Create a difference loop from a cumulative matrix

Hello

I have this data filed that I imported into a 831x1 matrix. The numbers in it are cumulative so I need to create a for loop that shows the difference between the row and the one before it so that I can have the pure added quantity in each row. If possible, I would like the first row to stay the same as it was in the imported file.

How do I do this?

[madhan ravi]

diff(matrix) % ?

[Mohamed Amine Mouajib]

The file is now a single column matrix. A single column cumulative array. I'm supposed to create a for loop that simply shows the difference between rows.

[madhan ravi]

Why are you sticking with a loop , is it your homework?

[Mohamed Amine Mouajib]

Yes. Nevermind I managed to do it.

Answer 1

[Raj]

Suppose 'A' is your 831x1 matrix. Use this:

B=diff(A);

Apure=[A(1);B];

If you are specific with "for loop" then you can use this:

Apure=zeros(831,1);

Apure(1)=A(1);

for i=2:831

Apure(i)=A(i)-A(i-1);

end

[Mohamed Amine Mouajib]

This doesn't work?

I ended up using

for m=1:length(B)

C=diff(B)

end

given that A was the file matrix, but B is A that starts with a zero column)

It worked

[madhan ravi]

@Mohammed: Why is there a loop in your code?, even without the loop you would get the same result. Basically you are just repeating the same calculation over and over again.

[Mohamed Amine Mouajib]

It's supposed to be a loop. The reason I added C is because I have to use it in a calculation.

Answer 2

[Mohamed Amine Mouajib]

Here's my final script for the whole problem

% Amine Mouajib

% Homeowork 4

% I ommited the text part of the fuel_totalizer.txt file

%in preferences, I changed the command window numerical format from short

% to long g

clc, clear all

disp('Problem 1')

A=[0;importdata('fuel_totalizer.txt')]

y=0

sec=size(A,1)

hour=sec/60

disp('Problem 2,3,&4')

for m=1:length(A)

B= diff(A) % Fuel consumption in each consecutive minute

C= diff(A/60) % Flow rate in lbs/sec for each minute

sum(diff(A)) % Total fuel consumed. Notice it's the same value as the last row of given data matrix

end

disp('Problem 5')

avg=sum(diff(A))/hour %avg fuel consumption for whole flight in lbs/hr

disp('Problem 6')

takeoffmin=1.5*mean(diff(A))% the minimum possible minute flow rate during takeoff in lbs/min

E=B(1:416)% First half of data

D=E(E>takeoffmin)

takeofftime=size(D,1)% in minutes

disp('bonus')

bns=max(C) % the maximum flow rate in lbs/sec

bns2=max(B)

Z=find(B == bns2) % which minute the maximum flow rate in lb/sec occured