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

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Ibro Tutic on 17 May 2017
Commented: Ibro Tutic on 17 May 2017
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?

Jan on 17 May 2017
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?
Ibro Tutic on 17 May 2017
The invalid column is always the 2nd one.

Stephen Cobeldick on 17 May 2017
Edited: Stephen Cobeldick on 17 May 2017
>> 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

Ibro Tutic on 17 May 2017
I need it to increment in .1, not 1.
Guillaume on 17 May 2017
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;
Ibro Tutic on 17 May 2017
That worked perfectly, thank you very much.