Delete selected rows in uitable

Delete selected rows in uitable - Matlab

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I built in GUI (using GUIDE) a uitable (4x5) that last row is logical so I can select lines to delete.

d = {'L1',1,10,true;'L2',2,20,true;'L3',3,30,false;'L4',4,40,true;'L4',5,50,false};

set(handles.outputTable,'Data',d)

I created a button to delete the selectd rows but I does not work:

function deleteButton_Callback(hObject, eventdata, handles)

% hObject    handle to deleteButton (see GCBO)

% eventdata  reserved - to be defined in a future version of MATLAB

% handles    structure with handles and user data (see GUIDATA)

dataTable = (get(handles.outputTable,'data'));



[m n] = size(dataTable);



disp(dataTable);



for i = 1:m

    if num2str(cell2mat(dataTable(i,4))) =='1'    

      dataTable(i,:)=;

    end

end



disp('Modifed table')

disp(dataTable);

How can I fix it so I get set the table again in the GUI?

asked Nov 16 '18 at 13:45

wrek

1148

(1) You first get the data from the uitable in your variable dataTable =>ok. Then (2) you modify the dataTable according to your needs => ok. Now you just need to complete the process: (3) push your dataTable back into the uitable, something like set(handles.outputTable,'data',dataTable).

– Hoki
Nov 16 '18 at 14:18

I get an error with the loop, the rows are not deleted. Any idea why?

– wrek
Nov 16 '18 at 14:28

you don't need num2str and cell2mat to just check for a boolean value in a cell array. You could try just checking for if dataTable{i,4} ; dataTable(i,:)=; end. Note the use of the curly braces: {...} instead of the parenthesis, when you want to retrieve the content of a given cell.

– Hoki
Nov 16 '18 at 14:42

add a comment |

I built in GUI (using GUIDE) a uitable (4x5) that last row is logical so I can select lines to delete.

d = {'L1',1,10,true;'L2',2,20,true;'L3',3,30,false;'L4',4,40,true;'L4',5,50,false};

set(handles.outputTable,'Data',d)

I created a button to delete the selectd rows but I does not work:

function deleteButton_Callback(hObject, eventdata, handles)

% hObject    handle to deleteButton (see GCBO)

% eventdata  reserved - to be defined in a future version of MATLAB

% handles    structure with handles and user data (see GUIDATA)

dataTable = (get(handles.outputTable,'data'));



[m n] = size(dataTable);



disp(dataTable);



for i = 1:m

    if num2str(cell2mat(dataTable(i,4))) =='1'    

      dataTable(i,:)=;

    end

end



disp('Modifed table')

disp(dataTable);

How can I fix it so I get set the table again in the GUI?

asked Nov 16 '18 at 13:45

wrek

1148

(1) You first get the data from the uitable in your variable dataTable =>ok. Then (2) you modify the dataTable according to your needs => ok. Now you just need to complete the process: (3) push your dataTable back into the uitable, something like set(handles.outputTable,'data',dataTable).

– Hoki
Nov 16 '18 at 14:18

I get an error with the loop, the rows are not deleted. Any idea why?

– wrek
Nov 16 '18 at 14:28

you don't need num2str and cell2mat to just check for a boolean value in a cell array. You could try just checking for if dataTable{i,4} ; dataTable(i,:)=; end. Note the use of the curly braces: {...} instead of the parenthesis, when you want to retrieve the content of a given cell.

– Hoki
Nov 16 '18 at 14:42

add a comment |

I built in GUI (using GUIDE) a uitable (4x5) that last row is logical so I can select lines to delete.

d = {'L1',1,10,true;'L2',2,20,true;'L3',3,30,false;'L4',4,40,true;'L4',5,50,false};

set(handles.outputTable,'Data',d)

I created a button to delete the selectd rows but I does not work:

function deleteButton_Callback(hObject, eventdata, handles)

% hObject    handle to deleteButton (see GCBO)

% eventdata  reserved - to be defined in a future version of MATLAB

% handles    structure with handles and user data (see GUIDATA)

dataTable = (get(handles.outputTable,'data'));



[m n] = size(dataTable);



disp(dataTable);



for i = 1:m

    if num2str(cell2mat(dataTable(i,4))) =='1'    

      dataTable(i,:)=;

    end

end



disp('Modifed table')

disp(dataTable);

How can I fix it so I get set the table again in the GUI?

asked Nov 16 '18 at 13:45

wrek

1148

I built in GUI (using GUIDE) a uitable (4x5) that last row is logical so I can select lines to delete.

d = {'L1',1,10,true;'L2',2,20,true;'L3',3,30,false;'L4',4,40,true;'L4',5,50,false};

set(handles.outputTable,'Data',d)

I created a button to delete the selectd rows but I does not work:

function deleteButton_Callback(hObject, eventdata, handles)

% hObject    handle to deleteButton (see GCBO)

% eventdata  reserved - to be defined in a future version of MATLAB

% handles    structure with handles and user data (see GUIDATA)

dataTable = (get(handles.outputTable,'data'));



[m n] = size(dataTable);



disp(dataTable);



for i = 1:m

    if num2str(cell2mat(dataTable(i,4))) =='1'    

      dataTable(i,:)=;

    end

end



disp('Modifed table')

disp(dataTable);

How can I fix it so I get set the table again in the GUI?

matlab

asked Nov 16 '18 at 13:45

wrek

1148

asked Nov 16 '18 at 13:45

wrek

1148

asked Nov 16 '18 at 13:45

wrek

1148

asked Nov 16 '18 at 13:45

wrek

1148

asked Nov 16 '18 at 13:45

wrek

1148

(1) You first get the data from the uitable in your variable dataTable =>ok. Then (2) you modify the dataTable according to your needs => ok. Now you just need to complete the process: (3) push your dataTable back into the uitable, something like set(handles.outputTable,'data',dataTable).

– Hoki
Nov 16 '18 at 14:18

I get an error with the loop, the rows are not deleted. Any idea why?

– wrek
Nov 16 '18 at 14:28

you don't need num2str and cell2mat to just check for a boolean value in a cell array. You could try just checking for if dataTable{i,4} ; dataTable(i,:)=; end. Note the use of the curly braces: {...} instead of the parenthesis, when you want to retrieve the content of a given cell.

– Hoki
Nov 16 '18 at 14:42

add a comment |

(1) You first get the data from the uitable in your variable dataTable =>ok. Then (2) you modify the dataTable according to your needs => ok. Now you just need to complete the process: (3) push your dataTable back into the uitable, something like set(handles.outputTable,'data',dataTable).

– Hoki
Nov 16 '18 at 14:18

I get an error with the loop, the rows are not deleted. Any idea why?

– wrek
Nov 16 '18 at 14:28

you don't need num2str and cell2mat to just check for a boolean value in a cell array. You could try just checking for if dataTable{i,4} ; dataTable(i,:)=; end. Note the use of the curly braces: {...} instead of the parenthesis, when you want to retrieve the content of a given cell.

– Hoki
Nov 16 '18 at 14:42

(1) You first get the data from the uitable in your variable dataTable =>ok. Then (2) you modify the dataTable according to your needs => ok. Now you just need to complete the process: (3) push your dataTable back into the uitable, something like set(handles.outputTable,'data',dataTable).

– Hoki
Nov 16 '18 at 14:18

I get an error with the loop, the rows are not deleted. Any idea why?

– wrek
Nov 16 '18 at 14:28

you don't need num2str and cell2mat to just check for a boolean value in a cell array. You could try just checking for if dataTable{i,4} ; dataTable(i,:)=; end. Note the use of the curly braces: {...} instead of the parenthesis, when you want to retrieve the content of a given cell.

– Hoki
Nov 16 '18 at 14:42

add a comment |

1 Answer
1

active

oldest

votes

This is wrong:

for i = 1:m

    if num2str(cell2mat(dataTable(i,4))) =='1'    

      dataTable(i,:)=;

    end

end

First of all, if num2str(cell2mat(dataTable(i,4))) =='1' is a convoluted equivalent of if dataTable{i,4}==1. You should learn to use curly braces {} to access the content of a cell array.

Then, it will work only if the counter is decreased.
See what happens:

Test if row n should be deleted

Delete line n; the content of row (n+1) have now moved to row n

Increment counter i from value n to n+1

The row now at position n has never been tested for deletion !

What was at row (n+1) is never tested, since the delete operation moves it backwards first, then the counter is incremented without testing again. The solution is to decrement the counter.

for i = m:-1:1

    if dataTable{i,4}   

      dataTable(i,:)=;

    end

end

The rows moved by the deletion operation have already been tested, so in the end it is certain that all lines will have been tested.

Now, the same can be obtained in a vectorized form with a single line:

dataTable = dataTable(cell2mat(dataTable(:,4))==0,:);

The whole function boils down to:

function deleteButton_Callback(hObject, eventdata, handles)

    dataTable = get(handles.outputTable,'data');

    % Do some checks to make sure that the values input by are correct %

    assert(all(cellfun(@isscalar,dataTable(:,4))), 'Last colum should contain scalars!');

    set(handles.outputTable,'data' , dataTable(cell2mat(dataTable(:,4))==0,:));

end

answered Nov 16 '18 at 15:25

Brice

1,416110

add a comment |

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1 Answer
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1 Answer
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oldest

votes

This is wrong:

for i = 1:m

    if num2str(cell2mat(dataTable(i,4))) =='1'    

      dataTable(i,:)=;

    end

end

First of all, if num2str(cell2mat(dataTable(i,4))) =='1' is a convoluted equivalent of if dataTable{i,4}==1. You should learn to use curly braces {} to access the content of a cell array.

Then, it will work only if the counter is decreased.
See what happens:

Test if row n should be deleted

Delete line n; the content of row (n+1) have now moved to row n

Increment counter i from value n to n+1

The row now at position n has never been tested for deletion !

What was at row (n+1) is never tested, since the delete operation moves it backwards first, then the counter is incremented without testing again. The solution is to decrement the counter.

for i = m:-1:1

    if dataTable{i,4}   

      dataTable(i,:)=;

    end

end

The rows moved by the deletion operation have already been tested, so in the end it is certain that all lines will have been tested.

Now, the same can be obtained in a vectorized form with a single line:

dataTable = dataTable(cell2mat(dataTable(:,4))==0,:);

The whole function boils down to:

function deleteButton_Callback(hObject, eventdata, handles)

    dataTable = get(handles.outputTable,'data');

    % Do some checks to make sure that the values input by are correct %

    assert(all(cellfun(@isscalar,dataTable(:,4))), 'Last colum should contain scalars!');

    set(handles.outputTable,'data' , dataTable(cell2mat(dataTable(:,4))==0,:));

end

answered Nov 16 '18 at 15:25

Brice

1,416110

add a comment |

This is wrong:

for i = 1:m

    if num2str(cell2mat(dataTable(i,4))) =='1'    

      dataTable(i,:)=;

    end

end

First of all, if num2str(cell2mat(dataTable(i,4))) =='1' is a convoluted equivalent of if dataTable{i,4}==1. You should learn to use curly braces {} to access the content of a cell array.

Then, it will work only if the counter is decreased.
See what happens:

Test if row n should be deleted

Delete line n; the content of row (n+1) have now moved to row n

Increment counter i from value n to n+1

The row now at position n has never been tested for deletion !

What was at row (n+1) is never tested, since the delete operation moves it backwards first, then the counter is incremented without testing again. The solution is to decrement the counter.

for i = m:-1:1

    if dataTable{i,4}   

      dataTable(i,:)=;

    end

end

The rows moved by the deletion operation have already been tested, so in the end it is certain that all lines will have been tested.

Now, the same can be obtained in a vectorized form with a single line:

dataTable = dataTable(cell2mat(dataTable(:,4))==0,:);

The whole function boils down to:

function deleteButton_Callback(hObject, eventdata, handles)

    dataTable = get(handles.outputTable,'data');

    % Do some checks to make sure that the values input by are correct %

    assert(all(cellfun(@isscalar,dataTable(:,4))), 'Last colum should contain scalars!');

    set(handles.outputTable,'data' , dataTable(cell2mat(dataTable(:,4))==0,:));

end

answered Nov 16 '18 at 15:25

Brice

1,416110

add a comment |

This is wrong:

for i = 1:m

    if num2str(cell2mat(dataTable(i,4))) =='1'    

      dataTable(i,:)=;

    end

end

First of all, if num2str(cell2mat(dataTable(i,4))) =='1' is a convoluted equivalent of if dataTable{i,4}==1. You should learn to use curly braces {} to access the content of a cell array.

Then, it will work only if the counter is decreased.
See what happens:

Test if row n should be deleted

Delete line n; the content of row (n+1) have now moved to row n

Increment counter i from value n to n+1

The row now at position n has never been tested for deletion !

What was at row (n+1) is never tested, since the delete operation moves it backwards first, then the counter is incremented without testing again. The solution is to decrement the counter.

for i = m:-1:1

    if dataTable{i,4}   

      dataTable(i,:)=;

    end

end

The rows moved by the deletion operation have already been tested, so in the end it is certain that all lines will have been tested.

Now, the same can be obtained in a vectorized form with a single line:

dataTable = dataTable(cell2mat(dataTable(:,4))==0,:);

The whole function boils down to:

function deleteButton_Callback(hObject, eventdata, handles)

    dataTable = get(handles.outputTable,'data');

    % Do some checks to make sure that the values input by are correct %

    assert(all(cellfun(@isscalar,dataTable(:,4))), 'Last colum should contain scalars!');

    set(handles.outputTable,'data' , dataTable(cell2mat(dataTable(:,4))==0,:));

end

answered Nov 16 '18 at 15:25

Brice

1,416110

This is wrong:

for i = 1:m

    if num2str(cell2mat(dataTable(i,4))) =='1'    

      dataTable(i,:)=;

    end

end

First of all, if num2str(cell2mat(dataTable(i,4))) =='1' is a convoluted equivalent of if dataTable{i,4}==1. You should learn to use curly braces {} to access the content of a cell array.

Then, it will work only if the counter is decreased.
See what happens:

Test if row n should be deleted

Delete line n; the content of row (n+1) have now moved to row n

Increment counter i from value n to n+1

The row now at position n has never been tested for deletion !

What was at row (n+1) is never tested, since the delete operation moves it backwards first, then the counter is incremented without testing again. The solution is to decrement the counter.

for i = m:-1:1

    if dataTable{i,4}   

      dataTable(i,:)=;

    end

end

The rows moved by the deletion operation have already been tested, so in the end it is certain that all lines will have been tested.

Now, the same can be obtained in a vectorized form with a single line:

dataTable = dataTable(cell2mat(dataTable(:,4))==0,:);

The whole function boils down to:

function deleteButton_Callback(hObject, eventdata, handles)

    dataTable = get(handles.outputTable,'data');

    % Do some checks to make sure that the values input by are correct %

    assert(all(cellfun(@isscalar,dataTable(:,4))), 'Last colum should contain scalars!');

    set(handles.outputTable,'data' , dataTable(cell2mat(dataTable(:,4))==0,:));

end

answered Nov 16 '18 at 15:25

Brice

1,416110

answered Nov 16 '18 at 15:25

Brice

1,416110

answered Nov 16 '18 at 15:25

Brice

1,416110

answered Nov 16 '18 at 15:25

Brice

1,416110

add a comment |

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