Problem 796. Removing rows from a matrix is easy - but what about inserting rows?

Created by Dirk Engel

Solve Later

{"relationships":[{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/document","targetMode":"","relationshipId":"rId1","target":"/matlab/document.xml"},{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/output","targetMode":"","relationshipId":"rId2","target":"/matlab/output.xml"}],"parts":[{"partUri":"/matlab/document.xml","relationship":[],"contentType":"application/vnd.mathworks.matlab.code.document+xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n<w:document xmlns:w=\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\"><w:body><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>Assume A is a 5-by-5 matrix. A([2,4],:) = [] is a quick way to remove rows 2 and 4. Can you find a quick way to insert rows into A?</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>You are given a n-by-m matrix A, a 1-by-p vector of indices IND, and either a scalar value (1-by-1) or a p-by-m matrix b to be inserted AFTER rows IND of A. In case of a scalar value it should fill the whole row. Note that IND may contain duplicate indices, which means that more than one row should be inserted after that specific row index.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>For example:</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"code\"/></w:pPr><w:r><w:t><![CDATA[   A = [1 1; \n        3 3; \n        4 4];\n   IND = [1 3];\n   b = [2 2; \n        5 5];]]></w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>should become</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"code\"/></w:pPr><w:r><w:t><![CDATA[   y = [1 1; \n        2 2; \n        3 3; \n        4 4; \n        5 5];]]></w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>And</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"code\"/></w:pPr><w:r><w:t><![CDATA[   A = [0 0; \n        1 1];\n   IND = [1 1 2];\n   b = NaN;]]></w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>should become</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"code\"/></w:pPr><w:r><w:t><![CDATA[   y = [0   0; \n        NaN NaN; \n        NaN NaN; \n        1 1; \n        NaN NaN];]]></w:t></w:r></w:p></w:body></w:document>"},{"partUri":"/matlab/output.xml","contentType":"text/xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"no\" ?><embeddedOutputs><metaData><evaluationState>manual</evaluationState><layoutState>code</layoutState><outputStatus>ready</outputStatus></metaData><outputArray type=\"array\"/><regionArray type=\"array\"/></embeddedOutputs>"}]}

Solve

Solution Stats

481 Solutions
127 Solvers

Last Solution submitted on Dec 16, 2020

Last 200 Solutions

Problem Comments

3 Comments

3 Comments

Sean de Wolski on 27 Jun 2012

Many of the test cases are wrong.

For example:

A = (1:2:10)';
IND = 1:5;
b = (2:2:10)';
y_correct = (1:10)';

y_correct should be [2:2:10,1:2:10]'

The zeros one has zeros that are the wrong size etc.

Dirk Engel on 27 Jun 2012

I corrected one error regarding isequal with NaNs, but all the other test cases are correct.
A = [1 3 5 7 9]; IND = [1 2 3 4 5]; b = [2 4 6 8 10]; definitely should result in [1 2 3 4 5 6 7 8 9 10]! You have to insert b(1) after A(1), b(2) after A(2) and so on.

Christian Schröder on 9 Jul 2019

Nice problem, but the specification could be improved: from the description, you'd think the entire matrix b is supposed to be inserted after each row in IND; it's only the first example that makes it clear that this is not the case.