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

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]

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]

The invalid column is always the 2nd one.

Answer 1

[Stephen23]

>> 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]

I need it to increment in .1, not 1.

[Guillaume]

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]

That worked perfectly, thank you very much.