indexing to different table

    Time          ICAO            age           A          ID    HPL     FLG
'12:59:45.466'  'OMA812'  0.50000000  5  1907317   135  228
'12:59:39.483'  'OMA812'  12.500000  5  1906896   135  228
'12:59:39.483'  'OMA812'  9.1999998  5  1906502   131  228
'12:59:39.483'  'OMA812'  8.5000000  5  1906421   131  228
'12:59:49.306'  'OMA812'  6.5999999  5  1907976   127  228
'12:59:39.483'  'OMA812'  7.5999999  5  1906277   127  228

I have the following table above, I would like to break up the table by HPL number. Move data for HPL 135 to one table, 131 to another table and 127 to another.

I like to put that in a loop so I can perform that for more HPL ids.

so far I have this but does not work

hplid = [];
hplid = unique(Join.spf_hplId);
latency_plots = [];
data_array=[];
for k=1: length(hplid)
    join_idx=Join.spf_hplId==hplid(k);
    %Join(k)=Join(join_idx,:);
    test.test=Join(join_idx,:);
    data_array=[data_array;test];
end

Thank you

[Ketan Bharucha]

any help would be much appreciated.

[Stephen23]

Usually keeping data together is a better idea than splitting it apart. Tables include easy ways of accessing only particular rows and columns, so it is not clear why you expect splitting up the data to be advantageous to you.

I suspect that this applies:

http://xyproblem.info/

and possibly this too:

https://www.mathworks.com/matlabcentral/answers/304528-tutorial-why-variables-should-not-be-named-dynamically-eval

[dpb]

Besides Stephen's well-considered points, look into 'Grouping Variables' as potentially being helpful here...

[Guillaume]

Agree with Stephen, splitting the table into multiple tables sounds like a very bad idea, so explain why you want to do that.

If it's so you can calculate some properties/statistics per HPL number, then this can be done a lot more easily on the whole table with rowfun or splitapply, e.g. to calculate the mean age per HPL:

rowfun(@mean, yourtable, 'InputVariables', 'age', 'GroupingVariables', 'HPL')

[Ketan Bharucha]

Thanks for the help!

There is a latency column as well which I didn't include in my original question. Basically I am looking to do is find 99%, 95% and 90% of the latency per HPL.

    Time          ICAO            latency           A          ID    HPL     FLG
'12:59:45.466'  'OMA812'  0.50000000  5  1907317   135  228
'12:59:39.483'  'OMA812'  12.500000  5  1906896   135  228
'12:59:39.483'  'OMA812'  9.1999998  5  1906502   131  228
'12:59:39.483'  'OMA812'  8.5000000  5  1906421   131  228
'12:59:49.306'  'OMA812'  6.5999999  5  1907976   127  228
'12:59:39.483'  'OMA812'  7.5999999  5  1906277   127  228

Answer 1

[Steven Lord]

Do you want to create individual variables with a name that somehow includes information about the HPL number it represents (in which case see Stephen's comment about dynamically named variables) or do you simply need to process each set of HPL values in turn? If the latter, use logical indexing.

% Create a sample table
rng default
n = 20;
X = randi([-10 10], n, 1);
Y = randi([0 4], n, 1);
processed = false(n, 1);
T = table(X, Y, processed);

% Get the list of unique values for the Y variable in the table
listOfUniqueY = unique(T.Y);

% Iterate over those unique values. The unique function returns a column vector
% but if I tranpose the result I can directly iterate over the values.
for k = listOfUniqueY.'

    % Select those rows of T that have Y equal to each unique value
    selectedRowsOfT = T.Y == k;
    TforY = T(selectedRowsOfT, :);

    % How many are there?
    fprintf('T has %d rows with Y value %d.\n', height(TforY), k);

    % Mark the selected rows as processed. Leave the semicolon off to display the table.
    T{selectedRowsOfT, 'processed'} = true
  end

Note I used the logical index vector both to extract data (to create TforY) and to assign (the line where I update the 'processed' variable.)

See this documentation page for more information on table indexing.

Answer 2

[Peter Perkins]

It sounds like what you ultimately want is to compute statistics within groups on variables in a table. If that's correct, here's what I'd do:

>> t = table(randn(10,1),randi(2,10,1),'VariableNames',{'latency' 'HPL'})
t =
  10×2 table
    latency     HPL
    ________    ___
     -2.2584     1 
      2.2294     2 
     0.33756     1 
      1.0001     2 
     -1.6642     2 
    -0.59003     2 
    -0.27806     2 
     0.42272     1 
     -1.6702     2 
     0.47163     1 
>> myStats = @(x) [min(x) median(x) max(x)];
>> tStats = varfun(myStats, t,'GroupingVariables','HPL')
tStats =
  2×3 table
    HPL    GroupCount              Fun_latency          
    ___    __________    _______________________________
     1         4         -2.2584     0.38014     0.47163
     2         6         -1.6702    -0.43405      2.2294

That creates one summary stats variable in the output table, with three columns. That's probably good if you have more than one data variable. Or, you could use rowfun to do essentially the same thing but create three separate variables in the output. Or you could just split them up by hand:

>> tStats.minLatency = tStats.Fun_latency(:,1);
>> tStats.medianLatency = tStats.Fun_latency(:,2);
>> tStats.maxLatency = tStats.Fun_latency(:,3);
>> tStats.Fun_latency = []
tStats =
  2×5 table
    HPL    GroupCount    minLatency    medianLatency    maxLatency
    ___    __________    __________    _____________    __________
     1         4          -2.2584         0.38014        0.47163  
     2         6          -1.6702        -0.43405         2.2294