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

Find slope for a row of data, not a column

nkormanik — Thu, 30 Jun 2016 05:22:31 GMT

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.

Nicholas

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

ron_horne — Thu, 30 Jun 2016 12:26:52 GMT

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.

best,

ron

// 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;

);

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

ms — Thu, 30 Jun 2016 14:29:54 GMT

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("Slope_Across_Columns.jmp",

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)));

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

txnelson — Thu, 30 Jun 2016 14:35:19 GMT

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)

biv=stacked<<Bivariate(invisible,

Y( :Data ),

X( :time ),

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

by(_1)

);

// 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 )

);