Re: if statement help

Hi Walter,
is it possible you could help me further?
It appears that the script is working as there are no erros
showing but now,the matlab says its 'busy' and after 10 mins
is still not showing and graph display or error, this is the
code i used:

i=2; newpressure(1:length(pressure))=0;
while i <= length(pressure)
if ((newpressure(i) <
newpressure(i-1))&&((newpressure(i-1)-newpressure(i))>=150))
newpressure(i:i+5)=pressure(i);
end
i=1+5;

for j=i:length(pressure)

newpressure(j)=(pressure(j)-pressure(j-1))+newpressure(j-1);
end
i=1+1;
end

i've then continued with my script:
case1 = (newpressure < 673.2);
case2 = (newpressure < 995.7);
case3 = (newpressure >= 995.7 & newpressure < 1149.2);
case4 = (newpressure >= 1149.2);
volume = zeros(size(newpressure));
volume(case1) = 0;
volume(case2) = ((newpressure(case2)-442.95)*5.365);
volume(case3) = ((newpressure(case3)-749.29)*12.034);
volume(case4) = ((newpressure(case4)-934.08)*22.371);


roberson@xxxxxxxxxxxxxxxxxx (Walter Roberson) wrote in
message <futmvj$ncp$1@xxxxxxxxxxxxxxxxxxxxxxx>...
In article <futm6u$6tc$1@xxxxxxxxxxxxxxxxxx>,
Adam Bright <abc_services_2000@xxxxxxxxxxx> wrote:
Hi, I've been trying to play around with it but I get the
following error:
??? Subscript indices must either be real positive integers
or logicals.

Error in ==> untitld at 3
if( ((A(i) < A(i-1))&&((A(i-1)-A(i))>=150)) )


A=[1500;1600;1793.00;1810.00;1827.00;1827.00;1845.00;1861.00;1876.00;1891.00;1686.00;1178.00;709.90;722.90;771.70;818.00;861.00;904.00;]
while(i < length(A)-5)
if( ((A(i) < A(i-1))&&((A(i-1)-A(i))>=150)) )

What is 'i' initialized to? There is no initialization
shown here.
Could i be 0? Could i be 1? If i is 1 then A(i-1) would be
A(0)
and 0 is not a real positive integer
--
"Okay, buzzwords only. Two syllables, tops." -- Laurie
Anderson

.

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