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

446 Solutions
163 Solvers

Last Solution submitted on Aug 06, 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.

Solution Comments

Solution 1180413

1 Comment

1 Comment

yurenchu on 5 May 2017

This works under the presumption that IND is already sorted (as is the case in the test suite).

Solution 193625

1 Comment

1 Comment

yurenchu on 5 May 2017

Nice trick! But it breaks when IND contains 0s (inserting rows at the top of the matrix).

Problem Recent Solvers163

Problem Tags

matrix manipulation

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

Solution Stats

Last 200 Solutions

Problem Comments

Solution Comments

Problem Recent Solvers163

Suggested Problems

More from this Author1

Problem Tags

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

Solution Stats

Last 200 Solutions

Problem Comments

Solution Comments

Problem Recent Solvers163

Suggested Problems

More from this Author1

Problem Tags

Players