How can I separate the data into two groups.?

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zhoug zho
zhoug zho el 21 de Mayo de 2021
Respondida: Kartikay Sapra el 21 de Mayo de 2021
How can I separate the data into two groups.
  1. If the first element of a column is largest number of that column, such type of columns are placed into one group.
  2. If the largest number is not at the first position of that column and is present as 2nd 3rd 4thor any other position of that column, then such columns are placed in one group.
I need to form two groups of this data. Screenshot of data is attached. I highlighted few coloumn's largest number.
Thanks in advance.
  3 comentarios
zhoug zho
zhoug zho el 21 de Mayo de 2021
Hi Rik, i am trying to understand but could not get, where to start.
Daniel Bengtson
Daniel Bengtson el 21 de Mayo de 2021
So you basically need to compare the first element in each column against the maximum value in each column. If they are equal then that column goes in the first group, otherwise the column would go in the second group.

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David Fletcher
David Fletcher el 21 de Mayo de 2021
Editada: David Fletcher el 21 de Mayo de 2021
%Get index of max values in each column
[val idx]=max(data)
%Create logical vector of max values where the max value is the first in the
%column
gp1_idx=(idx==1)
%Use logical vector to index out columns for one group
group1=data(:,gp1_idx)
%Use (not)logical vector to index out columns of the other group
group2=data(:,~gp1_idx)
  2 comentarios
zhoug zho
zhoug zho el 21 de Mayo de 2021
got it. thanks
David Fletcher
David Fletcher el 21 de Mayo de 2021
Editada: David Fletcher el 21 de Mayo de 2021
If you are ignoring zeros why have you got 0 marked as the max in the first column? Not sure I get it.

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Kartikay Sapra
Kartikay Sapra el 21 de Mayo de 2021
data = [-1 -2 -1; 0 -1 0; 0 0 0]
data(data==0) = NaN
[maxVal ids] = max(data)
data(isnan(data))=0
part_a = data(1:end,ids==1)
part_b = data(1:end,~(ids==1))
When finding the maximum, just keep the indices of non-zero values by making zeros 'NaN'. After that, NaN values dont interfere with max/min so after retrieving the indices, change NaN back to 0.

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