Scilab

Question

我正在尝试在 scilab 中创建一个简单的 GUI，仅用于显示函数 a'sin(x) 的图形，其中滑块驱动 "a" 的值。我似乎没有找到一种方法来仅显示与 "a" 当前值对应的曲线...它只是向现有曲线添加新曲线。我希望有一个简单的解决方案，因为我目前无法进一步了解（我只有一些 Mathematica 经验）。我使用的非常简单的代码是：

// This GUI file is generated by guibuilder version 3.0
//////////
f=figure('figure_position',[400,50],'figure_size',
[656,581],'auto_resize','on','background',[33],'figure_name','Graphic 
window number %d');
//////////
delmenu(f.figure_id,gettext('File'))
delmenu(f.figure_id,gettext('?'))
delmenu(f.figure_id,gettext('Tools'))
toolbar(f.figure_id,'off')
handles.dummy = 0;
handles.slider=uicontrol(f,'unit','normalized','BackgroundColor',
[-1,-1,-1],'Enable','on','FontAngle','normal','FontName','Tahoma',
'FontSize',[12],'FontUnits','points','FontWeight','normal',
'ForegroundColor',[-1,-1,-1],'HorizontalAlignment','left','ListboxTop',
[],'Max',[1],'Min',[0],'Position',[0.0359375,0.78125,0.2625,0.09375],
'Relief','default','SliderStep',
[0.01,0.1],'String','slider1','Style','slider','Value',
[0],'VerticalAlignment','middle','Visible','on','Tag','slider',
'Callback','slider_callback(handles)')
handles.graph= newaxes();handles.graph.margins = [ 0 0 0 
0];handles.graph.axes_bounds = [0.31875,0.01875,0.6640625,0.925];

//////////
// Callbacks are defined as below. Please do not delete the comments as 
it will be used in coming version
//////////

function slider_callback(handles)
//Write your callback for  slider  here
x=0:0.1:10;
a=handles.slider.Value;
plot(x,a*sin(x));
endfunction

预先感谢您的帮助。

Answer 1

默认情况下每个图形指令将其结果添加到当前显示如果要在显示新的图形指令之前自动清除图形，可以将轴的自动清除属性设置为 "on"。另一种解决方案是使用删除功能删除以前的绘图。

如果不希望显示刻度可以使用 xpoly 函数而不是 plot 函数

// This GUI file is generated by guibuilder version 3.0
//////////
f=figure('figure_position',[400,50],'figure_size',...
[656,581],'auto_resize','on','background',[33],'figure_name','Graphic window number %d');
//////////
delmenu(f.figure_id,gettext('File'))
delmenu(f.figure_id,gettext('?'))
delmenu(f.figure_id,gettext('Tools'))
toolbar(f.figure_id,'off')
handles.dummy = 0;
handles.slider=uicontrol(f,'unit','normalized','BackgroundColor',...
[-1,-1,-1],'Enable','on','FontAngle','normal','FontName','Tahoma',...
'FontSize',[12],'FontUnits','points','FontWeight','normal',...
'ForegroundColor',[-1,-1,-1],'HorizontalAlignment','left','ListboxTop',...
[],'Max',[1],'Min',[0],'Position',[0.0359375,0.78125,0.2625,0.09375],...
'Relief','default','SliderStep',...
[0.01,0.1],'String','slider1','Style','slider','Value',...
[0],'VerticalAlignment','middle','Visible','on','Tag','slider',...
'Callback','slider_callback(handles)')
handles.graph= newaxes();
handles.graph.margins = [ 0 0 0 0];
handles.graph.axes_bounds = [0.31875,0.01875,0.6640625,0.925];
handles.graph.data_bounds=[0 -1.2;10 1.2];
handles.graph.auto_clear="on";
//////////
// Callbacks are defined as below. Please do not delete the comments as 
//it will be used in coming version
//////////

function slider_callback(handles)
//Write your callback for  slider  here
x=0:0.1:10;
a=handles.slider.Value;
drawlater()
//delete(handles.graph.children)
xpoly(x,a*sin(x))
handles.graph.data_bounds=[0 -1.2;10 1.2];
drawnow
//plot(x,a*sin(x));
endfunction

Answer 2

您可以访问图形的折线并直接更改数据 (x,y)。在某处放置一个空（或虚拟）图（例如靠近 uicontrols 的定义）。

然后在回调函数中获取当前axes对象：

axes1=gca();// get the current axes

假设你只绘制了一条线，那么你的折线由 axes1.children.children 寻址（如果你绘制多条线，它们由 axes1.children.children(1), axes1.children.children(2 ) 等等）。

x 和 y 数据位于折线的 "data" 中，因此您只需更新此数据：

axes1.children.children.data=xymatrix([x1,y1;x2,y2;x3,y3;...xn,yn]);

您的回调函数可能如下所示：

function slider_callback(handles)
//Write your callback for  slider  here
x=zeros(101,2);//define matrix with 101 rows and 2 columns
x(:,1)=0:0.1:10;//fill first column with x axis values
a=handles.slider.Value;//get slider value
x(:,2)=a*sin(x(:,1));//fill second column with y axis values
axes1=gca();//get current axes
axes1.children.children.data=x;//update polyline data with new xy data
endfunction

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