Using logic to select specific values in a matrix

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Kristine
Kristine am 22 Nov. 2025
Kommentiert: Stephen23 am 23 Nov. 2025
Hi y'all,
I am trying to create a second table from my original table based on taking all the values in column B, that fulfill the requirement for column A.
Example data:
Edit: attached file
I want to create a second table that only includes row 2 as long as the first row = 1.
This is what I tried:
A = SmallerSample
idnt = A(:,1) == 1 ;
new = A(ind1, 2);
The error I get:
Error using ()
A table row subscript must be a numeric array containing real positive
integers, a logical array, a character vector, a string array, a cell array of
character vectors, or a pattern scalar.
Error in Graph_Bead_Distributions
new = A(ind1, 2);

Akzeptierte Antwort

Walter Roberson
Walter Roberson am 22 Nov. 2025
A = SmallerSample
assuming that A is a table
idnt = A(:,1) == 1 ;
() subscripting of a table yields a table.
== has, somewhat recently, become defined for tables -- and the result is a table.
So the idnt variable is going to be a table.
new = A(idnt, 2);
You are then trying to index A using a table as the first subscript.
You should have used
idnt = A.(1) == 1;
or
idnt = A{:,1} == 1;

Weitere Antworten (1)

John D'Errico
John D'Errico am 22 Nov. 2025
Bearbeitet: John D'Errico am 22 Nov. 2025
Well, you are doomed to failure, as long as you don't use the right variable names as you created them!
A = [1, 0.1234
1, 0.1345
1, 0.1456
2, 0.1567
2, 0.1678
2, 0.1789];
idnt = A(:,1) == 1 ;
You defined the variable idnt. But then you tried to use idn1. Which is a DIFFERENT variable name. Apparently you also had the variable idn1 defined in your workspace, as otherwise, you would have gotten a different error.
new = A(idnt, 2)
new = 3×1
0.1234 0.1345 0.1456
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But using the name you defined, it does indeed work. So you were not really doomed to fail. You just need to be more careful in your code. Look more carefully at the error message next time, as it suggested there was problem with the variable ind1, which is not in your workspace as a valid variable to use as an index.
  2 Kommentare
Kristine
Kristine am 22 Nov. 2025
Bearbeitet: Kristine am 22 Nov. 2025
I did misswrite it, but in my workspace both variables were equivalent, so I still get an error once correcting the mistake:
>> Graph_Bead_Distributions
Error using () (line 132)
A table row subscript must be a numeric array containing real positive integers,
a logical array, a character vector, a string array, a cell array of character
vectors, or a pattern scalar.
Error in Graph_Bead_Distributions (line 26)
new = A(idnt, 2);
Kristine
Kristine am 22 Nov. 2025
I think the example table I provided is not as helpful. The actual table I am using is where the issue lies.

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