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Find slope for a row of data, not a column

Data are by subject over time, across a row.  See the following example:

columns _1 _2 _3 _4 _5 _8

subject(001) data -0.069965723 0.492749371 0.955245597 1.346963522 1.701118239 2.523141195

I have thousands of such rows where the slope is needed.

All I really need is the slope for each series of numbers in the row.

A new additional column following each series should contain the slope value for that particular series.

I've asked on a SAS forum if SAS can handle this.  Wondering if JMP can??  Any help would be greatly appreciated.


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Super User

Re: Find slope for a row of data, not a column

hi nkormanik​,

the answer is yes.

in general you would need to transpose the data and do the calculations. assuming the data table is too big for that i think it is easier to transpose a matrix. therefore, the process in general involves extracting the data into a matrix, transposing it and calculating the slopes. have a look at this. it does all the dfferent steps for illustration purposes. it is not very robust (if you have missing values) or anything funny.



// create a data table

dt = new table ("data in rows",

      add rows (100),

      new column ("Subject", formula (row())),

      new column ("slope"),

      new column ("R2"),

      add multiple columns ("column", 20, numeric)



// set values for the data columns

for (i=4, i<=23, i++,

      :column (i) << formula ((i/2) + Random Normal()) << eval formula << delete formula ;


// start working

// get the table as a matrix

datamatrix = dt<<GetAllColumnsAsMatrix;

// remove the columns not used for regression calculations (columns 1-4)

datamatrix [ 0 , 1::3] = [];

// transpose the matrix

data = datamatrix`;

// make an x variable for time.

X = (1 :: ncol(datamatrix))`;

// add a column of 1's for the intercept term

X = J(Nrow(X),1) || X;

//now we have all the data we need for the calculations in matrix form

for (irow=1, irow<=nrows(dt), irow++, // for each row

// extract individual data one by one

Y = data [0,  irow ];

// compute least squares estimates

β = Inv(X`*X)*X`*Y;

dt:slope[irow] = β[2];

// compute R2

N = NRows(Y);

Ybar = Mean(Y);

R2 = (β`*X`*Y - N*Ybar^2)/(Y`*Y - N*Ybar^2);

dt:R2[irow] = R2;


Super User ms
Super User

Re: Find slope for a row of data, not a column

Here is an example of a column formula to calculate the slope across columns. It is not clear what the X values are, but I assume they are defined by the column names as in your example. As you can see the the X matrix is now hard coded. The formula could be made more general if the X values were held in e.g. the first row (or in a table variable).

// Example table

dt = New Table("",

    Add Rows(2),

    New Column("Subject", Character, Values({"001", "002"})),

    New Column("_1", Values([-0.069965723, 2])),

    New Column("_2", Values([0.492749371, 4])),

    New Column("_3", Values([0.955245597, 6])),

    New Column("_4", Values([1.346963522, 8])),

    New Column("_5", Values([1.701118239, 10])),

    New Column("_8", Values([2.523141195, 16]))


//Formula expression

F = Eval Expr(

    Local({i = 1, n = Expr(N Col(Current Data Table()) - 1), X = J(n, 1, 1) || [1, 2, 3, 4, 5, 8], Y},

        Y = Matrix(Repeat({i++ ; Column(i)[]}, n));

        (Inverse(X` * X) * X` * Y)[2];



//Add formula column

dt << New Column("Slope", Numeric, Formula(Name Expr(F)));

Super User

Re: Find slope for a row of data, not a column

Here is another method.  It has only one advantage, in that the script just duplicates what you could do running JMP interactively.  I have attached a sample data table that I developed the script around.

// Set a pointer to the sample data

names default to here(1);

dt = current data table();

// Stack the 6 data columns from each row

stacked = dt << Stack(

       columns( :_3, :_4, :_5, :_6, :_7, :_8 ),

       Source Label Column( "Label" ),

       Stacked Data Column( "Data" )


// Create a new column that represents the time factor

stacked << new column("time",formula(Num(Substr(:Label, 2, 1)) - 2));

// Run the analysis on the data, by each row(student)


       Y( :Data ),

       X( :time ),

       Fit Line( {Line Color( {213, 72, 87} )} ),



// Create a data table from the estimates output

dtslope=report(biv[1])["Parameter Estimates"][1]<<make combined data table;

// Get rid of the intercept data

dtslope<<select where(:term == "Intercept");

dtslope<<delete rows;

// Select the columns that are not wanted and delete them

dtslope:x<<set selected(1);

dtslope:y<<set selected(1);

dtslope:term<<set selected(1);

dtslope:Name("~bias")<<set selected(1);

dtslope:std error<<set selected(1);

dtslope:t ratio<<set selected(1);

dtslope:Name("prob>|t|")<<set selected(1);

dtslope<<delete columns;

// Rename the Estimate column to Slope

dtslope:Estimate<<set name("Slope");

// Add the slope to the original data table

dt<< Update(

       With( dtslope ),

       Match Columns( :_1 = :_1 )