How can i use interval with decimal at the end of a count in for loop. e.g. t=1:1:18.7

How can i use interval with decimal at the end of a count in for loop. e.g. t=1:1:18.7

intervals = 18.7;
for t=1:intervals;
    x(t+1)=Pd*t;
end

Answer 1

[Walter Roberson]

Your existing code works, executing with t set to 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, and 18. Then adding 1 to that would exceed 18.7 so there are no more iterations. The behavior is well defined.

What you need to watch out for is that if you are using a non-integer increment: https://www.mathworks.com/help/matlab/ref/colon.html

"x = j:i:k creates a regularly-spaced vector x using i as the increment between elements. The vector elements are roughly equal to [j,j+i,j+2*i,...,j+m*i] where m = fix((k-j)/i). However, if i is not an integer, then floating point arithmetic plays a role in determining whether colon includes the endpoint k in the vector, since k might not be exactly equal to j+m*i"

[Gali Musa]

Thank you for the useful suggestion, what i'm trying to do is to calculate the area under a curve at each iteration, i found out when ever i got decimal answers for the intervals (i.e intervals=hrs/Pd where hrs=8640 and Pd=456) it truncate before the last iteration (18th) and stop at 8208hrs. though the last one has decimal. is there a way to use this. I tried all the codes suggested and no one solved the problem.

[Stephen23]

@Gali Musa: do you want the last step to possibly have a non-standard step size? If "yes", then you will have to add this on the end as by definition the colon operator uses the same step size between all of its output values (and clearly an integer begin and step will not lead to a non-integer end value).

[Walter Roberson]

It sounds to me as if you might be using a non-integer step size, and that you are getting hit by "floating point arithmetic plays a role in determining whether colon includes the endpoint k in the vector"

If you have a non-integer step size intended to break an interval up into a fixed number of steps, then there are two patterns you can use:

1) Use linspace() like Star Strider shows. Just remember that the count you give to linspace() includes the first and last points, so for example linspace(0,1,3) does not break the interval [0, 1] into thirds and instead gives you [0, 1/2, 1]. The equivalent to 0:1/N:1 is not linspace(0,1,N): it is linspace(0,1,N+1)

2) Use integer indices in the for loop and divide them by the appropriate factor before using them. For example,

for tidx = 0 : 3
   t = tidx ./ 3;
   ...
   output(tidx+1) = ...
end

to have t be set to the four values 0/3, 1/3, 2/3, 3/3

[Daniel Shub]

I asked about a related issue a while back:

https://www.mathworks.com/matlabcentral/answers/53912-floating-point-accurarcy-and-colon

I don't think I would describe colon as being well defined and definitely not well documented. It seems to do different things depending on if everything is an integer:

http://web.archive.org/web/20120213165003/http://www.mathworks.com/support/solutions/en/data/1-4FLI96/index.html

From your quote of the documentation, I would expect 0:0.01:1 to end at approximately 0.99 and not 1.

[Stephen23]

"From your quote of the documentation, I would expect 0:0.01:1 to end at approximately 0.99 and not 1."

Why? 0.1 is represented by a binary value slightly larger than 0.1:

>> sprintf('%.20f\n',0.1)
ans = 0.10000000000000000555

and so it makes perfect sense that the end point would include 1, even taking into account floating point error. The whole point is that floating point error should not be relied up to behave in particular ways (e.g. always rounding down, as you assumed). Especially considering that calculations may also use guard digits, and different platforms may returns different results of the same precision.

[Jan]

@Daniel: According to the documentation colon creates the 1st half by addition to the start value, and the 2nd half by subtracting from the final value. Therefore the last value is exactly the specified number. I think this is exactly defined and documented.

Answer 2

[Star Strider]

It depends on what you want to do. The second value in your colon (link) operator syntax is the step-size. Your ‘t’ vector will go from 1 to 18 and stop, because the next step-size is 1, so the vector will not exceed the end value, 18.7.

Indices in MATLAB must be integers greater than zero.

I am not certain what you want to do. One option would be to use the linspace function to create the vector, then step through the vector’s length.

If ‘Pd’ is a scalar, this will work:

tv = linspace(1, 18.7, 20);                             % Create Vector Of Length ‘20’ From ‘1’ To ‘18.7’
intervals = 1:length(tv);
for t = intervals
    x(t+1)=Pd*tv(t);
end

[Walter Roberson]

I often express this as the general framework

tvals = .....  %list of actual values
num_tvals = length(tvals);
output = zeros(1, num_tvals);
for tidx = 1 : num_tvals;
  t = tvals(tidx);
  ...
  output(tidx) = ...
end

[Jan]

I prefer Walter's suggestion. It uses the faster "a:b" indexing in the for loop and allows to pre-allocate the output.

Answer 3

[John BG]

Hi Gali Musa

this is John BG <mailto:[e-mail] [e-mail]>

May be you would like to consider using an additional reference vector

nt

nt contains t vector indices and it's used as the for counter, not t or intervals directly

N=20
t = linspace(1, 18.7, N);
nt=[1:1:N];          % reference vector
dt=abs(t(1)-t(2))    % basic interval 

for k=nt
    x(nt(k)+1)=Pd*t(nt(k));
end

if you find this answer useful would you please be so kind to consider marking my answer as Accepted Answer?

To any other reader, if you find this answer useful please consider clicking on the thumbs-up vote link

thanks in advance

John BG

<mailto:[e-mail] [e-mail]>

[Walter Roberson]

t(k) is going to generally be non-integer. t(t(k)) is therefore generally going to be an error.

[John BG]

Thanks Walter, corrected, dt not used, but just in case it's needed for instance approximated differentiation.

[Walter Roberson]

Why bother taking

t(nt(k)

Why not just

t(k)

?

[John BG]

it looks like x is not just a vector but a function that does take non integers.

Therefore it is possible x(t) with for instance t=18.7

Gali attempts to use t at the same time as

• for loop counter
• x function input
• values in equation including x(t) and other terms

I agree that an integer index helps solve the question, so I suggested to add a reference vector.

[Walter Roberson]

On the other hand, if x were a function such that you could meaningfully say x(18.7), then you would have to construct it in the form

x = symfun(SOMETHING)

or in the form

x(SYMBOLIC_VARIABLE_NAME) = expression

where SYMBOLIC_VARIABLE_NAME would have to evaluate to a scalar symbolic variable name, not to a numeric value and not to a symbolic expression or symbolic numeric value. So, you could once do

syms t
x(t) = Pd*(t-1)

and then x(18.7) would be completely valid. But you would not be able to do this in a for loop with numeric or symbolic-numeric t values.

There could be places where it could be useful to define

x = @(t) Pd*(t-1)

provided that x was only being used as a formula and not as a vector.

[Gali Musa]

sorry for the late reply i thought i will receive a notification for the update. Could you explain how to use the recent code sent. however

intervals = hrs/Pd;
hrs=8640;
Pd= 456;    %%%this value changes for differnt environment, ranges (72hrs to 480hrs).

when i use the values like for Pd=72,120,240 and 480hrs it calculates the areas correctly, however when i use Pd= 456,408 and so on............ it truncate the area. is there a way to add up to the 8640hrs when ever it has a decimal answer so that the last iteration could be an integer.

[Stephen23]

"Could you explain how to use the recent code sent."

@Gali Musa: who sent you some code? Where to? What was the code?

[Gali Musa]

i'm referring to the symbolic variable expression sent by Walter Roverson. actually i want to calculate the area under a curve and the following are the period (Pd) and the hours of operation is (hrs). using

intervals = hrs/Pd;
hrs=8640;
%%%the value of Pd changes for different environment, ranges (72 to 480) using Pd= 456;  then intervals= 8640/456=18.95.

intervals = 18.95;
for t=1:intervals;
    x(t+1)=Pd*t;
end

is there a way to either add up to the 8640hrs when ever it has a decimal answer so that the last iteration could be an integer or to truncate it at exactly 8640hrs. based on our previous discussion its not possible to count 19 iteration since it is 18.95 and is not a whole number.

[Stephen23]

"is there a way to either add up to the 8640hrs when ever it has a decimal answer so that the last iteration could be an integer or to truncate it at exactly 8640hrs. based on our previous discussion its not possible to count 19 iteration since it is 18.95 and is not a whole number."

All of these things are possible, as the previous discussions explain:

• to truncate down to 18: use colon.
• all steps use (likely) non-integer step size: use linspace.
• only last step uses non-integer step size: fiddly, but possible.
• etc...

If your really want the 19 iterations and are happy if the last step is not an integer then you need to tell us this. So far you have not made it clear which possible solution you want. The only thing missing is you telling us what you want.

[Gali Musa]

i tried the possible cases and i found out it does not exactly suite my objective. i also think of this is another important step to solve the problem. IS IT POSSIBLE TO HAVE A CODE THAT CAN PICK ONLY THE NON DECIMAL ITERATION FOR THE CALCULATION. example where Pd= 72:480, hrs=8640, intervals=hrs/Pd. Thank you for your support Thank you

[Jan]

@Gali: The discussion is strange. You have been asked for explaining, what you want to achieve. Several persons offered some help, but as long as it is not clear, what you want to achieve and obtain as result, the suggestions are unlikely to match exactly.

Please, Gali, post, what you want as result. Do you want to create a vector, or a function? Should the steps be equidistant or do you want to append one partial step for the last element only?

[Stephen23]

"IS IT POSSIBLE TO HAVE A CODE THAT CAN PICK ONLY THE NON DECIMAL ITERATION FOR THE CALCULATION"

Yes.

[Gali Musa]

this exactly want to do now

hrs= 8640;
Pd= 72:480
intervals =hrs/Pd;
for t = 1:intervals
x(t+1) = Pd*t
end

i want the code to pick only the non-decimal of the ratio of

intervals = hrs/Pd = 8640/(72 to 480)

could you please help me. Thank you COUNT

for i =1:length(Pd)
    if Pd(i,1) = decimal
        cnt =0;
    elseif Pd(i,1) =integer
        cnt=1
    end
end
end

Please help me on how to pick only integer. Thank you

[Walter Roberson]

for i =1:length(Pd)
  cnt(i) = mod(Pd(i,1), 1) ~= 0;
end

or equivalently,

cnt = mod(Pd, 1) ~= 0;

[Gali Musa]

i tried it but i noticed its still counting all the Pd. Could you please advice

[Jan]

See my answer. You can find out more details by your own. Check the value of mod(Pd, 1), then try mod(Pd, 1) ~= 0. I assume you want: sum(mod(Pd, 1) == 0, not ~=. This is a tiny detail and with some own effort it should be possible to solve it by your own.

Answer 4

[Jan]

Get the number of integers in the vector Pd:

sum(mod(Pd, 1) == 0)

After all these comments and questions for clarifications I still do not know, if this is the actual question or not.