How to replace a column with an incremented vector starting at -10?

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Ibro Tutic
Ibro Tutic 2017-5-17
评论： Ibro Tutic 2017-5-17
I have some time data and I am trying to replace an invalid column with data starting at -10 and increases in increments of .1 up to the length of the dataset. How would I go about this?
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Jan
Jan 2017-5-17
Please post some code, which creates a dataset like yours. "some time data" is not enough to guess this detail reliably. How can the "invalid" columns be identified?

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Stephen23
Stephen23 2017-5-17
编辑：Stephen23 2017-5-17
>> M = rand(12,6); % example data
>> col = 4; % specify which column
>> M(:,col) = (1:size(M,1))-11
M =
0.53477 0.48090 0.36142 -10.00000 0.83279 0.27853
0.16247 0.14570 0.25672 -9.00000 0.80810 0.03114
0.05016 0.80648 0.86410 -8.00000 0.38718 0.34066
0.66991 0.61923 0.74592 -7.00000 0.23169 0.98405
0.14010 0.76240 0.09419 -6.00000 0.23363 0.02649
0.18001 0.00460 0.03549 -5.00000 0.60986 0.28603
0.92897 0.72864 0.67928 -4.00000 0.70401 0.01101
0.24724 0.90098 0.25873 -3.00000 0.64998 0.94363
0.98211 0.64409 0.32472 -2.00000 0.37861 0.60439
0.87888 0.29503 0.64717 -1.00000 0.46550 0.89163
0.00736 0.21508 0.01827 0.00000 0.46535 0.53023
0.55577 0.28285 0.53249 1.00000 0.77927 0.53609
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Guillaume
Guillaume 2017-5-17
Well, it's trivial to change:
M(:, col) = startvalue + (0:size(M, 1)-1) * increment;
So:
M(:, col) = -10 + (0:size(M, 1)-1) * 0.1;

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