Problem 2953. Split array into pieces according to corresponding array

Created by Michael Haderlein

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>An array x of length n has values 1...m with each value appearing at least once. The values are sorted (n&gt;=m).</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>A second array y of the same size should be split into elements of a cell array c such that the first element of c contains all elements of y at the positions where x=1 and so on:</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"code\"/></w:pPr><w:r><w:t><![CDATA[x=   [ 1  1  1  2  2  3  3  3  3];\ny=   [10 52 12 58 21 43 89 27 98];\nc{1}=[10 52 12];\nc{2}=[         58 21];\nc{3}=[               43 89 27 98];]]></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

70 Solutions
39 Solvers

Last Solution submitted on Sep 11, 2020

Last 200 Solutions

Problem Comments

Problem Recent Solvers39

Problem Tags

Community Treasure Hunt

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

Start Hunting!