Cody

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

Solution 2413567

Submitted on 29 May 2020 by Siân Chilcott
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### Test Suite

Test Status Code Input and Output
1   Pass
A = 1; IND = 1; b = 1; y_correct = [1;1]; assert(isequal(insertRows(A, IND, b), y_correct))

y = 1 1 sa_ind = 1 1 sort_it = 1 1 A_b_Index = 1 2 y = 1 1

2   Pass
A = [0 0; 1 1]; IND = [1 1 1 2]; b = NaN; y_correct = [0 0;NaN NaN;NaN NaN;NaN NaN;1 1;NaN NaN]; assert(isequalwithequalnans(insertRows(A, IND, b), y_correct))

b = NaN NaN NaN NaN b = NaN NaN NaN NaN NaN NaN NaN NaN y = 0 0 1 1 NaN NaN NaN NaN NaN NaN NaN NaN sa_ind = 1 2 1 1 1 2 sort_it = 1 1 1 1 2 2 A_b_Index = 1 3 4 5 2 6 y = 0 0 NaN NaN NaN NaN NaN NaN 1 1 NaN NaN

3   Pass
A = [1 1; 3 3; 4 4]; IND = [1 3]; b = [2 2; 5 5]; y_correct = [1 1;2 2;3 3;4 4;5 5]; assert(isequal(insertRows(A, IND, b), y_correct))

y = 1 1 3 3 4 4 2 2 5 5 sa_ind = 1 2 3 1 3 sort_it = 1 1 2 3 3 A_b_Index = 1 4 2 3 5 y = 1 1 2 2 3 3 4 4 5 5

4   Pass
A = (1:2:10)'; IND = 1:5; b = (2:2:10)'; y_correct = (1:10)'; assert(isequal(insertRows(A, IND, b), y_correct))

y = 1 3 5 7 9 2 4 6 8 10 sa_ind = 1 2 3 4 5 1 2 3 4 5 sort_it = 1 1 2 2 3 3 4 4 5 5 A_b_Index = 1 6 2 7 3 8 4 9 5 10 y = 1 2 3 4 5 6 7 8 9 10

5   Pass
A = zeros(10,5); IND = [3,7]; b = rand(2,5); y_correct = [zeros(3,5); b(1,:); zeros(4,5); b(2,:); zeros(3,5)]; assert(isequal(insertRows(A, IND, b), y_correct))

y = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.7704 0.9485 0.1684 0.7429 0.6909 0.5289 0.8512 0.8597 0.8871 0.2723 sa_ind = 1 2 3 4 5 6 7 8 9 10 3 7 sort_it = 1 2 3 3 4 5 6 7 7 8 9 10 A_b_Index = 1 2 3 11 4 5 6 7 12 8 9 10 y = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.7704 0.9485 0.1684 0.7429 0.6909 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.5289 0.8512 0.8597 0.8871 0.2723 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

6   Pass
assert(isempty(strfind(evalc('type insertRows'), 'regexp')));