isa not recognizing Line or Figure class

2 次查看(过去 30 天)
I wanted to test whether input data of a function is a Line object using something like:
>>
x=-2*pi:.01:2*pi;HLine=plot(sin(2*x));
However, now isa(HLine,'Line') returns false! So far, isa has faithfully recognized my self-made classes. Why doesn't it recognize Line?

采纳的回答

Guillaume
Guillaume 2018-8-10
>> x=-2*pi:.01:2*pi;HLine=plot(sin(2*x));
>> class(HLine)
ans =
'matlab.graphics.chart.primitive.Line'
>> isa(HLine, 'matlab.graphics.chart.primitive.Line')
ans =
logical
1
You need to use the fully qualified name of the class

更多回答(1 个)

KSSV
KSSV 2018-8-10
x=-2*pi:.01:2*pi;
HLine=plot(sin(2*x));
strcmp(HLine.Type,'line')
  2 个评论
Guillaume
Guillaume 2018-8-10
Note that the Type property of a graphics object is not relevant for isa.
KSSV
KSSV 2018-8-10
Yes....thats why I used strcmp to check is it a Line type.

请先登录,再进行评论。

类别

Help CenterFile Exchange 中查找有关 Data Type Identification 的更多信息

标签

产品


版本

R2018a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by