how to specify linewidth for one line in a multi line plot?

1 Kommentar

Ahmer Ashraf
Ahmer Ashraf am 23 Feb. 2020
I'm bound on linewidth property..I'm completing my assignment but this property is not going me forward.

Melden Sie sich an, um zu kommentieren.

 Akzeptierte Antwort

John BG
John BG am 7 Jan. 2017
Bearbeitet: John BG am 7 Jan. 2017

3 Stimmen

Noam
I kindly ask to have the following answer marked as Accepted Answer:
x = 1:1:10;
L=5 % amount of lines
y = randi([1 10], L,10);
Lwidth=[1 2 3 4 5]
figure(1);hold all
for k=1:1:L
plot(x,y(k,:),'LineWidth',Lwidth(k))
end
grid on
or
figure(1);hold all
for k=1:1:L
h=plot(x,y(k,:))
h.LineWidth=Lwidth(k)
end
grid on
.
appreciating time and attention
John BG

4 Kommentare

Star Strider
Star Strider am 7 Jan. 2017
That is exactly the essence of my Answer.
Thank you for plagiarizing it.
Noam
Noam am 7 Jan. 2017
Thank you all.
For two differnt graphs in nature I prefere Strider's, for a family of graphs i'd go with BG's. I used the hold trick before and was hoping for a more elegant solution.
Noam
John BG
John BG am 7 Jan. 2017
Dear Start Strider, here no one has plagiarised your answer.
Your answer has a constant line width of 2.
The question is asking for a variable line width.
I supply an answer that has a variable line width: 5 lines with different LineWidth parameters.
do you see the difference?
Star Strider
Star Strider am 7 Jan. 2017
There is no essential difference!
My Answer shows how to do what Noam wants to do. It should be easy for Noam to use my code and expand on it as necessary. ‘Multiline’ implies more than one line, and my code does exactly what was asked.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Star Strider
Star Strider am 6 Jan. 2017

4 Stimmen

Use the hold function and plot them individually:
x = 1:10;
y1 = rand(1, 10);
y2 = 1+rand(1,10);
figure(1)
plot(x, y1)
hold on
plot(x, y2, 'LineWidth',2)
hold off
grid

2 Kommentare

Noam
Noam am 6 Jan. 2017
Thamks
Star Strider
Star Strider am 6 Jan. 2017
My pleasure.
If my Answer solved your problem, please Accept it!

Melden Sie sich an, um zu kommentieren.

Produkte

Tags

Gefragt:

am 6 Jan. 2017

Kommentiert:

am 23 Feb. 2020

Community Treasure Hunt

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

Start Hunting!

Translated by