Coverting Timetable to Column Vector

Hi everyone,

I am working with .edf files and am using edfread, which is giving me a timetable output, when I would like a column vector.

We are recording voltage data in 1-hour increments at a sampling frequency of 1 KHz with 6 channels, each channel is unique.

I am using the option 'DataRecordOutputType' set to 'vector', which just makes this weirder.

Here's the code:

data = edfread('filename.edf','DataRecordOutputType' ,'vector');

and the output is:

I can solve this via doing the following:

data = edfread('filename.edf','DataRecordOutputType' ,'vector');

abf_like = [];

new_data = cell2mat(data{1,1});

abf_like = new_data;

for i = 2:size(data, 1)

new_data = cell2mat(data{i,1});

abf_like = vertcat(abf_like, new_data);

end

Which outputs my desired array (3600500 x 1). But this takes forever (understandably so), and there are 6 channels per edf file and many files. What am I missing with edfread, or what can I do to speed this up?

Thanks for any insight you can offer.

edit: here's something that works for what I need. Still would like some alternatives, if there are any:

clear all; close all; clc

edf = edfread('filename.edf');

abf_like = edf_to_abf(edf);

function out = edf_to_abf(edf)

edf = timetable2table(edf); edf = table2cell(edf);

edf = edf(:, 2:end);

abf_like = zeros(3600500,1);

abf_like = table(abf_like);

for j = 1:6

single_data = edf(:,j);

abf_like{:,j} = unpacker(single_data);

clear single_data

end

out = table2array(abf_like);

end

function out = unpacker(data)

abf_like(:,1) = cell2mat(data(1,1));

r = 501;

for i = 2:size(data, 1)

abf_like(r:r+499,1) = cell2mat(data(i,1));

r = r + 500;

end

out = abf_like;

end

Answer 1

[Peter Perkins]

Sam, it looks like you are ending up with a 7201x6 timetable whose variables are cell arrays, and each cell contains a 500x1 double vector. And it sounds like you want six (or one?) 7201*500x1 vector.I can't speak to how you are ending up with that timetable without seeing the file, but getting that should not take a long time:

>> tt = timetable(Size=[7201 2],VariableTypes=repmat("cell",1,2),StartTime=seconds(0),TimeStep=seconds(.5));
>> tt{:,:} = {rand(500,1)}
tt =
7201×2 timetable
Time            Var1              Var2     
__________    ______________    ______________
0 sec         {500×1 double}    {500×1 double}
0.5 sec       {500×1 double}    {500×1 double}
1 sec         {500×1 double}    {500×1 double}
1.5 sec       {500×1 double}    {500×1 double}
2 sec         {500×1 double}    {500×1 double}
2.5 sec       {500×1 double}    {500×1 double}
:               :                 :       
3597.5 sec    {500×1 double}    {500×1 double}
3598 sec      {500×1 double}    {500×1 double}
3598.5 sec    {500×1 double}    {500×1 double}
3599 sec      {500×1 double}    {500×1 double}
3599.5 sec    {500×1 double}    {500×1 double}
3600 sec      {500×1 double}    {500×1 double}
Display all 7201 rows.

>> tic, var1 = vertcat(tt.Var1{:}); toc
Elapsed time is 0.030527 seconds.
>> size(var1)
ans =
     3600500           1

[Sam]

@Peter Perkins This is exactly what I want to do! What you accomplished in one line too me way too many...

I have follow-up question, though. The timetable I am working with (tt by your example) has multiple columns, which all need have the same thing done to them (the goal is to produce a 3600500 x 6 matrix). Your method works on a single column, but how would you automate it to loop through the columns?

Using your example (which only has two columns, obviously), I have tried:

newMatrix = zeros(3600500, 2);

variableNames = tt.Properties.VariableName;

for i = 1:2

newMatrix(:, i) = vertcat(tt.(variableNames(i){:}));

end

along with some other things. All in all, the use of the dot oeprator appears to be what is making this work. How would you automate it to sweep across the columns?

Thank you for your help, it is really appreciated.

[Peter Perkins]

varfun to the rescue!

>> tt2 = varfun(@(varN)vertcat(varN{:}),tt,OutputFormat='table')
tt2 =
  3600500×2 table
    Fun_Var1    Fun_Var2
    ________    ________
     0.86119     0.86119
     0.97724     0.97724
     0.44452     0.44452
    0.034243    0.034243
     0.52377     0.52377
    0.031639    0.031639
       :           :    
     0.21357     0.21357
     0.43393     0.43393
    0.033077    0.033077
     0.92932     0.92932
     0.10604     0.10604
     0.26629     0.26629
	Display all 3600500 rows.

(The two vars are the same because I was lazy when creating the test data.) And then

>> X = tt2.Variables;
>> X(1:5,:)
ans =
      0.86119      0.86119
      0.97724      0.97724
      0.44452      0.44452
     0.034243     0.034243
      0.52377      0.52377

[Sam]

@Peter Perkins You are absolutely amazing! Thank you very much. I never would have figured that out.