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resize and fill table

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Sanley Guerrier
Sanley Guerrier 2023 年 7 月 17 日
コメント済み: Voss 2023 年 7 月 17 日
Hello, How can I resize the table, I mean fill out missing values in column A so that the pattern repeat itself. For example, 0 1 2 3 0 1 2 3.........
And then replace the corresponding rows to these values by NaN in column B and C. After resizind it, the table will have 97 rows it currently has 86 rows.
Thank you!

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Voss
Voss 2023 年 7 月 17 日
編集済み: Voss 2023 年 7 月 17 日
Ran in:
Here's one way, using a table:
t = readtable('file.xlsx');
disp(t);
QHR A B ___ _____ _____ 0 -6.67 -6.01 1 -6.73 -6.03 2 -6.71 -6.05 3 -6.73 -6.03 0 -6.8 -6.01 1 -6.86 -6.14 2 -6.97 -6.27 3 -7.13 -6.36 1 -7.13 -6.46 2 -7.09 -6.44 3 -7.12 -6.52 0 -7.12 -6.49 1 -7.09 -6.57 2 -7.09 -6.55 3 -7.07 -6.51 1 -7.02 -6.52 2 -6.93 -6.48 3 -6.86 -6.41 0 -6.82 -6.43 1 -6.82 -6.43 2 -6.79 -6.33 3 -6.78 -6.36 1 -6.78 -6.3 2 -6.74 -6.35 3 -6.75 -6.33 0 -6.7 -6.25 1 -6.67 -6.28 2 -6.58 -6.22 3 -6.54 -6.18 1 -6.48 -6.15 2 -6.33 -6.04 0 -6.25 -5.94 1 -6.15 -5.92 2 -6.02 -5.76 3 -5.93 -5.73 0 -5.57 -5.47 1 -5.43 -5.28 2 -5.08 -5.16 3 -4.81 -4.92 0 -4.68 -4.73 1 -4.52 -4.55 2 -4.47 -4.5 3 -4.42 -4.42 1 -4.21 -4.27 2 -4.01 -4.11 3 -4.03 -4.08 0 -4.06 -4.02 1 -4.05 -4.03 2 -4.01 -4.04 3 -3.98 -3.93 1 -4.05 -3.99 2 -3.87 -3.91 3 -3.88 -3.85 0 -3.97 -3.82 1 -3.91 -3.86 2 -4.03 -3.91 3 -4.05 -3.93 1 -4.31 -4.07 2 -4.23 -4.05 3 -4.25 -4.12 0 -4.29 -4.1 1 -4.29 -4.15 2 -4.23 -4.15 3 -4.25 -4.1 1 -4.25 -4.09 2 -4.13 -4.03 3 -4.1 -4.02 0 -4.06 -3.94 1 -4.12 -3.97 2 -4.03 -3.96 0 -3.96 -3.97 1 -3.92 -3.84 2 -3.95 -3.91 3 -3.99 -3.84 0 -3.95 -3.82 1 -3.93 -3.84 2 -3.87 -3.89 3 -3.84 -3.85 1 -3.85 -3.77 2 -3.78 -3.72 3 -3.81 -3.72 0 -3.76 -3.71 1 -3.77 -3.74 2 -3.74 -3.7 3 -3.69 -3.72
[N,M] = size(t);
ii = 2;
single_row_table = array2table(NaN(1,M),'VariableNames',t.Properties.VariableNames);
while ii <= N
if t{ii,1}-t{ii-1,1} ~= 1 && (t{ii,1} ~= 0 || t{ii-1,1} ~= 3)
single_row_table{1,1} = mod(t{ii-1,1}+1,4);
t = [t(1:ii-1,:); ...
single_row_table; ...
t(ii:end,:)];
N = N+1;
end
ii = ii+1;
end
disp(t);
QHR A B ___ _____ _____ 0 -6.67 -6.01 1 -6.73 -6.03 2 -6.71 -6.05 3 -6.73 -6.03 0 -6.8 -6.01 1 -6.86 -6.14 2 -6.97 -6.27 3 -7.13 -6.36 0 NaN NaN 1 -7.13 -6.46 2 -7.09 -6.44 3 -7.12 -6.52 0 -7.12 -6.49 1 -7.09 -6.57 2 -7.09 -6.55 3 -7.07 -6.51 0 NaN NaN 1 -7.02 -6.52 2 -6.93 -6.48 3 -6.86 -6.41 0 -6.82 -6.43 1 -6.82 -6.43 2 -6.79 -6.33 3 -6.78 -6.36 0 NaN NaN 1 -6.78 -6.3 2 -6.74 -6.35 3 -6.75 -6.33 0 -6.7 -6.25 1 -6.67 -6.28 2 -6.58 -6.22 3 -6.54 -6.18 0 NaN NaN 1 -6.48 -6.15 2 -6.33 -6.04 3 NaN NaN 0 -6.25 -5.94 1 -6.15 -5.92 2 -6.02 -5.76 3 -5.93 -5.73 0 -5.57 -5.47 1 -5.43 -5.28 2 -5.08 -5.16 3 -4.81 -4.92 0 -4.68 -4.73 1 -4.52 -4.55 2 -4.47 -4.5 3 -4.42 -4.42 0 NaN NaN 1 -4.21 -4.27 2 -4.01 -4.11 3 -4.03 -4.08 0 -4.06 -4.02 1 -4.05 -4.03 2 -4.01 -4.04 3 -3.98 -3.93 0 NaN NaN 1 -4.05 -3.99 2 -3.87 -3.91 3 -3.88 -3.85 0 -3.97 -3.82 1 -3.91 -3.86 2 -4.03 -3.91 3 -4.05 -3.93 0 NaN NaN 1 -4.31 -4.07 2 -4.23 -4.05 3 -4.25 -4.12 0 -4.29 -4.1 1 -4.29 -4.15 2 -4.23 -4.15 3 -4.25 -4.1 0 NaN NaN 1 -4.25 -4.09 2 -4.13 -4.03 3 -4.1 -4.02 0 -4.06 -3.94 1 -4.12 -3.97 2 -4.03 -3.96 3 NaN NaN 0 -3.96 -3.97 1 -3.92 -3.84 2 -3.95 -3.91 3 -3.99 -3.84 0 -3.95 -3.82 1 -3.93 -3.84 2 -3.87 -3.89 3 -3.84 -3.85 0 NaN NaN 1 -3.85 -3.77 2 -3.78 -3.72 3 -3.81 -3.72 0 -3.76 -3.71 1 -3.77 -3.74 2 -3.74 -3.7 3 -3.69 -3.72
Here's the same method, but using a matrix:
m = readmatrix('file.xlsx');
disp(m);
0 -6.6700 -6.0100 1.0000 -6.7300 -6.0300 2.0000 -6.7100 -6.0500 3.0000 -6.7300 -6.0300 0 -6.8000 -6.0100 1.0000 -6.8600 -6.1400 2.0000 -6.9700 -6.2700 3.0000 -7.1300 -6.3600 1.0000 -7.1300 -6.4600 2.0000 -7.0900 -6.4400 3.0000 -7.1200 -6.5200 0 -7.1200 -6.4900 1.0000 -7.0900 -6.5700 2.0000 -7.0900 -6.5500 3.0000 -7.0700 -6.5100 1.0000 -7.0200 -6.5200 2.0000 -6.9300 -6.4800 3.0000 -6.8600 -6.4100 0 -6.8200 -6.4300 1.0000 -6.8200 -6.4300 2.0000 -6.7900 -6.3300 3.0000 -6.7800 -6.3600 1.0000 -6.7800 -6.3000 2.0000 -6.7400 -6.3500 3.0000 -6.7500 -6.3300 0 -6.7000 -6.2500 1.0000 -6.6700 -6.2800 2.0000 -6.5800 -6.2200 3.0000 -6.5400 -6.1800 1.0000 -6.4800 -6.1500 2.0000 -6.3300 -6.0400 0 -6.2500 -5.9400 1.0000 -6.1500 -5.9200 2.0000 -6.0200 -5.7600 3.0000 -5.9300 -5.7300 0 -5.5700 -5.4700 1.0000 -5.4300 -5.2800 2.0000 -5.0800 -5.1600 3.0000 -4.8100 -4.9200 0 -4.6800 -4.7300 1.0000 -4.5200 -4.5500 2.0000 -4.4700 -4.5000 3.0000 -4.4200 -4.4200 1.0000 -4.2100 -4.2700 2.0000 -4.0100 -4.1100 3.0000 -4.0300 -4.0800 0 -4.0600 -4.0200 1.0000 -4.0500 -4.0300 2.0000 -4.0100 -4.0400 3.0000 -3.9800 -3.9300 1.0000 -4.0500 -3.9900 2.0000 -3.8700 -3.9100 3.0000 -3.8800 -3.8500 0 -3.9700 -3.8200 1.0000 -3.9100 -3.8600 2.0000 -4.0300 -3.9100 3.0000 -4.0500 -3.9300 1.0000 -4.3100 -4.0700 2.0000 -4.2300 -4.0500 3.0000 -4.2500 -4.1200 0 -4.2900 -4.1000 1.0000 -4.2900 -4.1500 2.0000 -4.2300 -4.1500 3.0000 -4.2500 -4.1000 1.0000 -4.2500 -4.0900 2.0000 -4.1300 -4.0300 3.0000 -4.1000 -4.0200 0 -4.0600 -3.9400 1.0000 -4.1200 -3.9700 2.0000 -4.0300 -3.9600 0 -3.9600 -3.9700 1.0000 -3.9200 -3.8400 2.0000 -3.9500 -3.9100 3.0000 -3.9900 -3.8400 0 -3.9500 -3.8200 1.0000 -3.9300 -3.8400 2.0000 -3.8700 -3.8900 3.0000 -3.8400 -3.8500 1.0000 -3.8500 -3.7700 2.0000 -3.7800 -3.7200 3.0000 -3.8100 -3.7200 0 -3.7600 -3.7100 1.0000 -3.7700 -3.7400 2.0000 -3.7400 -3.7000 3.0000 -3.6900 -3.7200
[N,M] = size(m);
ii = 2;
while ii <= N
if m(ii,1)-m(ii-1,1) ~= 1 && (m(ii,1) ~= 0 || m(ii-1,1) ~= 3)
m = [m(1:ii-1,:); ...
mod(m(ii-1,1)+1,4) NaN(1,M-1); ...
m(ii:end,:)];
N = N+1;
end
ii = ii+1;
end
disp(m);
0 -6.6700 -6.0100 1.0000 -6.7300 -6.0300 2.0000 -6.7100 -6.0500 3.0000 -6.7300 -6.0300 0 -6.8000 -6.0100 1.0000 -6.8600 -6.1400 2.0000 -6.9700 -6.2700 3.0000 -7.1300 -6.3600 0 NaN NaN 1.0000 -7.1300 -6.4600 2.0000 -7.0900 -6.4400 3.0000 -7.1200 -6.5200 0 -7.1200 -6.4900 1.0000 -7.0900 -6.5700 2.0000 -7.0900 -6.5500 3.0000 -7.0700 -6.5100 0 NaN NaN 1.0000 -7.0200 -6.5200 2.0000 -6.9300 -6.4800 3.0000 -6.8600 -6.4100 0 -6.8200 -6.4300 1.0000 -6.8200 -6.4300 2.0000 -6.7900 -6.3300 3.0000 -6.7800 -6.3600 0 NaN NaN 1.0000 -6.7800 -6.3000 2.0000 -6.7400 -6.3500 3.0000 -6.7500 -6.3300 0 -6.7000 -6.2500 1.0000 -6.6700 -6.2800 2.0000 -6.5800 -6.2200 3.0000 -6.5400 -6.1800 0 NaN NaN 1.0000 -6.4800 -6.1500 2.0000 -6.3300 -6.0400 3.0000 NaN NaN 0 -6.2500 -5.9400 1.0000 -6.1500 -5.9200 2.0000 -6.0200 -5.7600 3.0000 -5.9300 -5.7300 0 -5.5700 -5.4700 1.0000 -5.4300 -5.2800 2.0000 -5.0800 -5.1600 3.0000 -4.8100 -4.9200 0 -4.6800 -4.7300 1.0000 -4.5200 -4.5500 2.0000 -4.4700 -4.5000 3.0000 -4.4200 -4.4200 0 NaN NaN 1.0000 -4.2100 -4.2700 2.0000 -4.0100 -4.1100 3.0000 -4.0300 -4.0800 0 -4.0600 -4.0200 1.0000 -4.0500 -4.0300 2.0000 -4.0100 -4.0400 3.0000 -3.9800 -3.9300 0 NaN NaN 1.0000 -4.0500 -3.9900 2.0000 -3.8700 -3.9100 3.0000 -3.8800 -3.8500 0 -3.9700 -3.8200 1.0000 -3.9100 -3.8600 2.0000 -4.0300 -3.9100 3.0000 -4.0500 -3.9300 0 NaN NaN 1.0000 -4.3100 -4.0700 2.0000 -4.2300 -4.0500 3.0000 -4.2500 -4.1200 0 -4.2900 -4.1000 1.0000 -4.2900 -4.1500 2.0000 -4.2300 -4.1500 3.0000 -4.2500 -4.1000 0 NaN NaN 1.0000 -4.2500 -4.0900 2.0000 -4.1300 -4.0300 3.0000 -4.1000 -4.0200 0 -4.0600 -3.9400 1.0000 -4.1200 -3.9700 2.0000 -4.0300 -3.9600 3.0000 NaN NaN 0 -3.9600 -3.9700 1.0000 -3.9200 -3.8400 2.0000 -3.9500 -3.9100 3.0000 -3.9900 -3.8400 0 -3.9500 -3.8200 1.0000 -3.9300 -3.8400 2.0000 -3.8700 -3.8900 3.0000 -3.8400 -3.8500 0 NaN NaN 1.0000 -3.8500 -3.7700 2.0000 -3.7800 -3.7200 3.0000 -3.8100 -3.7200 0 -3.7600 -3.7100 1.0000 -3.7700 -3.7400 2.0000 -3.7400 -3.7000 3.0000 -3.6900 -3.7200
  4 件のコメント
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Sanley Guerrier
Sanley Guerrier 2023 年 7 月 17 日
Excellent, thank you very much!
Voss
Voss 2023 年 7 月 17 日
You're welcome!

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その他の回答 (1 件)

Image Analyst
Image Analyst 2023 年 7 月 17 日
Ran in:
Does this work for you?
data = readmatrix('file.xlsx');
% Note: sometimes 0's are missing from colum 1 for some reason.
% Is this the way it's supposed to be???
[rows, columns] = size(data)
rows = 85
columns = 3
finalRow = 97;
data(end : finalRow, 2:3) = nan;
% Fill up tail of column 1 with 0;1;2;3;0;1;2;3; etc.
for row = rows+1 : finalRow
data(row, 1) = mod(row+2, 4);
end
% Optional: Convert from array to table
t = table(data(:, 1), data(:, 2), data(:, 3), 'VariableNames', {'QHR', 'A', 'B'})
t = 97×3 table
QHR A B ___ _____ _____ 0 -6.67 -6.01 1 -6.73 -6.03 2 -6.71 -6.05 3 -6.73 -6.03 0 -6.8 -6.01 1 -6.86 -6.14 2 -6.97 -6.27 3 -7.13 -6.36 1 -7.13 -6.46 2 -7.09 -6.44 3 -7.12 -6.52 0 -7.12 -6.49 1 -7.09 -6.57 2 -7.09 -6.55 3 -7.07 -6.51 1 -7.02 -6.52
  1 件のコメント
Sanley Guerrier
Sanley Guerrier 2023 年 7 月 17 日
Thank you,
The main point is to fill in the missing number from column 1 so the pattern continue as 0 1 2 3 0 1 2 3 0 1 2 3 until the end.

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