How to delete a row of a table in a loop while indices are changing

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Johnny 2016-11-28

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评论： Johnny 2016-11-29

I would like to do a comparison of a value with a previous value of a variable which appears after every sixth values. And if this variable is the same, I would like to delete the whole row. But when I delete, the indices change and it no longer appears after sixth values. How do I delete a row in a loop when the indices are changing?

for i=7:length(table)
  if isequal(table(i,9),table(i-6),9)
     table(i,:)=[];
  end
end

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Answer 1

Ryan Smith 2016-11-28

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Your provided code is a little unclear in what you want to compare. Your conditional should always return 0, as it is:

if isequal(table(i,9),table(i-6))

Assuming you mean:

if isequal(table(i,9),table(i-6,9))

Furthermore, table(i,9) = []; just deletes a single cell of table, not the entire row.

Nonetheless, you might could try unique(). https://de.mathworks.com/help/matlab/ref/unique.html

But unique will compare your entire list/table. It might be best to just use a second array:

A = table(1:6,:);
j = 7;
for i=7:length(table)
  if table(i,9) ~= table(i-6,9)
    A(j,:)=table(i,:);
    j = j + 1;
  end
end

Hope this helps!

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Johnny 2016-11-28

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Thank you for the swift answer and sorry for errors in my code. However, unique is not what I want to do. I want to compare each entry with the next entry, and if it is the same i would like to delete the entire row. I have six variables, that's why i wanna do the comparison with the step of six. I am using

if true
   if isequal(table(i,9),table(i-6,9))
end

because table cells will not support operations of type '==' or '~='. Hence I could not test your solution. I feel like I should write again my approach without errors this time :).

for i=7:length(table)
  % condition testing the equlity with the previous entry
   if isequal(table(i,9),table(i-6))
   % deleting the entire column if the condition is true
   table(i,:)=[];
end

but as I said, after the first row is deleted the indices are modified , what's done is no longer what I would like to achieve. What do you think about that?

Johnny 2016-11-28

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Ok, so with some modifications I got what I want.

A=table(1:6,:);
j = 7;
for i=7:length(table)
if isequal(table(i,9), table(i-6,9))
else
  A(j,:)=table(i,:);
  j = j + 1;
end
end
Thanks mate!

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Answer 2

Guillaume 2016-11-28

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From your comments, I'm going go out on a limb and assume that your (very badly named) table variable is actually not a table but a cell array. It can't be a table since the length function is not defined for tables.

Even for matrix and cell array, where length is defined, you shouldn't be using it since it will return either the number of rows or columns depending on which is greater. Use size with an explicit dimension.

If you want to use a loop to delete your rows, you have to store which rows you want to delete in the loop and perform the deletion at the end of the loop.

%somevariable: your badly named table variable renamed to something better
deleterow = false(size(somevariable, 1)  %allocate array which tells us which row to delete
for row = 7: size(somevariable, 1)   %use size instead of length
    %to access the content of a cell of a cell array use {} not (). 
    %you can then compare with == assuming the content is scalar
    if somevariable{i, 9} == somevariable{i-6, 9}  %if not scalar then use isequal
        deleterow(row) = true;
    end
end
somevariable(deleterow, :) = [];   %and delete all rows at once

If the content of column 9 is scalar, you don't even need a loop:

deleterow = cell2mat(somevariable(1:end-7)) == cell2mat(somevariable(7:end));  
somevariable(deleterow, :) = [];

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Johnny 2016-11-29

Thank you for the correction of the terms, it is indeed a cell array. However, your approach was very slow and it did not work. It gave me in the end a cell array with only first six entries and deleted all the values.

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