concatenate repeating digit to the end of a column in numeric matrix

I have a large numeric matrix that has truncated the 3rd decimal place. I would like to include the 3rd decimal place in the binDepth column. I am sure this is just a loop that I need to run but nothing is working for me, (I am also new at this)

binDepth binCount binDepth binCount

-0.11 170 -0.112 170

-0.11 178 -0.111 178

-0.11 161 -0.110 161

-0.10 167 -0.109 167

-0.10 191 ... ...

-0.10 156

-0.10 177

-0.10 174

-0.10 173

-0.10 179

-0.10 165

-0.10 178

-0.10 212

-0.09 191

-0.09 185

-0.09 191

[David Hill]

You are going to have to explain more. What is the input? Provide an example of your desired output from a known input.

[Andrew Forbes]

Sorry, input is the 2 column table on the left, the desired output is the 2 column table on the right.

[Mathieu NOE]

hello

sorry, but from where are we getting the missing 3rd decimal ??

[Walter Roberson]

Don't use "format bank" ?

Or don't use round() with 2 decimal places?

Or change sprintf() or fprintf() or compose() to specify at %.3f format instead of a %.2f format ?

[Andrew Forbes]

The output from a 3rd party data set automatically truncated the data. The output is a 2 column text file. I am measuring differences to the millimeter so I I need to add in the 3rd decimal place. The output text file has 10 iterations of each measurement (e.g -1.09, -1.09, -1.09, -1.09, -1.09, -1.09, -1.09, -1.09, -1.09, -1.09) before moving to -1.08 etc… It is a loop i need to run to concatenate the additional digit (9, 8, 7, 6,5, 4, 3, 2, 1, 0…) When the number is negative it will be in descending order, (9, 8, 7, 6….) when it becomes a positive number it will be in ascending order (1, 2 ,3,…)

Answer 1

[Mathieu NOE]

hello

so my 2 cents suggestion, build on @DGM input (tx to him ! )

it just needed an additionnal correction term to be perfect (hopefully)

clearvars;
clc

%% data extraction
[out, delimiter, headerlines]  = importdata('data.txt');
header = out.textdata;
data = out.data;
data1 = data(:,1);
data2 = data(:,2);
samples = length(data1);

myvector = (data1(1):0.001:data1(end)).'; % transpose for col vector
ind = find(diff(data1)>0); % find the index of the first "staircase" edge (this will be used later)
corr = data1(ind(1)) - myvector(ind(1)); % correction term
myvector = myvector + corr; % correction term added

Answer 2

[DGM]

Are you sure the data is being truncated, or is it rounded? Guessing the wrong answer means your alignment of additional digits will be off by 5.

If the column vector has known start and endpoints and a fixed stepsize, it would be easier and more certain to just rebuild it from scratch.

myvector = (startpoint:0.001:endpoint).'; % transpose for col vector