setdiff for two matrices

7 Ansichten (letzte 30 Tage)
Mahmoud Zeydabadinezhad
Mahmoud Zeydabadinezhad am 18 Mär. 2016
Kommentiert: Shuhao Cao am 7 Mai 2020
Hi,
I have two large matrices with the same number of rows but a different number of columns. Let's say A is the larger matrix and B is the smaller one.
Values in each row of the matrix B are a subset of the values of the corresponding row in A. The goal is to have another matrix C which in each row contains the values from the corresponding row of A which are not in corresponding row of B. In other words, I want to use the setdiff(A,B) command, but it should act row by row. Any idea how can I do this without looping?
Thank you!

Antworten (2)

Guillaume
Guillaume am 18 Mär. 2016
Bearbeitet: Guillaume am 18 Mär. 2016
I'm afraid there's no other way than looping over the rows either explicitly with a for loop, or with cellfun:
cell2mat(cellfun(@setdiff, num2cell(A, 2), num2cell(B, 2), 'UniformOutput', false))
Note that the above assumes that setdiff returns the same number of elements for each row. Otherwise, you'll have to get rid of the cell2mat call and keep the result as a cell array.
Also, note that an explicit loop may be faster as you wouldn't have to split the inputs into cell arrays of rows.
  1 Kommentar
Mahmoud Zeydabadinezhad
Mahmoud Zeydabadinezhad am 18 Mär. 2016
That's unfortunate as looping over for millions of times takes absolute ages to complete.

Melden Sie sich an, um zu kommentieren.


Fangjun Jiang
Fangjun Jiang am 18 Mär. 2016
setdiff(a,b,'rows') ??
  4 Kommentare
Fangjun Jiang
Fangjun Jiang am 24 Apr. 2018
just use setdiff(A(k,:), B(k,:)) and loops through all rows. What is the point to struggle to avoid a loop?
Shuhao Cao
Shuhao Cao am 7 Mai 2020
Because loop is slow for large matrices (especially the sparse matrices).

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Matrix Indexing finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by