Appending vectors in a loop

Data1  =
   4
   5
   6
   7
   8
   9
  10
  11
  12
  -1
   4
   5
   6
   7
   8
   9
  10
  11
  12
  -1
   4
   5
   6
   7
   8
   9
  10
  11
  12
  -1
   4
   5
  6
   7
   8
   9
  10
  11
  12
  -1

Data2= Columns 1 through 2

    [ 4]    [0.0119]   
    [ 5]    [0.0119]   
    [ 6]    [0.0143]   
    [ 7]    [0.0143]
    [ 8]    [0.0187]   
    [ 9]    [0.0256]   
    [10]    [0.0273]   
    [11]    [0.0119]   
    [12]    [0.0143]

I want to be able to take in values from data2 for column 2 using the corresponding values in column 1 and data1 . I would use the following command:

find(ismember(cell2mat(data2),data1)==1);

The problem, if you look closely at the first set, is that there is a -1. I need to force it’s value to be zero. Thus I would have to append the vector.

If the command about is within a loop, how do I append data2 without prompting an error?

[Azzi Abdelmalek]

What are you expecting as result, you can make your example shorter

[T]

I am expecting:

Data2=

[ 4]    [0.0119]   
  [ 5]    [0.0119]   
  [ 6]    [0.0143]   
  [ 7]    [0.0143]
  [ 8]    [0.0187]   
  [ 9]    [0.0256]   
  [10]    [0.0273]   
  [11]    [0.0119]   
  [12]    [0.0143]  
  [-1]    [ 0 ]

and I want to be able to retrieve the second column in Data1 with each corresponding value in column 1 of Data2.

Answer 1

[Azzi Abdelmalek]

out=Data2(ismember(cell2mat(Data2(:,1)),Data1),:)

% to change values of Data1 from -1 to 0

Data1(Data1==-1)=0

[T]

So if I use

negativeID = [Data2; [{1},{0}]

That gives me

Data2=
  [ 4]    [0.0119]   
  [ 5]    [0.0119]   
  [ 6]    [0.0143]   
  [ 7]    [0.0143]
  [ 8]    [0.0187]   
  [ 9]    [0.0256]   
  [10]    [0.0273]   
  [11]    [0.0119]   
  [12]    [0.0143]  
  [-1]    [ 0 ]

However, I get an error when I use

out=negativeID(ismember(cell2mat(negativeID(:,1)),Data1),:)

It says All contets of the input cell array must be of the same data type.

[Azzi Abdelmalek]

You are missing a bracket

negativeID = [Data2; [{1},{0}]]

[T]

That was a typo but it still gives me the same result.

Nevertheless,

if I have

negativeID(:,1) = 
  [ 4]    
  [ 5]     
  [ 6]     
  [ 7]     
  [ 8]    
  [ 9]       
  [10]       
  [11]       
  [12]       
  [-1]     

and use cell2mat(negativeID(:,1))

then I get the error I mentioned, why would it work without the -1???

[T]

The problem is that with 4-12 is a different data type and -1 is just an integer when I appended the vector. How do I find out what data type the cell is?

[Azzi Abdelmalek]

Copy and past this code

Data1 ={ 4
 5
 6
 7
 8
 9
10
11
12
-1
 4
 5
 6
 7
 8
 9
10
11
12
-1
 4
 5
 6
 7
 8
 9
10
11
12
-1
 4
 5
6
 7
 8
 9
10
11
12
-1}
Data2={[ 4]    [0.0119]   
[ 5]    [0.0119]   
[ 6]    [0.0143]   
[ 7]    [0.0143]
[ 8]    [0.0187]   
[ 9]    [0.0256]   
[10]    [0.0273]   
[11]    [0.0119]   
[12]    [0.0143]  
[-1]    [ 0 ]}
negativeID = [Data2; [{1},{0}]]
a=negativeID(:,1);
out=Data2(ismember([a{:}],[Data1{:}]),:)

[T]

That works but I cannot get the indices for each value.

find(Data2(ismember([a{:}],[Data1{:}]),:)) says Undefined function or method 'find' for input arguments of type 'cell'.

[Azzi Abdelmalek]

find(ismember([a{:}],[Data1{:}]))

[T]

find(ismember([a{:}],[Data1{:}])==1) is what I needed, thanks.

Answer 2

[Andrei Bobrov]

Data2 =  [...
          4       0.0119
          5       0.0119
          6       0.0143
          7       0.0143
          8       0.0187
          9       0.0256
         10       0.0273
         11       0.0119
         12       0.0143]

out = Data1*[1 0];
[l,ii] = ismember(Data1,Data2(:,1));
out(l,2) = Data2(ii(l),2);

[T]

Where did you take into account -1?

[Andrei Bobrov]

in out = Data1*[1 0];