I want the equation of a graph data.
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M=[0 0
93.07876 0.224404
186.1575 0.476858
262.5298 0.729313
303.1026 0.869565
310.2625 1.290323
319.8091 1.598878
329.3556 1.486676
341.2888 1.318373
353.222 1.542777
362.7685 1.851332
372.315 2.103787
377.0883 2.412342
381.8616 2.664797
386.6348 2.889201
391.4081 3.169705
396.1814 3.45021
400.9547 3.702665
408.1146 3.98317
410.5012 4.235624
415.2745 4.460028
420.0477 4.712482
424.821 4.964937
446.3007 5.021038
467.7804 5.105189
477.327 5.329593
484.4869 5.553997
494.0334 5.778401
501.1933 5.946704
508.3532 6.143058
515.5131 6.367461
522.673 6.563815
529.8329 6.788219
536.9928 7.040673
544.1527 7.265077
548.926 7.517532
556.0859 7.798036
565.6325 8.078541
572.7924 8.387097
579.9523 8.695652
589.4988 9.004208
594.2721 9.284712
601.432 9.565217
606.2053 9.817672
613.3652 10.07013
625.2983 9.901823
637.2315 9.677419
649.1647 9.509116
663.4845 9.340813
675.4177 9.11641
684.9642 9.340813
692.1241 9.565217
696.8974 9.817672
704.0573 10.07013
706.4439 10.18233
718.3771 10.26648
720.7637 10.49088
723.1504 10.63114
725.537 10.79944
727.9236 10.96774
730.3103 11.1641
732.6969 11.41655
735.0835 11.64095
737.4702 11.86536
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742.2434 12.37027
744.6301 12.62272
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751.79 13.09958
756.5632 13.29593
761.3365 13.54839
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775.6563 14.2216
780.4296 14.446
785.2029 14.69846
789.9761 14.89481
792.3628 15.11921
797.136 15.28752
801.9093 15.45582
806.6826 15.68022
811.4558 15.84853
813.8425 16.10098
818.6158 16.32539
828.1623 16.63394
830.5489 16.91445
835.3222 17.13885
840.0955 17.36325
844.8687 17.64376
852.0286 17.89621
875.895 17.89621
890.2148 17.89621
902.148 17.89621
918.8544 17.89621
926.0143 17.89621
933.1742 17.75596
940.3341 17.58766
947.494 17.3913
952.2673 17.223
954.6539 17.0547
964.2005 16.91445
971.3604 16.69004
978.5203 16.46564
985.6802 16.24123
995.2267 15.98878
1002.387 15.79243
1009.547 15.53997
1016.706 15.31557
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1035.8 14.5582
1038.186 14.36185
1040.573 14.1655
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1045.346 13.74474
1047.733 13.52034
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1071.599 11.5568
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1112.172 9.649369
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1114.558 9.088359
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1121.718 9.453015
1121.718 9.621318
1121.718 9.789621
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1124.105 10.21038
1126.492 10.40673
1133.652 10.40673
1136.038 10.23843
1138.425 10.09818
1145.585 9.929874
1145.585 9.70547
1147.971 9.509116
1147.971 9.284712
1147.971 9.14446
1152.745 8.948107
1155.131 8.723703
1157.518 8.499299
1162.291 8.274895
1167.064 7.99439
1171.838 7.798036
1171.838 7.573633
1171.838 7.293128
1176.611 7.040673
1178.998 6.788219
1181.384 6.591865
1183.771 6.367461
1186.158 6.171108
1188.544 5.974755
1190.931 5.778401
1195.704 5.525947
1198.091 5.357644
1200.477 5.217391
1200.477 5.077139
1202.864 5.13324
1202.864 5.301543
1205.251 5.441795
1202.864 5.553997
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1202.864 6.030856
1202.864 6.227209
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1205.251 6.84432
1205.251 7.040673
1207.637 7.237027
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1210.024 7.826087
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1214.797 7.713885
1214.797 7.573633
1217.184 7.461431
1217.184 7.293128
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1243.437 6.058906
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1252.983 5.189341
1255.37 4.908836
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1260.143 4.460028
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1267.303 4.01122
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1276.85 3.955119
1281.623 4.123422
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1295.943 4.431978
1303.103 4.291725
1305.489 4.123422
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1310.263 3.786816
1315.036 3.562412
1315.036 3.281907
1315.036 3.057504
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1317.422 2.524544
1319.809 2.215989
1322.196 1.991585
1322.196 1.71108
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1324.582 1.234222
1324.582 1.009818
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1334.129 1.262272
1338.902 1.430575
1343.675 1.654979
1348.449 1.823282
1353.222 1.963534
1357.995 2.103787
1362.768 2.215989
1367.542 2.328191
1372.315 2.131837
1372.315 1.935484
1379.475 1.767181
1381.862 1.598878
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1386.635 1.262272
1389.021 1.093969
1403.341 1.12202
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1439.141 1.290323
1460.621 1.374474
1477.327 1.430575
1486.874 1.430575
498.807 1.234222
1510.74 1.037868
1520.286 0.785414
1529.833 0.56101
1539.379 0.392707
1548.926 0.252454
553.699 0.112202
1558.473 0 ];
plot(M(:,1),M(:,2),'o')
4 Kommentare
Sam Chak
am 20 Jan. 2024
I've noticed that in your previous post, you might have missed exploring the application of certain curve-fitting tools in MATLAB to derive the equation. It seems your primary goal is obtaining the equation model that best fits the data, rather than delving into the intricacies of the curve-fitting tools themselves.
If I'm not mistaken, your future inquiries might also revolve around acquiring the equation rather than understanding the tools. Could we perhaps initiate the process by plotting the graph using the plot() function and analyzing the pattern? This approach could showcase your minimal effort while still achieving the desired outcome.
plot(x, y)
Antworten (2)
Sulaymon Eshkabilov
am 20 Jan. 2024
Here is the answer code that I posted for your prevous post that works in the same way:
D = load('Patel_Data.txt');
[x, IDX] = sort(D(:,1));
y = D(IDX, 2);
yy = smoothdata(y, 'gaussian', 5);
plot(x,yy, 'b--', 'LineWidth', 2)
hold on
FUN = @(a, x) a(1) * exp(-((x - a(2)) / a(3)).^2); % Gaussian model
a0 = [15; 777; 333]; % Initial guess values for a
MODEL = fitnlm(x, yy, FUN, a0)
plot(x, MODEL.Fitted, 'm-', 'LineWidth', 2);
legend('Data', 'Gaussian Fit Model')
grid on
xlabel('Time, [s]')
ylabel('Depth, [m]')
2 Kommentare
Star Strider
am 20 Jan. 2024
‘@Alexsha give the solution but how he did, he did not told.’
He does not use MATLAB. He uses a different (very expensive) optimisation software system.
Steven Lord
am 20 Jan. 2024
so please help me in getting the equation of the above data.
What makes you assume that there is a unique equation that fits that data? [That there is a "the equation"?]
x = 0:5;
y = zeros(size(x));
plot(x, y, 'o')
What is the unique equation that connects those six data points? Is it:
xfinergrid = 0:(1/16):5;
figure
plot(x, y, 'o');
p = 0;
hold on
plot(xfinergrid, polyval(p, xfinergrid), '-')
title('y = 0')
Or is it this?
figure
plot(x, y, 'o')
hold on
plot(xfinergrid, sinpi(xfinergrid), '-')
title('y = sin(\pi x)')
I can come up with an infinite number of curves that pass through those six points. So which one is "the right" one? Without more information it is literally impossible to decide.
1 Kommentar
Sam Chak
am 20 Jan. 2024
@Steven Lord, In the past, I queried several students who mentioned that their professors or supervisors had directed them to conduct experiments and gather data. Subsequently, they were tasked with identifying suitable equations to fit the data and publishing their findings as novel contributions.
Surprisingly, many of these students had not undergone any formal training in curve-fitting (including myself). They held the belief that specialized functions in MATLAB or Origin Pro (a software app renowned for its advanced graphing and data analysis capabilities) existed, enabling them to input the data into the application, which would then 'intelligently' provide the equation and coefficients that best fit the data.
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