# How to know the date to which a group (findgroups) corresponds?

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Angelavtc 2020 年 9 月 6 日

Hello!
I have a timetable (TT) where one column is a Date (dd/mm/yyyy format) and another is an Hour, I applied the following code to classify them by groups:
groups_TT = findgroups(TT.Date, TT.Hour);
This gives me a unique number for each day and hour. My question is, how can I get an identifier from this number that tells me the Date it corresponds to? The unique number is not as useful for me to perform my analysis. I need to know the date that this number corresponds to.
I attach a small example. Here my code will give me a unique identifier (from 1 to 6) and I would like to create an additional column that tells me what date each one corresponds to (if possible another table with two columns: unique id and date)
Angela
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KSSV 2020 年 9 月 6 日

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### 採用された回答

Adam Danz 2020 年 9 月 6 日

findgroups()
Use the 2nd output [G,ID] = findgroups(A)
G(i) belongs to ID(G(i),:). For example,
A = ["D" "B" "B" "A" "D" "D" "D"];
[G,ID] = findgroups(A)
% G =
% 3 2 2 1 3 3 3
% ID =
% 1×3 string array
% "A" "B" "D"
So, G(1) shows that A(1) is in group #3. Group #3 is ID(3) or ID(G(1)) which is "D".
To find all samples in your data that belong to a specific group,
isGroupMember = G==find(ID=="D"); % or strcmp(ID,"D")
% or, G==3
isGroupMember is a logical vector the same size as A and contains 1s (True) where A=="D".
unique()
Note that [ID,~,G] = unique(A) produces the same outputs (but G is a column vector)!
Use [ID,~,G] = unique(A,'rows') when A is a matrix and the columns contain the grouping variables.
% ID =
% 1×3 string array
% "A" "B" "D"
% G =
% 3
% 2
% 2
% 1
% 3
% 3
% 3

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