The graph you provided is upside down
That's irrelevant. It does not affect the veracity of the plot.
But as it seems to bother you; I've posted the same graph flipped vertically (with a couple of additions) below.
Correct Graph
Your link leads to a page that reads: This image or video is currently unavailable.
Please provide your spreadsheet that you used to generate it,
I didn't use a spreadsheet. This is a Perl site. I used Perl :)
I am very confused how you could even make that happen. You'd have to have multiplied all y values with a -1 to cause such a thing.
Or; plot the data on a medium that has the origin top left with Y running top to bottom. As is the convention with most computer graphics.
The 3 curves presented here are not to scale and lack axes. As such they misrepresent the meaningfulness of the first derivative in peak detection.
Since all the Y values on the three curves are related numerically, their absolute values are irrelevant. Hence scales are superfluous.
The X values are the same, drawn to the same scale and offset for all three plots.
Your conclusion that upside down derivatives are not useful in peak detection is in fact correct. I assure you that a correctly oriented first derivative is useful for this type of problem.
Sorry, but that makes no sense at all. Since the important points are where the curves transition across the Y=0 line. Whether that transition occurs going from above to below or below to above doesn't change anything one iota.
If you redraw it to scale with axes you will notice that any time the first derivative crosses the x axis (aka it is zero) there will be a peak.
If that were the case, I would not have posted. Take another, closer look at the graph.
The additional black horizontal line is the x-axis (Y=0) of both the 1st (green) and 2nd (blue) derivative plots. The additional black vertical lines mark the peaks and troughs in the data plot (red).
Note how the 1st derivative plot (green) doesn't transition 0 at all for the first two turning points, and is substantially inaccurate for turning points four and six; slightly inaccurate for the fifth; leaving just 2: the third and seventh that it hits accurately.
You may now be wondering how the black vertical lines were drawn.
If you subtract consecutive data points and compare the results to 0:
my @deltas = map{ ( 0 <=> $y[$_-1] - $y[$_] ) } 1 .. $#y;
You get a dataset like this:
-1 1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 1 1 1 1 1 1 1 -1 -1 -1 -1 1 1 +1 1 1 1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 1 1 1 1 1 1
You then plot a vertical line at the X value everwhere the sign changes:
$deltas[$_-1] != $deltas[$_] and $im->line( $x[$_], 0, $x[$_], 800, 0 +) for 1 .. $#x-1;
And there you have your maxima and minima simply, directly, and accurately.
With a very small perl program (less than 200 lines) I have a peak detector that works pretty well for this type of psuedo-sinusoidal data.
So verbose!? :)
Here is my Perl code, a whole 40 lines, that plots the above linked graph:
#! perl -slw use strict; use Data::Dump qw[ pp ]; use GD; use constant { WHITE => unpack( 'N', pack 'CCCC', 0, 255, 255, 255 ), RED => unpack( 'N', pack 'CCCC', 0, 255, 0, 0 ), GREEN => unpack( 'N', pack 'CCCC', 0, 0, 255, 0 ), BLUE => unpack( 'N', pack 'CCCC', 0, 0, 0, 255 ), }; my( @x, @y, @yd1, @yd2 ); ( $x[@x], $y[@y], $yd1[@yd1], $yd2[@yd2] ) = map{ $_ //= 0 } split whi +le <DATA>; chomp @yd2; $_ = ( $_ -4 ) * 1000 for @x; $_ /= 6 for @y; $_ = $_ / 320 + 400 for @yd1; $_ = $_ / 8000 + 400 for @yd2; my $im = GD::Image->new( 1000, 800, 1 ); $im->filledRectangle( 0, 0, 1000, 800, WHITE ); $im->line( 0, 400, 1000, 400, 0 ); $im->line( $x[$_-1], $y[$_-1], $x[$_], $y[$_], RED ) for 1 .. $#x; $im->line( $x[$_-1], $yd1[$_-1], $x[$_], $yd1[$_], GREEN ) for 2 .. $# +x-1; $im->line( $x[$_-1], $yd2[$_-1], $x[$_], $yd2[$_], BLUE ) for 3 .. $#x +-2; my @deltas = map{ ( 0 <=> $y[$_-1] - $y[$_] ) } 1 .. $#y; $deltas[$_-1] != $deltas[$_] and $im->line( $x[$_], 0, $x[$_], 800, 0 +) for 1 .. $#x-1; $im->flipVertical; open PNG, '>:raw', "$0.png" or die $!; print PNG $im->png; close PNG; system 1, "$0.png";
And the dataset taken directly from salva's post above:
In reply to Re^7: Any idea for predicting the peak points in the graph by perl
by BrowserUk
in thread Any idea for predicting the peak points in the graph by perl
by Anonymous Monk
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