Distinguish the different curves and identify the points corresponding to each way

I have this figure but it is difficult to distinguish the different curves and identify the points corresponding to each approach. Is there any other way to represent each method?

Thanks in advance

[Sam Chak]

Hi @rana, could you provide the data and plot it in MATLAB?

[rana]

it's big dataset. Can you please help me with how to represent them in a differnt way so it will be clear?

Answer 1

[Star Strider]

Another approach, using the stackedplot (introduced in R2018b) function —

T1 = array2table(10*(rand(100,5)+(4:-1:0)), VariableNames=["MACS","Static","Random","RL-Based","DL-Based"])

T1 = 100x5 table
     MACS     Static    Random    RL-Based    DL-Based
    ______    ______    ______    ________    ________

     45.87    33.169    26.318     14.936       1.5879
    46.375    37.681    29.883     13.997       1.1667
    49.351    36.993    29.863      16.13       5.1319
     48.95    30.412    25.398     14.294        3.092
    47.133    31.487    26.569     14.454       9.5756
    46.329    35.209    21.752     11.727       2.6868
    42.347    36.462    22.361     19.173       3.2923
    43.382    31.707    20.989     17.864       9.5463
    44.295    32.014    23.593     18.068        7.918
    41.683      32.8    27.471     14.716        3.087
    49.056    34.589    29.483     14.623       8.8817
    40.568    38.024    23.899     13.012       1.5231
    43.661    36.536    25.815     15.613       1.1264
    49.711    36.383    22.347     10.501       6.5348
    48.582    38.039    20.514     17.983     0.071536
     40.37    34.064     21.76     16.704       7.2876

figure
stackedplot(T1)
xlabel("Number of Iterations")
title("Client Selection Across FL Methods")

.

[Sam Chak]

Hi @rana, take @Star Strider's approach. More compact to use the stackedplot() function.

[rana]

Thanks but unforuntally i'm using matlab basic(Online) and doesn't supporting this function. Is there any alternative way?

[Star Strider]

I just now opened MATLAB Online and ran my exact code there successfully.

The stackedplot function is part of core MATLAB and does not require any Toolboxes. However, it allows only what you have a license for, so if you have R2018b or later, you should have the function. (I have R2024b, so it works for me.)

One alterenative (if you have an earlier version that R2018b) is to run it here with your data, and then do a screencap of the resulting plot. Upload your data to a comment to your Question here using the ‘paperclip’ icon, then run my code with it.

.

[rana]

it works thank you so much

[Star Strider]

As always, my pleasure!

Answer 2

[Sam Chak]

Hi @rana

I now understand that you are seeking advice on graphically displaying the data. Here is an example of what I consider to be good and bad figures.

y1  = randi([60, 80], 1, 100);
y2  = randi([10, 50], 1, 100);
y3  = randi([15, 55], 1, 100);
y4  = randi([30, 70], 1, 100);
y5  = randi([35, 75], 1, 100);
x   = linspace(1, 100);

figure(1)
plot(x, y1, 'r'), hold on
plot(x, y2, 'g'), plot(x, y3, 'b'), plot(x, y4, 'm'), plot(x, y5, 'c'), hold off
title('BAD: Chartjunk')

figure(2)
tL  = tiledlayout(5, 1, 'TileSpacing', 'Compact');

nexttile
plot(x, y1, 'r', 'linewidth', 2.5), hold on
plot(x, y2, 'g'), plot(x, y3, 'b'), plot(x, y4, 'm'), plot(x, y5, 'c'), hold off, ylabel('Method 1')

nexttile
plot(x, y2, 'g', 'linewidth', 2.5), hold on
plot(x, y1, 'r'), plot(x, y3, 'b'), plot(x, y4, 'm'), plot(x, y5, 'c'), hold off, ylabel('Method 2')

nexttile
plot(x, y3, 'b', 'linewidth', 2.5), hold on
plot(x, y1, 'r'), plot(x, y2, 'g'), plot(x, y4, 'm'), plot(x, y5, 'c'), hold off, ylabel('Method 3')

nexttile
plot(x, y4, 'm', 'linewidth', 2.5), hold on
plot(x, y1, 'r'), plot(x, y2, 'g'), plot(x, y3, 'b'), plot(x, y5, 'c'), hold off, ylabel('Method 4')

nexttile
plot(x, y5, 'c', 'linewidth', 2.5), hold on
plot(x, y1, 'r'), plot(x, y2, 'g'), plot(x, y3, 'b'), plot(x, y4, 'm'), hold off, ylabel('Method 5')

title(tL, 'BETTER: Split into 5 plots')

[John D'Errico]

Nice idea.

[rana]

Thanks