Substitue values with text

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Leonor Vieira Dias
Leonor Vieira Dias 2021 年 1 月 29 日
コメント済み: Leonor Vieira Dias 2021 年 1 月 29 日
Hello,
I have a table of data. I wanted to create a column that would have written "early" or "late" according to the difference between two values.
So far I have been able to differentiate the values to get 1 or 0 whether the condition was met or not. I wanted to have text instead.
Can we subsitute 1 by "late" and 0 by "early"? Is there an easier way to do this, e.g. using if loop?
Many thanks
T.Final = (T.Early<T.Delay)

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Iuliu Ardelean
Iuliu Ardelean 2021 年 1 月 29 日
編集済み: Iuliu Ardelean 2021 年 1 月 29 日
Maybe this works for you:
early = [10 20 30];
delay = [11 19 31];
final = strings(1, 3); % create an empty string array with 1 line and 3 columns
final(early<delay) = "late";
final(early>=delay) = "early";
  3 件のコメント
Leonor Vieira Dias
Leonor Vieira Dias 2021 年 1 月 29 日
I got it, I forgot to put a T in front of it. Thank you very much!
Iuliu Ardelean
Iuliu Ardelean 2021 年 1 月 29 日
awesome

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その他の回答 (1 件)

dpb
dpb 2021 年 1 月 29 日
% create sample data table
T=table(randi([30,70],10,1)+rand(10,1),'VariableNames',{'Early'});
T.Final=T.Early+(randi([-20,20],10,1)+rand(10,1));
% the desired conditional categorical variable
STR=categorical(["Early";"Late"]);
% the engine
T.Final=STR((T.Early<T.Delay)+1);
A given realization of the table data here resulted in:
>> T.Final=STR((T.Early<T.Delay)+1)
T =
10×3 table
Early Delay Final
______ ______ _____
40.327 26.39 Early
68.584 58.236 Early
69.197 75.902 Late
37.909 25.636 Early
36.492 46.388 Late
33.04 35.818 Late
58.207 51.744 Early
69.131 74.646 Late
58.487 69.468 Late
69.074 79.39 Late
>>
  1 件のコメント
Leonor Vieira Dias
Leonor Vieira Dias 2021 年 1 月 29 日
Thank you!

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