How to get rows with all similar columns and adjust matrix with shorter length to that of longer length

Question

Sachin Hegde 2023-10-12

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2032534-how-to-get-rows-with-all-similar-columns-and-adjust-matrix-with-shorter-length-to-that-of-longer-len

编辑： Dyuman Joshi 2023-10-12

Hello,

I have two example arrays here (Time vectors in format [Y M D H M S]) as follows (my original data has 73200 sample points)

A = [2023 6 29 7 8 9; 2023 6 29 7 8 10; 2023 6 29 7 8 11; 2023 6 29 7 8 12; 2023 6 29 7 8 18; 2023 6 29 7 8 19; 2023 6 29 7 8 20; 2023 6 29 7 8 21; 2023 6 29 7 8 22; 2023 6 29 7 8 23; 2023 6 29 7 8 24]

B = [2023 6 29 7 8 22.5; 2023 6 29 7 8 23; 2023 6 29 7 8 24]

And a data vector for matrix B:

B_data = [12 21 21] (To note, A matrix also has a data vector but it is not the part of the problem)

What i am planning to do is to is to allign both the vectors to get same length i.e. to fill in the missing time data in B and to make it same length as A.

Here was my effort: I tried to find the rows with common columns and for these rows i kept the origianl data of B, and for the missing rows i added those from A.

I used (ismember(B,A,rows)) to get the index.

What is confusing to me is that when i use (ismember(B,A,rows)) and (ismember(A,B,rows)), i am not getting the same number of elements.

Because of this the end result i want is not correct. Can someon help me out here. I thank you in advance.

2 个评论
显示无隐藏无

Dyuman Joshi 2023-10-12

编辑：Dyuman Joshi 2023-10-12

"Here was my effort: I tried to find the rows with common columns and for these rows i kept the origianl data of B, and for the missing rows i added those from A."

"Because of this the end result i want is not correct."

You are assuming that A and B will exactly have size(A,1)-size(B,1) different rows.

What if A and B have less than size(A,1)-size(B,1) different rows? or more than that? What should be the output then?

Given the data, what is the expected result? And what is the logic/criteria behind achieveing that result?

Edit - Changed the incorrect numel(x) to size(x,1) for number of rows

Sachin Hegde 2023-10-12

Hi,

Let me add some more information, There are no repeated rows in A , and the same goes for B. This eliminates the different rows being more than numel(A) - numel (B).

Also i am not assuming different rows = numel(A) - numel (B). This can be seen in the example as well. In th elast column, A has a value of 22 and B 22.5, and because of this the number of different rows = 9 ~= numel(A) - numel(B).

In this case i would still keep with A. For any row of A which is not in B shall be replaced with that of A, and any extra row of B which is not in A shall be deleted in order to make it same length as A.

i hope this clears up the problem even more.

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Sulaymon Eshkabilov 2023-10-12

在 MATLAB Online 中打开

If understood correctly or presuming it :), is this what you are try to get:

A = [2023	6	29	7	8	9; 
    2023	6	29	7	8	10;
2023	6	29	7	8	11; 
2023	6	29	7	8	12; 
2023	6	29	7	8	18; 
2023	6	29	7	8	19; 
2023	6	29	7	8	20; 
2023	6	29	7	8	21; 
2023	6	29	7	8	22; 
2023	6	29	7	8	23; 
2023	6	29	7	8	24]
A = 11×6
        2023           6          29           7           8           9
        2023           6          29           7           8          10
        2023           6          29           7           8          11
        2023           6          29           7           8          12
        2023           6          29           7           8          18
        2023           6          29           7           8          19
        2023           6          29           7           8          20
        2023           6          29           7           8          21
        2023           6          29           7           8          22
        2023           6          29           7           8          23
B = [2023	6	29	7	8	22.5; 
    2023	6	29	7	8	23; 
    2023	6	29	7	8	24]
B = 3×6
1.0e+03 *

    2.0230    0.0060    0.0290    0.0070    0.0080    0.0225
    2.0230    0.0060    0.0290    0.0070    0.0080    0.0230
    2.0230    0.0060    0.0290    0.0070    0.0080    0.0240
ID_miss=ismember(A,B,'rows')         % This is what finds out what rows are missing in B
ID_miss = 11×1 logical array
   0
   0
   0
   0
   0
   0
   0
   0
   0
   1
   1
IDX = find(ID_miss==0);
B(IDX, :)=A(IDX,:)                   % Fills up with the data in A like equating B to A      
B = 9×6
        2023           6          29           7           8           9
        2023           6          29           7           8          10
        2023           6          29           7           8          11
        2023           6          29           7           8          12
        2023           6          29           7           8          18
        2023           6          29           7           8          19
        2023           6          29           7           8          20
        2023           6          29           7           8          21
        2023           6          29           7           8          22

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Dyuman Joshi 2023-10-12

The size of the final output should be the same as the size of A.

"What i am planning to do is to is to allign both the vectors to get same length i.e. to fill in the missing time data in B and to make it same length as A."

请先登录，再进行评论。