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

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Gali Musa 2017-8-19

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评论： Jan 2017-8-30

采纳的回答： Walter Roberson

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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

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Walter Roberson 2017-8-19

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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"

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Stephen23 2017-8-29

编辑：Stephen23 2017-8-29

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"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 2017-8-30

@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.

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Star Strider 2017-8-19

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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

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Walter Roberson 2017-8-19

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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 2017-8-20

编辑：Jan 2017-8-21

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

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John BG 2017-8-19

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编辑：John BG 2017-8-20

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Hi Gali Musa

this is John BG <mailto:jgb2012@sky.com jgb2012@sky.com>

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

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:jgb2012@sky.com jgb2012@sky.com>

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Gali Musa 2017-8-30

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

Jan 2017-8-30

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.

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