MATLAB Answers


linear interpolation of matrix.

Asked by Patrick
on 18 Nov 2012


I'm new to matlab so this might be very simple but I would appreciate the help..

I have a matrix of 32 values across and 2209 down. Each line represents one hour of rainfall so these values represent values of rainfall at 1.875min timesteps. I need to use linear interpolation to make this either 2min or 5 min timesteps (a matrix of 30 values across or 12 values across).

for example, trying to convert the 32x2209 into 30x2209.

The new first value would be 0.9375 * the original first value.

The new second value would be (0.0625 * the original first value) + ((0.9375 - 0.0625) * the original second value)

The new third value would be ((2 * 0.0625) * the original second value) + ((0.9375 - (2 * 0.0625)) * the original third value)

... and so on.

Is there an easy way to do this in matlab?

Many thanks in advance to anyone that can help.




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3 Answers

Answer by Teja Muppirala
on 19 Nov 2012
 Accepted answer

This is a method that will preserve area.

%Step 0. Just making some random data to work with...

R = rand(2209,32);

%Step 1. Get the cumulative area

cumulativeArea = [zeros(size(R,1),1) cumsum(R,2)];

%Step 2. Interpolate the accumulated area on the new time intervals

t0 = 0:(1/32):1; %Old
t1 = 0:(1/30):1; %New
cumulativeAreaNew = interp1( t0, cumulativeArea', t1 );

%Step 3. Differentiate the result.

Rnew = diff( cumulativeAreaNew )';

% Just a check:

max(abs(sum(R,2) - sum(Rnew,2))) %Very small

  1 Comment

Thanks Teja,

This appears to work perfectly. I don't understand all the commands you have used here but comparing the inputs and the outputs it appears to be doing mathematically what I would hope for!


Answer by Matt J
on 18 Nov 2012

Use the INTERP1 command


2 minute interpolation:

t  = 60/32*(1:32) % base timeline (monotonous increasing)
y    = 32 values in row
newT = 2:2:60 % new 2 minute timestep line, assuming end at 60min.
newY = interp1(t,y,newT)

Thanks Jurgen,

I am embarrassed to say I explained this really badly/incorrectly in my original question. I am looking to conserve all the mass from my original matrix in a smaller matrix.. decreasing from 32 columns to 30 columns.

therefore the new first value would be original first value + (2/30)*original second value)

the new second value would be (28/30)*original second value + (4/30)*original third value.


i'm just checking does the interp1 function actually achieve this?

if so, is there a way to adapt your above commands to do all 2209 lines of my matrix at once rather than 1 row?

Thanks in advance,


Answer by Ilham Hardy
on 19 Nov 2012

A beginner approach,

%1.875 is the default sample rate
n_col_def = (60/1.875);
%row amount
n_row_def = 2209; 
%create time vector of original data (start from t = 1.875, hop 1.875,until data end)
time_1 = (60/n_col_def):(60/n_col_def):(60/n_col_def)*n_row_def*n_col_def; 
%your matrix name
data_1 = rand(n_col_def,n_row_def); 
%transpose the matrix to get correct reshape
data_1_tr = data_1'; 
%reshape matrix to vector (1*n)
data_1_res = reshape(data_1_tr,1,n_row_def*n_col_def); 
% same row amount with 30 columns require lemgth n_time_2 to be specific
n_col_new = (60/2);
n_time_new = n_col_new*n_row_def;
%t_start = 2,sample time 2 minutes . [2,4,6...,]
time_2 = 2:2:(n_time_new*2); 
%interpolate the data based on the new time vector (of 5 minutes)
data_2 = interp1(time_1,data_1_res,time_2,'linear'); 
% reshape vector to matrix
data_2_res = reshape(data_2,n_col_new,[]);
data_2 = data_2_res';


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