Appending vectors in a loop

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T
T 2013-8-17
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?
  2 个评论
Azzi Abdelmalek
Azzi Abdelmalek 2013-8-17
What are you expecting as result, you can make your example shorter
T
T 2013-8-18
编辑：T 2013-8-18
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.

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采纳的回答

Azzi Abdelmalek
Azzi Abdelmalek 2013-8-17
编辑：Azzi Abdelmalek 2013-8-17
out=Data2(ismember(cell2mat(Data2(:,1)),Data1),:)
% to change values of Data1 from -1 to 0
Data1(Data1==-1)=0
  8 个评论
显示 6更早的评论
T
T 2013-8-19
编辑：T 2013-8-19
find(ismember([a{:}],[Data1{:}])==1) is what I needed, thanks.

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更多回答（1 个）

Andrei Bobrov
Andrei Bobrov 2013-8-18
编辑：Andrei Bobrov 2013-8-18
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);
  2 个评论
T
T 2013-8-18
Where did you take into account -1?

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