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Find slope for a row of data, not a column

Jun 29, 2016 10:22 PM
(4276 views)

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

3 REPLIES 3

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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.

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;

**)**;

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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**(**"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**)))**;

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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)**

**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 )**

**);**

Jim