Find number of row in a table.
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fyza affandi
el 24 de Dic. de 2018
I have the following table
tblB =
Symbol Count
______ _____
1 25
3 8
2 7
5 7
4 6
10 3
8 2
11 1
12 1
14 1
15 1
16 1
21 1
which is based on the matrix E1 below
E1 =
1 1 4 5 1 1 3 1
1 1 2 5 1 1 1 1
3 4 1 3 5 10 5 4
4 2 3 4 5 8 4 1
1 2 12 16 1 1 3 3
1 2 10 14 1 1 1 1
1 1 10 8 2 5 3 15
5 2 1 3 2 21 1 11
How can I find the number of row for the symbol of the matrix.?
For example for E1(1,1) is from row 1 of tblB and so on....
row = [1 1 5 4 1 1 2 1......]
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Image Analyst
el 24 de Dic. de 2018
Editada: Image Analyst
el 24 de Dic. de 2018
E1(1,1) is 1. So you want to find every row of tblB that has equals E1(1,1)?
Like
rowsWith1 = tblB{:, 1} == E1(1,1) | tblB{:, 2} == E1(1,1);
Please clarify. How are you getting the columns (values) in your row vector called "row" from tblB and E1(1,1)?
Respuesta aceptada
dpb
el 24 de Dic. de 2018
Preliminaries...
E1 =[1 1 4 5 1 1 3 1
1 1 2 5 1 1 1 1
3 4 1 3 5 10 5 4
4 2 3 4 5 8 4 1
1 2 12 16 1 1 3 3
1 2 10 14 1 1 1 1
1 1 10 8 2 5 3 15
5 2 1 3 2 21 1 11];
u=unique(E1);
n=histc(E1(:),u);
[n,ix]=sort(n,'descend');
t=table(u(ix),n,'VariableNames',{'Symbol','Count'});
Now the magic...
>> arrayfun(@(x) find(x==t.Symbol),E1)
ans =
1 1 5 4 1 1 2 1
1 1 3 4 1 1 1 1
2 5 1 2 4 6 4 5
5 3 2 5 4 7 5 1
1 3 9 12 1 1 2 2
1 3 6 10 1 1 1 1
1 1 6 7 3 4 2 11
4 3 1 2 3 13 1 8
>>
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dpb
el 24 de Dic. de 2018
Editada: dpb
el 25 de Dic. de 2018
No problem...glad to help. :)
BTW, be aware that the data of the table t is embedded in the anonymous function when the function is defined. Thus if the data in the table change, you must also redefine the function...that happens automagically as written but if you were to define a separate function handle first and then change the data, then the existing function handle wouldn't know anything about that new data...
Más respuestas (1)
KSSV
el 24 de Dic. de 2018
YOu have many functions to get that information. Read about find, ismember and logical sign ==. YOu can access the whole column of table tblB using.
tblB.(1)
tblB.(2)
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