Undefined unary operator '-' for input arguments of type 'string error

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Elnaz P
Elnaz P 2020년 7월 7일
댓글: Elnaz P 2020년 7월 8일
I am going to show x axis in scatter plot by text label like below with positive and negative signs however I get this error
"Undefined unary operator '-' for input arguments of type 'string' "
Can someone please assist in rectifying?
n= ["-240-260";"-220-240";"-200-220";"-180-200";"-160-180";"-140-160";"-120-140";"-100-120";"-80-100";"-60-80";"-40-60";"-20- 40";"0-20";"0+20";"+20+40";"+40+60";"+60+80";"+80+100";"+100+120";"+120+140";"+140+160";"+160+180";"+180+200";"+200+220";"+220+240"];
y1 = [0,0,0,0,0,0,0,0,0,0,0,0.04,0.32,0.56,0.08,0,0,0,0,0,0,0,0,0,0];
y2 = [0,0.05,0.05,0.05,0,0.1,0,0,0.2,0.3,0.1,0.1,0.05,0,0,0,0,0,0,0,0,0,0,0,0];
x=categorical(n);
plot(x,y1,'r--o', x,y2,'k--o');
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Rik
Rik 2020년 7월 7일
편집: Rik 2020년 7월 7일
You have told Matlab those strings are categories, so Matlab treated them as categories. Do you want to treat them as bins of a histogram instead?
Elnaz P
Elnaz P 2020년 7월 7일
Yes it is histogram. I am going to sort x axis from "-260-240" to "+220+240". I did x=sort (categorical(n)); but doesn't work. Please see attached photo.

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채택된 답변

Steven Lord
Steven Lord 2020년 7월 7일
When you call categorical with one input, the list of categories in that categorical is the sorted unique values in the input. From the documentation page:
"B = categorical(A) creates a categorical array from the array A. The categories of B are the sorted unique values from A."
Call categorical with two inputs to keep the categories in your desired order.
"B = categorical(A,valueset) creates one category for each value in valueset. The categories of B are in the same order as the values of valueset"
x = categorical(n, n);
plot(x,y1,'r--o', x,y2,'k--o');
  댓글 수: 2
Elnaz P
Elnaz P 2020년 7월 8일
it works, thanks a lot
Elnaz P
Elnaz P 2020년 7월 8일
Hi,
Could I ask another question about change origin point of y axis to (0-20)? I gonna shift Y axis to point (0-20) but I don't have any idea.
ax.XAxisLocation = 'origin';
ax.YAxisLocation = 'origin';
origin = (0-20);

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추가 답변 (1개)

Alan Stevens
Alan Stevens 2020년 7월 7일
Why not just:
y1 = [0,0,0,0,0,0,0,0,0,0,0,0.04,0.32,0.56,0.08,0,0,0,0,0,0,0,0,0,0];
y2 = [0,0.05,0.05,0.05,0,0.1,0,0,0.2,0.3,0.1,0.1,0.05,0,0,0,0,0,0,0,0,0,0,0,0];
x=-250:20:230;
plot(x,y1,'r--o', x,y2,'k--o');
This plots the "y" points in the middle of your "x" bands.
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Elnaz P
Elnaz P 2020년 7월 7일
Thanks for your help,
but please consider I have
x= ["-240-260";"-220-240";"-200-220";"-180-200";"-160-180";"-140-160";"-120-140";"-100-120";"-80-100";"-60-80";"-40-60";"-20- 40";"0-20";"0+20";"+20+40";"+40+60";"+60+80";"+80+100";"+100+120";"+120+140";"+140+160";"+160+180";"+180+200";"+200+220";"+220+240"];
not exactly x=-250:20:230;
Alan Stevens
Alan Stevens 2020년 7월 7일
편집: Alan Stevens 2020년 7월 7일
Where does it differ?

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