How to create an average of 2 lines on a single graph
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Hi,
I have two lines which represent magnetic intensity data collected along a transect back and forth, hence the 2 lines, producing this figure;
% Single-line Magnetic Anomaly graph, Charlie Hillary 2021
% load top and bottom data files
top = load('top_sensor_data.txt')
% calculate the temporal gradient
dtopdt = (top(end,3)-top(1,3))/(top(end,2)-top(1,2)) ;
% correct the data
top(:,3) = top(:,3) - dtopdt*(top(:,2)-top(1,2)) ;
% plot
figure
plot(top(1:end-1,1),top(1:end-1,3)-48e3,'k-') ;
xlabel('Distance / m')
ylabel('Upper B / nT ') ;
title('Top @ 0.95m')
I need to create an average of the 2 lines. I have a rough understanding of how to do so. I think I need to create a loop for each Y value(s) at a given X value, and create an average of the two, by adding and dividing by two. The data is a 3 column array, with column 2 being used to process column 1 (x) and column 3 (y).
Just need a point in the right direction, thanks!
Charlie
4 件のコメント
dpb
2021 年 3 月 7 日
編集済み: dpb
2021 年 3 月 7 日
"...if you run this code"
With what for input data?
plot(top(1:end-1,1),top(1:end-1,3)-48e3,'k-') ;
is the only plot command in the code; unless the first column does represent a triangle waveform of
[0:Xmax fliplr[0:Xmax]]
but we can't know that, if so.
What's with the magic number of 48000? Especially after the earlier comment that you "corrected" it???
All in all, is pretty confusing to the outsider... :)
採用された回答
dpb
2021 年 3 月 7 日
OK, I see...the data are collected on a up- and down-slope symmetric ramp..
Try something like--
top=readmatrix('top_sensor_data.txt');
top=reshape(top,length(top)/2,[]);
top(:,[2:2:end])=flipud(top(:,2:2:end));
top=array2table(top(:,[1 3:end]),'VariableNames',{'Distance','V1_Out','V1_In','B_Out','B_In'});
>> head(top)
ans =
8×5 table
Distance V1_Out V1_In B_Out B_In
________ ______ ______ ________ ________
15.00 0.50 495.00 48700.67 48699.92
16.00 2.25 499.09 48701.12 48699.82
17.00 3.75 486.75 48700.58 48699.33
18.00 5.25 485.75 48701.99 48693.28
19.00 7.00 484.75 48702.83 48703.04
20.00 8.50 483.75 48703.13 48702.60
21.00 10.00 482.75 48703.30 48701.68
22.00 11.50 482.00 48703.92 48703.99
hAx(1)=subplot(2,1,1);
plot(top.Distance,[top.B_Out top.B_In])
legend('B Out','B In','Location','southwest')
xlim([0 265])
hAx(2)=subplot(2,1,2);
plot(top.Distance,detrend([top.B_Out top.B_In]))
xlim([0 265])
mnB=mean([top.B_Out top.B_In],'all')
>> mnB =
48685.42
>> mean([top.B_Out top.B_In])
ans =
48686.39 48684.44
>> diff(ans)
ans =
-1.95
>>
yyaxis right
hL=plot(top.Distance,[top.B_Out top.B_In]-mnB);
legend('B Out Detrended','B In Detrended','B Out Less Mean','B In Less Mean','Location','southwest')
This shows a very nearly flat profile going across as the small bias between the detrended and mean-subtracted signals show--what's the best thing to do I've no klew having no knowledge of what the data are (nor the field if did :) ).
Oh -- I realize I completely got carried away with the pretty figures and totally forget the actual Q? -- alhough by now it ought to be apparent how to do the mean of the two--
top.B_Avg=mean([top.B_Out top.B_In],2);
>> head(top)
ans =
8×6 table
Distance V1_Out V1_In B_Out B_In B_Avg
________ ______ ______ ________ ________ ________
15.00 0.50 495.00 48700.67 48699.92 48700.29
16.00 2.25 499.09 48701.12 48699.82 48700.47
17.00 3.75 486.75 48700.58 48699.33 48699.96
18.00 5.25 485.75 48701.99 48693.28 48697.64
19.00 7.00 484.75 48702.83 48703.04 48702.94
20.00 8.50 483.75 48703.13 48702.60 48702.86
21.00 10.00 482.75 48703.30 48701.68 48702.49
22.00 11.50 482.00 48703.92 48703.99 48703.95
>>
Not knowing what to do with the second variable, I simply carried it along for the ride...
2 件のコメント
dpb
2021 年 3 月 8 日
Thanks for the background...interesting to know what one is looking at!
Why not remove the actual mean instead of something approximating it?
And, detrend does that along with compensating for a trend; not knowing the reason(s) behind any trend there might be; I already cleared the data but iirc the slope is pretty small; that's indicative in how well the two overlay when plot detrend(B) over B-mn(B).
その他の回答 (1 件)
David Goodmanson
2021 年 3 月 8 日
編集済み: David Goodmanson
2021 年 3 月 8 日
Hi Charley
since the second half of the distance array has exactly the same values as the first half (only reversed), you can use just the first half of the distance array for plotting purposes. After you make the data correction to top3, then
dist = top(1:245,1);
B1 = top(1:245,3);
B2 = flip(top(246:end,3));
Bmean = mean([B1 B2],2);
plot(dist,B1,'b',dist,B2,'b',dist,Bmean,'r')
xlabel('Distance / m')
ylabel('Upper B / nT ') ;
title('Top @ 0.95m')
Here B1 and B2 are column vectors. The square brackets concatenate them side-by-side, creating a 245x2 matrix. The '2' in the mean command says to take the mean across rows, rather than down columns which is the default. So you get the mean of the two curves.
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