Thread Subject: if statement help

Hi I'm having trouble composing an if statement to say if
the difference between the next value and the current value
is gretaer or equal to 150 & is less then the next value,
then for this value and the next 5 values, to use only the
previous value.however I then also need it so that after the
5 values using the previous value; any values after this use
the difference between the two values + the initial previosu
value. (if that makes any sense).

For example:
1793.00 1793.00
1810.00 1810.00
1827.00 1827.00
1845.00 1845.00
1861.00 1861.00
1876.00 1876.00
1891.00 1891
1686.00 1891
1178.00 1891
709.90 1891
722.90 1891
771.70 1891
818.00 1937.3 ((818-771.7)+1891)
861.00 1980.3 ((861-818)+1937.3)
904.00 2023.3 ((904-861)+1980.3)

The same process might need to occur further down the array
several times, so it needs to be some kind of statement
which can control it in the above manner.So far I've managed
to write an if statement without knowing how to get it to
apply to the future values and getting the difference:

for i = 2:length(pressure),
    if ((pressure(i) <
pressure(i-1))&&((pressure(i-1)-pressure(i))>=150))
        ((pressure(i) = pressure(i-1))&&...
            (pressure(i+1) = pressure(i-1))&&(pressure(i+2)
= pressure(i-1))&&(pressure(i+3)...
            = pressure(i-1))&&(pressure(i+4) =
pressure(i-1))&&(pressure(i+5) = pressure(i-1)));
    end,
end

thanks for your help

"Adam Bright" <abc_services_2000@yahoo.co.uk> writes:

> Hi I'm having trouble composing an if statement to say if
> the difference between the next value and the current value
> is gretaer or equal to 150 & is less then the next value,
> then for this value and the next 5 values, to use only the
> previous value.however I then also need it so that after the
> 5 values using the previous value; any values after this use
> the difference between the two values + the initial previosu
> value. (if that makes any sense).
>
> For example:
> 1793.00 1793.00
> 1810.00 1810.00
> 1827.00 1827.00
> 1845.00 1845.00
> 1861.00 1861.00
> 1876.00 1876.00
> 1891.00 1891
> 1686.00 1891
> 1178.00 1891
> 709.90 1891
> 722.90 1891
> 771.70 1891
> 818.00 1937.3 ((818-771.7)+1891)
> 861.00 1980.3 ((861-818)+1937.3)
> 904.00 2023.3 ((904-861)+1980.3)
>
> The same process might need to occur further down the array
> several times, so it needs to be some kind of statement
> which can control it in the above manner.So far I've managed
> to write an if statement without knowing how to get it to
> apply to the future values and getting the difference:
>
> for i = 2:length(pressure),
> if ((pressure(i) <
> pressure(i-1))&&((pressure(i-1)-pressure(i))>=150))
> ((pressure(i) = pressure(i-1))&&...
> (pressure(i+1) = pressure(i-1))&&(pressure(i+2)
> = pressure(i-1))&&(pressure(i+3)...
> = pressure(i-1))&&(pressure(i+4) =
> pressure(i-1))&&(pressure(i+5) = pressure(i-1)));
> end,
> end

I would suggest a "while" loop instead, so you can modify the index
inside the loop to take care of that group of 5.

while(i < length(pressure)-5)
  if( <your condition> )
    for j=0:4
      pressure(i+j) = <old value>;
    end
    i = i + 5;
  else
    apply adjustment to current value
    i = i + 1;
  end

end

In other words, "i" always points to the output value.

-Peter

thanks peter for your reply,I kind of see how this would
work,however I dont understand why you have given j=0.4 the
difference between the two will not always be the same,
however I will always want the current value and the next 5
values to use the 1st previous value and thereafter to use
this value + the differences gained .

Please could you explain further so i understand it
better.many thanks once again

"Adam Bright" <abc_services_2000@yahoo.co.uk> writes:

> thanks peter for your reply,I kind of see how this would
> work,however I dont understand why you have given j=0.4 the
> difference between the two will not always be the same,
> however I will always want the current value and the next 5
> values to use the 1st previous value and thereafter to use
> this value + the differences gained .
>
> Please could you explain further so i understand it
> better.many thanks once again

Please quote from previous posts for context, and reply to the post you
are actually replying to, so that threading works correctly.

I never used 0.4. for j=0:4 is a loop of j from 0 to 4. It is a way to
make the "next 5 values" the same as the input.

Copy of the code:

while(i < length(pressure)-5)
  if( <your condition> )
    for j=0:4
      pressure(i+j) = <old value>;
    end
    i = i + 5;
  else
    apply adjustment to current value
    i = i + 1;
  end
end


-Peter

I'm sorry Peter,i'm extremely grateful for your help,but I
can't get it to work.

while(i < length(pressure)-5)
  if( ((pressure(i) <
pressure(i-1))&&((pressure(i-1)-pressure(i))>=150)) )
    for j=0:4
      pressure(i+j) = pressure(i-1);
    end
    i = i + 5;
  else
    apply adjustment to current value
    i = i + 1;
  end
end

I'm not too sure what I would put where you have written
apply adjustment to current value, would this be
((pressure(i+6)-pressure(i+5)) + pressure(i))?
i've tried a few variations but have'nt got it to work.


Peter Boettcher <boettcher@ll.mit.edu> wrote in message
<muytzhqkjwb.fsf@G99-Boettcher.llan.ll.mit.edu>...
> "Adam Bright" <abc_services_2000@yahoo.co.uk> writes:
>
> > thanks peter for your reply,I kind of see how this would
> > work,however I dont understand why you have given j=0.4 the
> > difference between the two will not always be the same,
> > however I will always want the current value and the next 5
> > values to use the 1st previous value and thereafter to use
> > this value + the differences gained .
> >
> > Please could you explain further so i understand it
> > better.many thanks once again
>
> Please quote from previous posts for context, and reply to
the post you
> are actually replying to, so that threading works correctly.
>
> I never used 0.4. for j=0:4 is a loop of j from 0 to 4.
It is a way to
> make the "next 5 values" the same as the input.
>
> Copy of the code:
>
> while(i < length(pressure)-5)
> if( <your condition> )
> for j=0:4
> pressure(i+j) = <old value>;
> end
> i = i + 5;
> else
> apply adjustment to current value
> i = i + 1;
> end
> end
>
>
> -Peter

"Adam Bright" <abc_services_2000@yahoo.co.uk> writes:

> I'm sorry Peter,i'm extremely grateful for your help,but I
> can't get it to work.
>
> while(i < length(pressure)-5)
> if( ((pressure(i) <
> pressure(i-1))&&((pressure(i-1)-pressure(i))>=150)) )
> for j=0:4
> pressure(i+j) = pressure(i-1);
> end
> i = i + 5;
> else
> apply adjustment to current value
> i = i + 1;
> end
> end
>
> I'm not too sure what I would put where you have written
> apply adjustment to current value, would this be
> ((pressure(i+6)-pressure(i+5)) + pressure(i))?
> i've tried a few variations but have'nt got it to work.

Sorry, I can't help further. Pretend you're a computer, remember that i
will always point at the current *output* value and step-by-step go
through it. It might help to do an example by hand, at all times
keeping track of the value of i, and figuring out what offsets against i
you need to apply to get to the numbers you need.

-Peter


> Peter Boettcher <boettcher@ll.mit.edu> wrote in message
> <muytzhqkjwb.fsf@G99-Boettcher.llan.ll.mit.edu>...
>> "Adam Bright" <abc_services_2000@yahoo.co.uk> writes:
>>
>> > thanks peter for your reply,I kind of see how this would
>> > work,however I dont understand why you have given j=0.4 the
>> > difference between the two will not always be the same,
>> > however I will always want the current value and the next 5
>> > values to use the 1st previous value and thereafter to use
>> > this value + the differences gained .
>> >
>> > Please could you explain further so i understand it
>> > better.many thanks once again
>>
>> Please quote from previous posts for context, and reply to
> the post you
>> are actually replying to, so that threading works correctly.
>>
>> I never used 0.4. for j=0:4 is a loop of j from 0 to 4.
> It is a way to
>> make the "next 5 values" the same as the input.
>>
>> Copy of the code:
>>
>> while(i < length(pressure)-5)
>> if( <your condition> )
>> for j=0:4
>> pressure(i+j) = <old value>;
>> end
>> i = i + 5;
>> else
>> apply adjustment to current value
>> i = i + 1;
>> end
>> end
>>
>>
>> -Peter
>

--

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)) )
    for j=0:4
      A(i+j) = A(i-1);
    end
    i = i + 5;
  else
    (A(i+1)-A(i))+A(i-1)
    i = i + 1;
  end
end


Peter Boettcher <boettcher@ll.mit.edu> wrote in message
<muymynhlt8x.fsf@G99-Boettcher.llan.ll.mit.edu>...
> "Adam Bright" <abc_services_2000@yahoo.co.uk> writes:
>
> > I'm sorry Peter,i'm extremely grateful for your help,but I
> > can't get it to work.
> >
> > while(i < length(pressure)-5)
> > if( ((pressure(i) <
> > pressure(i-1))&&((pressure(i-1)-pressure(i))>=150)) )
> > for j=0:4
> > pressure(i+j) = pressure(i-1);
> > end
> > i = i + 5;
> > else
> > apply adjustment to current value
> > i = i + 1;
> > end
> > end
> >
> > I'm not too sure what I would put where you have written
> > apply adjustment to current value, would this be
> > ((pressure(i+6)-pressure(i+5)) + pressure(i))?
> > i've tried a few variations but have'nt got it to work.
>
> Sorry, I can't help further. Pretend you're a computer,
remember that i
> will always point at the current *output* value and
step-by-step go
> through it. It might help to do an example by hand, at
all times
> keeping track of the value of i, and figuring out what
offsets against i
> you need to apply to get to the numbers you need.
>
> -Peter
>
>
> > Peter Boettcher <boettcher@ll.mit.edu> wrote in message
> > <muytzhqkjwb.fsf@G99-Boettcher.llan.ll.mit.edu>...
> >> "Adam Bright" <abc_services_2000@yahoo.co.uk> writes:
> >>
> >> > thanks peter for your reply,I kind of see how this would
> >> > work,however I dont understand why you have given
j=0.4 the
> >> > difference between the two will not always be the same,
> >> > however I will always want the current value and the
next 5
> >> > values to use the 1st previous value and thereafter
to use
> >> > this value + the differences gained .
> >> >
> >> > Please could you explain further so i understand it
> >> > better.many thanks once again
> >>
> >> Please quote from previous posts for context, and reply to
> > the post you
> >> are actually replying to, so that threading works
correctly.
> >>
> >> I never used 0.4. for j=0:4 is a loop of j from 0 to 4.
> > It is a way to
> >> make the "next 5 values" the same as the input.
> >>
> >> Copy of the code:
> >>
> >> while(i < length(pressure)-5)
> >> if( <your condition> )
> >> for j=0:4
> >> pressure(i+j) = <old value>;
> >> end
> >> i = i + 5;
> >> else
> >> apply adjustment to current value
> >> i = i + 1;
> >> end
> >> end
> >>
> >>
> >> -Peter
> >
>
> --

In article <futm6u$6tc$1@fred.mathworks.com>,
Adam Bright <abc_services_2000@yahoo.co.uk> 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

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@ibd.nrc-cnrc.gc.ca (Walter Roberson) wrote in
message <futmvj$ncp$1@canopus.cc.umanitoba.ca>...
> In article <futm6u$6tc$1@fred.mathworks.com>,
> Adam Bright <abc_services_2000@yahoo.co.uk> 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