# Creating a new column in a table which is true or false

조회 수: 2(최근 30일)
Luca 2021년 5월 6일
댓글: Luca 2021년 5월 9일
Hi,
I have to tables. One is called subsetx the other one is called subsety. In subsetx are two columns one is called ID and the other one SIC. It looks like this:
ID SIC
10000 3990
10000 3990
.
.
.
10001 4920
.
.
.
10002 6020
In subsety i have two rows with value 3990 4920.
I would like to create a new column in subsetx of logical type (true or false)
If variable SIC matches one of the numbers in subsety I would like to get a 1 if not i would like to get a 0.
The proble is you cant compare table with diffrent numbers of row.
I did attached the two subsets.
How can I solve this problem ?

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### 채택된 답변

Steven Lord 2021년 5월 6일
rng default
id = (1:10).';
SIC = randi(3, 10, 1);
t = table(id, SIC)
t = 10×2 table
id SIC __ ___ 1 3 2 3 3 1 4 3 5 2 6 1 7 1 8 2 9 3 10 3
t.is1or2 = ismember(t.SIC, [1 2])
t = 10×3 table
id SIC is1or2 __ ___ ______ 1 3 false 2 3 false 3 1 true 4 3 false 5 2 true 6 1 true 7 1 true 8 2 true 9 3 false 10 3 false
##### 댓글 수: 1표시숨기기 없음
Luca 2021년 5월 9일
Thank you very much it worked.

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### 추가 답변(1개)

David Fletcher 2021년 5월 6일
편집: David Fletcher 2021년 5월 6일
vals=subsety{:,1};
subsetx{:,3}=subsetx.SIC==vals(2,1)|subsetx.SIC==vals(1,1);
##### 댓글 수: 1표시숨기기 없음
Luca 2021년 5월 9일
Thank you it worked

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R2021a

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