I Would like to create a variable array based upon user inputs.

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CodeJunkie
CodeJunkie 2019-1-28
评论： CodeJunkie 2019-1-29
I have a variable that the user inputs called Type allocated as such:
n=input('How many measurements are you determining uncertainty upon? ');
Type=size(n)';
Uo=1;
while Uo<=n;
if Uo==1;
Type(Uo)=input('What type of measurement is your first measurement? (Input 1,2,or 3 for Pressure,Temperature, and Flow Rate Respectively) ');
Uo=Uo+1;
elseif Uo>1;
Type(Uo)=input('What type of measurement is your next measurement? (Input 1,2,or 3 for Pressure,Temperature, and Flow Rate Respectively) ');
Uo=Uo+1;
end
end
I would like to use these values to create another variable matrix later in my code. I tried the following:
Factor=cell(n:1)
for iter = 1:n
if Type == 1
Factor{iter}=1
elseif Type == 2
Factor{iter}=1
elseif Type ==3
Factor{iter}= 4
elseif Type < 1
break
elseif Type > 3
break
end
end
Could someone shine some light into what I am doing wrong here? or is there a far better method I should be using to do this?
Thanks ahead of time for any help you can lend!
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Adam Danz
Adam Danz 2019-1-28
Also, could you provide some examples of inputs and expected outputs?
CodeJunkie
CodeJunkie 2019-1-29
Sorry if I was a bit unclear basically the expected output I would like is something along the lines of if Type= [1,2,2,3,3,2,1] I would like to set a Variable factor to read type upon each iteration and return some factor. e.g. Factor=[5,6,6,8,8,6,5]. I was also hoping to make the factor possibly an equation so if I could keep it in matrix form that may be ideal rather than the cell form its currently in as one of the comments pointed out

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Jan
Jan 2019-1-29
The code contains several strange commands.
Type=size(n)'
Are you sure? Without an explanation in the comment the readers can only guess, what you want. The question "How many ..." sound like you expect a scalar input, e.g. 5. Then size(5) replies [1,1] and transposing this is [1;1].
Later you fill the elements 1 to n of Type.
Factor=cell(n:1)
Do you mean:
Factor=cell(n, 1)
n:1 is the vector from n to 1, so most likely a scalar, if n>1. Then you create a {n x n} cell matrix.
When Type is a vector, them command if Type == 3 might not do, what you expect. if requires a scalar condition, therefore an all() is inserted implicitly. I guess you mean:
if Type(iter) == 3
  1 个评论
CodeJunkie
CodeJunkie 2019-1-29
yes i meant (n,1) sorry that was a typo in the code I wrote into here. you are also correct I should specify Type per iteration it seems my issue was that simple I appreciate your help vehemently!

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