Create a for cycle over an array
显示 更早的评论
In my case I have X different job with Y different duration
D = [2 3 4 4 5 6 7 3 ];
so job X=1 will last Y=2.
I need to write a code that give me the following result:
tin1= 0 , tfin1=2, tin2= 2 , tfin2=2+3=5, tin3= 5 , tfin3=5+4=9, tin4=9 , tfin4=9+4=13........ So "tinx" means the inital time of job x, while "tfinx" means the final time of job x. if job one finish at 2, the job two will finish at 2 plus the duration of the job 2 (3) obtaining 5, and so on.
I want to write a code that is able to generate all this value
Does someone help me? the length of D should be variable so I would like to obtain a code that is independent from its length
2 个评论
Jan
2019-9-4
I've cleaned the tags. It is not useful to use a pile of tags, which have no relation to your problem.
Steven Lord
2019-9-4
I'm trimming more of the tags.
采纳的回答
Do not create a bunch of variables with an index hidden in the names: See TUTORIAL: Why and how to avoid Eval
I assume all you need is:
D = [2 3 4 4 5 6 7 3 ]
T = cumsum([0, D])
Now e.g. T(1) and T(2) contain everything you want. This is much better than tin(1) and tfin(1). If you really need 2 variables:
Tini = T(1:end-1);
Tfin = T(2:end);
5 个评论
luca
2019-9-4
Fabio Freschi
2019-9-4
why?
Len = size(D, 2);
s = 0;
for k = 1:Len
tini(k) = s;
tfin(k) = D(k) + s;
s = tfin(k);
end
Again: Do not hide indices in the names of the variables, but use indices for this job.
Or:
Len = numel(D);
tini = zeros(1, Len); % Pre-allocation
tfin = zeros(1, Len);
s = 0;
for k = 1:Len
tini(k) = s;
s = s + D(k);
tfin(k) = s;
end
luca
2019-9-4
luca
2019-9-4
更多回答（2 个）
darova
2019-9-4
It is a job for cumsum() function:
D = [2 3 4 4 5 6 7 3 ];
tin1 = [0 2 5 9 13 18 24 31];
tin2 = [2 5 9 13 18 24 31 34];]
Fabio Freschi
2019-9-4
Try this
>> tfin = cumsum(D)
tfin =
2 5 9 13 18 24 31 34
>> tin = [0 tfin(1:end-1)]
tin =
0 2 5 9 13 18 24 31
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