フィルターのクリア

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

3 ビュー (過去 30 日間)
Cai Chin
Cai Chin 2020 年 12 月 30 日
回答済み: Image Analyst 2020 年 12 月 30 日
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.

採用された回答

Ameer Hamza
Ameer Hamza 2020 年 12 月 30 日
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))

その他の回答 (1 件)

Image Analyst
Image Analyst 2020 年 12 月 30 日
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

カテゴリ

Help Center および File ExchangeImage Data についてさらに検索

タグ

Community Treasure Hunt

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

Start Hunting!

Translated by