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How do I replace elements in a vector with other vectors?

조회 수: 1(최근 30일)
Cai Chin
Cai Chin 2020년 12월 30일
답변: Image Analyst 2020년 12월 30일
Hi, I am currently trying to replace the elements in a vector called 'seq1' with other vectors. seq1 has 12 numbers ranging from 1 to 9 and I am trying to replace each number with vectors of other numbers as follows:
seq1 = [1 2 3 4 5 6 7 8 9 2 4 5];
rectColor1 = [0.1 0.1 0.1];
rectColor2 = [0.15 0.15 0.15];
rectColor3 = [0.22 0.22 0.22];
rectColor4 = [0.30 0.30 0.30];
rectColor5 = [0.40 0.40 0.40];
rectColor6 = [0.55 0.55 0.55];
rectColor7 = [0.7 0.7 0.7];
rectColor8 = [0.85 0.85 0.85];
rectColor9 = [1 1 1];
for i = 1:numel(seq1)
if seq1(i) == 1
seq1(i) = rectColor1;
elseif seq1(i) == 2
seq1(i) = rectColor2;
elseif seq1(i) == 3
seq1(i) = rectColor3;
elseif seq1(i) == 4
seq1(i) = rectColor4;
elseif seq1(i) == 5
seq1(i) = rectColor5;
elseif seq1(i) == 6
seq1(i) = rectColor6;
elseif seq1(i) == 7
seq1(i) = rectColor7;
elseif seq1(i) == 8
seq1(i) = rectColor8;
elseif seq1(i) == 9
seq1(i) = rectColor9;
end
end
However, I keep getting the following error:
Unable to perform assignment because the left and right sides have a different number of elements.
How do I resolve this issue? Thanks in advance.

채택된 답변

Ameer Hamza
Ameer Hamza 2020년 12월 30일
Don't create variable name dynamically like rectColor1, rectColor2, .. It always makes code more difficult. Following shows how you can easily solve the problem using a cell array
seq1 = [1 2 3 4 5 6 7 8 9 2 4 5];
rectColor{1} = [0.1 0.1 0.1];
rectColor{2} = [0.15 0.15 0.15];
rectColor{3} = [0.22 0.22 0.22];
rectColor{4} = [0.30 0.30 0.30];
rectColor{5} = [0.40 0.40 0.40];
rectColor{6} = [0.55 0.55 0.55];
rectColor{7} = [0.7 0.7 0.7];
rectColor{8} = [0.85 0.85 0.85];
rectColor{9} = [1 1 1];
seq1 = cell2mat(rectColor(seq1))

추가 답변(1개)

Image Analyst
Image Analyst 2020년 12월 30일
Perhaps this is what you want, where there are 3 rows in the output where each column has the color identified by seq1:
seq1 = [1 2 3 4 5 6 7 8 9 2 4 5]; % 1-by-12
% Define 9 different colors:
rectColor = zeros(3, 9); % Initialize to 3x9 shape.
rectColor(:, 1) = [0.1 0.1 0.1]';
rectColor(:, 2) = [0.15 0.15 0.15]';
rectColor(:, 3) = [0.22 0.22 0.22]';
rectColor(:, 4) = [0.30 0.30 0.30]';
rectColor(:, 5) = [0.40 0.40 0.40]';
rectColor(:, 6) = [0.55 0.55 0.55]';
rectColor(:, 7) = [0.7 0.7 0.7]';
rectColor(:, 8) = [0.85 0.85 0.85]';
rectColor(:, 9) = [1 1 1]';
output = zeros(3, length(seq1)); % Initialize to proper shape, 3-by-12.
% For each column paste in the proper color:
output = rectColor(:, seq1)
You get a 3-by-12 matrix where each column is a color:
output =
Columns 1 through 9
0.1 0.15 0.22 0.3 0.4 0.55 0.7 0.85 1
0.1 0.15 0.22 0.3 0.4 0.55 0.7 0.85 1
0.1 0.15 0.22 0.3 0.4 0.55 0.7 0.85 1
Columns 10 through 12
0.15 0.3 0.4
0.15 0.3 0.4
0.15 0.3 0.4

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