Extract or keep rows based on an array.
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I have been trying multible things and researching on the MathWorks site for hours for what should be a very simple thing I believe. So, I have a table of data that contains three columns. In column one there is are values that are repeated but each row is unique. For example:
A = [1 'cat' 24; 1 'dog' 2; 2 'horse' 10; 2 'dog' 3; 3 'sheep' 45; 3 'cow' 203; 4 'cow' 233; 5 'pig' 23; 5 'walrus' 2; 6 'horse' 4]
And I have a vector of values B = [2 4 5]
I want to pull out or keep only the rows where the value in column one of A is the same as any one of the values in the vector B.
In short, want to parse down A from the above to a leaner version with only the data I need. A = [2 'horse' 10; 2 'dog' 3; 4 'cow' 233; 5 'pig' 23; 5 'walrus' 2]
My real example is 3,142 rows in size what should parse down to around 861.
回答（2 个）
Geoff Hayes
2017-8-1
Kellie - if we can assume that your A is constructed as a cell array like
A = {1 'cat' 24; 1 'dog' 2; 2 'horse' 0; 2 'dog' 3; 3 'sheep' 45; 3 'cow' 203; 4 'cow' 233; 5 'pig' 23; 5 'walrus' 2; 6 'horse' 4}
then we can find those rows whose first column element is in the set defined by B by evaluating
A(cell2mat(arrayfun(@(x)ismember(x,B), cell2mat(A(:,1)), 'UniformOutput', false)),:)
which returns
[2] 'horse' [ 0]
[2] 'dog' [ 3]
[4] 'cow' [233]
[5] 'pig' [ 23]
[5] 'walrus' [ 2]
In the above line of code, we extract the first column of A and convert it from a cell array into a matrix
cell2mat(A(:,1))
We then use arrayfun to evaluate each element of this column to see if it is a member of B
arrayfun(@(x)ismember(x,B), cell2mat(A(:,1)), 'UniformOutput', false)
Each call to ismember returns a logical zero (not a member) or one (is a member). We then convert this to a matrix and extract those rows of A that are members of B (so the logical zero means the row will be ignored).
Try the above and see what happens!
3 个评论
Kellie Anton
2017-8-1
Kellie Anton
2017-8-1
Geoff Hayes
2017-8-1
Kellie - it wasn't clear what your A was (to me) so I assumed that it could be written as a cell array. You can try converting your table into a cell array with table2cell and then trying the above code.
Andrei Bobrov
2017-8-1
A = {1 'cat' 24; 1 'dog' 2; 2 'horse' 10; 2 'dog' 3; 3 'sheep' 45; 3 'cow' 203; 4 'cow' 233; 5 'pig' 23; 5 'walrus' 2; 6 'horse' 4};
B = [2 4 5];
out = A(ismember([A{:,1}],B),:);
8 个评论
Kellie Anton
2017-8-1
Kellie Anton
2017-8-1
编辑：Kellie Anton
2017-8-1
Andrei Bobrov
2017-8-1
编辑：Andrei Bobrov
2017-8-1
@Kellie Anton: Please attach your data's example (mat - file).
>> A = {1 'cat' 24; 1 'dog' 2; 2 'horse' 10; 2 'dog' 3; 3 'sheep' 45; 3 'cow' 203; 4 'cow' 233; 5 'pig' 23; 5 'walrus' 2; 6 'horse' 4};
B = [2 4 5];
out = A(ismember([A{:,1}],B),:)
out =
5×3 cell array
[2] 'horse' [ 10]
[2] 'dog' [ 3]
[4] 'cow' [233]
[5] 'pig' [ 23]
[5] 'walrus' [ 2]
>>
Kellie Anton
2017-8-1
Andrei Bobrov
2017-8-1
out = poverty(ismember(str2double(poverty.State),rfstates),:)
Kellie Anton
2017-8-1
Kellie Anton
2017-8-1
Andrei Bobrov
2017-8-1
Here it is customary to "accepted" the answer that solve your problem...
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