","F","G","I","P","Q"]; %similarly for Course2 and Course3
or use compose on the existing data -
Course1=['A','B','C','E','F','G','I','P','Q'];
Course2=['B','E','F','H','K','Q','R','S','T','U','V','Z'];
Course3=['C','E','G','H','J','K','O','Q','Z'];
Course1=compose("%s",Course1')';
Course2=compose("%s",Course2')';
Course3=compose("%s",Course3')'
Course3 = 1×9 string array
"C" "E" "G" "H" "J" "K" "O" "Q" "Z"
Course1orCourse2 = union(Course1,Course2)
Course1orCourse2 = 1×17 string array
"A" "B" "C" "E" "F" "G" "H" "I" "K" "P" "Q" "R" "S" "T" "U" "V" "Z"
Course1andCourse2 = intersect(Course1,Course2)
Course1andCourse2 = 1×4 string array
"B" "E" "F" "Q"

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