Search Results

Search found 832 results on 34 pages for 'plot'.

Page 14/34 | < Previous Page | 10 11 12 13 14 15 16 17 18 19 20 21  | Next Page >

  • Bing Maps Integrated With ASP.NET Pivot Grid v2010 vol 1

    Check out this slick demo which shows how sales data from the ASPxPivotGrid is plotted and displayed using the Bing.com maps service. The Bing Maps service provides you the capability to plot data geographically on a map. For example, this ASPxPivotGrid shows the quantity of products sold per country: We can plot this data on to a map because the Bing maps services provides developers with a JavaScript API to display maps, locate countries and businesses and create pushpin indicators! Now, we...Did you know that DotNetSlackers also publishes .net articles written by top known .net Authors? We already have over 80 articles in several categories including Silverlight. Take a look: here.

    Read the article

  • Free/opensource application for charting stock prices?

    - by Homunculus Reticulli
    I am looking for a free or FOSS software application for SIMPLY charting stock prices. I am not interested in any of the other nonsense typically bundled with such packages (technical analysis, back testing, tracking etc, etc). All I want to do is the following: Import file from CSV and plot on the chart Ability to scroll the chart left/right (zoom feature would be nice to) Ability to draw straight line (between 2 points) on the plot Ability to plot the graph for different resolutions (for e.g weekly, monthly - or some other custom resolution that I want) print the displayed graph (I can always use screen capture if printing is too much to ask) Thats all I want to do. I am not interested in anything else. I would have thought I could have found something by now. I would have written my own tool (I still will do that at a later stage), but I am a bit short of time at the moment, so I just want something that will do all of the above. Can anyone recommend a package. Last but not the least, I am running on Linux (and would prefer to do so - BUT if I have to, I can run on ahem - you know, Windows)

    Read the article

  • Create matplotlib legend out of the figure

    - by Werner
    I added the legend this way: leg = fig.legend((l0,l1,l2,l3,l4,l5,l6), ('0 Cl : r2, slope, origin', '1 Cl :'+str(r1b)+' , '+str(m1)+' , '+str(b1), '2 Cl :'+str(r2b)+' , '+str(m2)+' , '+str(b2), '3 Cl :'+str(r3b)+' , '+str(m3)+' , '+str(b3), '4 Cl :'+str(r4b)+' , '+str(m4)+' , '+str(b4), '5 Cl :'+str(r5b)+' , '+str(m5)+' , '+str(b5), '6 Cl :'+str(r6b)+' , '+str(m6)+' , '+str(b6), ), 'upper right') but the legend appears inside the plot. How can I tell matplotlib to put it to the right of the plot and at the right?

    Read the article

  • Tools for visualizing and implementing elliptic curve cryptography

    - by LL
    I need to create a program which will show how elliptic curve cryptography works. I was considering using Java Swing to create the GUI, but the main problem is what tool to use to plot the elliptic curve itself, and how to integrate that with Java Swing. I would like the plot to be included in the user interface and also allow changes to be made to it. Can you suggest any tools that would help with this?

    Read the article

  • fit a ellipse in Python given a set of points xi=(xi,yi)

    - by Gianni
    I am computing a series of index from a 2D points (x,y). One index is the ratio between minor and major axis. To fit the ellipse i am using the following post when i run these function the final results looks strange because the center and the axis length are not in scale with the 2D points center = [ 560415.53298363+0.j 6368878.84576771+0.j] angle of rotation = (-0.0528033467597-5.55111512313e-17j) axes = [0.00000000-557.21553487j 6817.76933256 +0.j] thanks in advance for help import numpy as np from numpy.linalg import eig, inv def fitEllipse(x,y): x = x[:,np.newaxis] y = y[:,np.newaxis] D = np.hstack((x*x, x*y, y*y, x, y, np.ones_like(x))) S = np.dot(D.T,D) C = np.zeros([6,6]) C[0,2] = C[2,0] = 2; C[1,1] = -1 E, V = eig(np.dot(inv(S), C)) n = np.argmax(np.abs(E)) a = V[:,n] return a def ellipse_center(a): b,c,d,f,g,a = a[1]/2, a[2], a[3]/2, a[4]/2, a[5], a[0] num = b*b-a*c x0=(c*d-b*f)/num y0=(a*f-b*d)/num return np.array([x0,y0]) def ellipse_angle_of_rotation( a ): b,c,d,f,g,a = a[1]/2, a[2], a[3]/2, a[4]/2, a[5], a[0] return 0.5*np.arctan(2*b/(a-c)) def ellipse_axis_length( a ): b,c,d,f,g,a = a[1]/2, a[2], a[3]/2, a[4]/2, a[5], a[0] up = 2*(a*f*f+c*d*d+g*b*b-2*b*d*f-a*c*g) down1=(b*b-a*c)*( (c-a)*np.sqrt(1+4*b*b/((a-c)*(a-c)))-(c+a)) down2=(b*b-a*c)*( (a-c)*np.sqrt(1+4*b*b/((a-c)*(a-c)))-(c+a)) res1=np.sqrt(up/down1) res2=np.sqrt(up/down2) return np.array([res1, res2]) if __name__ == '__main__': points = [(560036.4495758876, 6362071.890493258), (560036.4495758876, 6362070.890493258), (560036.9495758876, 6362070.890493258), (560036.9495758876, 6362070.390493258), (560037.4495758876, 6362070.390493258), (560037.4495758876, 6362064.890493258), (560036.4495758876, 6362064.890493258), (560036.4495758876, 6362063.390493258), (560035.4495758876, 6362063.390493258), (560035.4495758876, 6362062.390493258), (560034.9495758876, 6362062.390493258), (560034.9495758876, 6362061.390493258), (560032.9495758876, 6362061.390493258), (560032.9495758876, 6362061.890493258), (560030.4495758876, 6362061.890493258), (560030.4495758876, 6362061.390493258), (560029.9495758876, 6362061.390493258), (560029.9495758876, 6362060.390493258), (560029.4495758876, 6362060.390493258), (560029.4495758876, 6362059.890493258), (560028.9495758876, 6362059.890493258), (560028.9495758876, 6362059.390493258), (560028.4495758876, 6362059.390493258), (560028.4495758876, 6362058.890493258), (560027.4495758876, 6362058.890493258), (560027.4495758876, 6362058.390493258), (560026.9495758876, 6362058.390493258), (560026.9495758876, 6362057.890493258), (560025.4495758876, 6362057.890493258), (560025.4495758876, 6362057.390493258), (560023.4495758876, 6362057.390493258), (560023.4495758876, 6362060.390493258), (560023.9495758876, 6362060.390493258), (560023.9495758876, 6362061.890493258), (560024.4495758876, 6362061.890493258), (560024.4495758876, 6362063.390493258), (560024.9495758876, 6362063.390493258), (560024.9495758876, 6362064.390493258), (560025.4495758876, 6362064.390493258), (560025.4495758876, 6362065.390493258), (560025.9495758876, 6362065.390493258), (560025.9495758876, 6362065.890493258), (560026.4495758876, 6362065.890493258), (560026.4495758876, 6362066.890493258), (560026.9495758876, 6362066.890493258), (560026.9495758876, 6362068.390493258), (560027.4495758876, 6362068.390493258), (560027.4495758876, 6362068.890493258), (560027.9495758876, 6362068.890493258), (560027.9495758876, 6362069.390493258), (560028.4495758876, 6362069.390493258), (560028.4495758876, 6362069.890493258), (560033.4495758876, 6362069.890493258), (560033.4495758876, 6362070.390493258), (560033.9495758876, 6362070.390493258), (560033.9495758876, 6362070.890493258), (560034.4495758876, 6362070.890493258), (560034.4495758876, 6362071.390493258), (560034.9495758876, 6362071.390493258), (560034.9495758876, 6362071.890493258), (560036.4495758876, 6362071.890493258)] a_points = np.array(points) x = a_points[:, 0] y = a_points[:, 1] from pylab import * plot(x,y) show() a = fitEllipse(x,y) center = ellipse_center(a) phi = ellipse_angle_of_rotation(a) axes = ellipse_axis_length(a) print "center = ", center print "angle of rotation = ", phi print "axes = ", axes from pylab import * plot(x,y) plot(center[0:1],center[1:], color = 'red') show() each vertex is a xi,y,i point plot of 2D point and center of fit ellipse

    Read the article

  • Mathematica, PDF Curves and Shading

    - by Venerable Garbage Collector
    I need to plot a normal distribution and then shade some specific region of it. Right now I'm doing this by creating a plot of the distribution and overlaying it with a RegionPlot. This is pretty convoluted and I'm certain there must be a more elegant way of doing it. I Googled, looked at the docs, found nothing. Help me SO! I guess Mathematica counts as programming? :D

    Read the article

  • Control the size of points in an R scatterplot?

    - by Nick
    In R, the plot() function takes a pch argument that controls the appearance of the points in the plot. I'm making scatterplots with tens of thousands of points and prefer a small, but not too small dot. Basically, I find pch='.' to be too small, but pch=19 to be too fat. Is there something in the middle or some way to scale the dots down somehow?

    Read the article

  • Gnuplot - splot matrix csv data

    - by Jakub Czaplicki
    How can I plot (a 3D plot) a matrix in Gnuplot having such data structure. I cannot find a way to use the first row and column as a x and y ticks (or to ignore them) ,5,6,7,8 1,-6.20,-6.35,-6.59,-6.02 2,-6.39,-6.52,-6.31,-6.00 3,-6.36,-6.48,-6.15,-5.90 4,-5.79,-5.91,-5.87,-5.46 Is the splot 'data.csv' matrix the correct parameter to use ?

    Read the article

  • Combining graphs in Mathematica

    - by pizziaolo
    Any idea how I can overlay the following two functions to compare them? ln[1]:= p1 = Plot[(E^((Pi/6)^(1/3)/x) (Pi/6)^(2/3))/((-1 + E^((Pi/6)^(1/3)/x))^2 x^2), {x, -2.2, 2.2}] ln[2]:= p2 = Plot[(E^((Pi/6)^(1/3)/t) (Pi/6)^(2/3))/((-1 + E^((Pi/6)^(1/3)/t))^2 t^2), {t, 0, 2.0}] When I try Show[p1,p2] it doesn't work

    Read the article

  • How do I set selection to Nothing when programming Excel using VBA?

    - by Curt
    When I create a graph after using range.copy and range.paste it leaves the paste range selected, and then when I create a graph a few lines later, it uses the selection as the first series in the plot. I can delete the series, but is there a more elegant way to do this? I tried Set selection = nothing but it won't let me set selection. I also tried selection.clear, but that just cleared the last cells that were selected, and still added an extra series to the plot. Curt

    Read the article

  • MATLAB clear current figure

    - by rlbond
    I want to clear MATLAB's global CurrentFigure property, because I need a plot that I make to not be overwritten if a careless user uses plot without opening a new figure. I tried set(0, 'CurrentFigure', []); But it doesn't seem to work. Is this impossible?

    Read the article

  • plotting stem with a continuous line

    - by Abruzzo Forte e Gentile
    Hi All I need to plot a stem plot of my signal using python and matplotlib. I saw the example and the code but the line connecting the black big dot and the x-axis is not a continous line. Do you know whether is possible and how to get a straight line instead? Thank you very much AFG #!/usr/bin/env python from pylab import * x = linspace(0.1, 2*pi, 10) markerline, stemlines, baseline = stem(x, cos(x), '-.') setp(markerline, 'markerfacecolor', 'b') setp(baseline, 'color','r', 'linewidth', 2) show()

    Read the article

  • Python: Set window focus on terminal

    - by janoliver
    Hey, I have a python application that opens some plots for me and then asks the user for input. The problem is, that after opening the plot, the focus isn't on the terminal anymore, so you have to click or tab to it manually. I would like to set the focus to the terminal window with python - is that possible? I'm using gnuplot.py, maybe there is an option to open the plot in the background? Thanks!

    Read the article

  • What does the symbol ::: mean in R

    - by Milktrader
    I came across this in the following context from B. Pfaff's "Analysis of Integrated and Cointegrated Time Series in R" ## Impulse response analysis of SVAR A-type model 1 args (vars ::: irf.svarest) 2 irf.svara <- irf (svar.A, impulse = ”y1 ” , 3 response = ”y2 ” , boot = FALSE) 4 args (vars ::: plot.varirf) 5 plot (irf.svara)

    Read the article

  • How to use an excel data-set for a multi-line ggplot in R?

    - by user1299887
    I have a data set in excel that I am trying to create a multiple line plot with on R. The data set contains 7 food groups and the calories consumed daily associated to the groups. As well, there is that set of data over 38 years (from 1970-2008) and I am attempting to use this data set to create a multiple line plot on R. I have tried for hours on end but can not seem to get R to recognize the variables within the data set.

    Read the article

  • Drawing random smooth lines contained in a square [migrated]

    - by Doug Mercer
    I'm trying to write a matlab function that creates random, smooth trajectories in a square of finite side length. Here is my current attempt at such a procedure: function [] = drawroutes( SideLength, v, t) %DRAWROUTES Summary of this function goes here % Detailed explanation goes here %Some parameters intended to help help keep the particles in the box RandAccel=.01; ConservAccel=0; speedlimit=.1; G=10^(-8); % %Initialize Matrices Ax=zeros(v,10*t); Ay=Ax; vx=Ax; vy=Ax; x=Ax; y=Ax; sx=zeros(v,1); sy=zeros(v,1); % %Define initial position in square x(:,1)=SideLength*.15*ones(v,1)+(SideLength*.7)*rand(v,1); y(:,1)=SideLength*.15*ones(v,1)+(SideLength*.7)*rand(v,1); % for i=2:10*t %Measure minimum particle distance component wise from boundary %for each vehicle BorderGravX=[abs(SideLength*ones(v,1)-x(:,i-1)),abs(x(:,i-1))]'; BorderGravY=[abs(SideLength*ones(v,1)-y(:,i-1)),abs(y(:,i-1))]'; rx=min(BorderGravX)'; ry=min(BorderGravY)'; % %Set the sign of the repulsive force for k=1:v if x(k,i)<.5*SideLength sx(k)=1; else sx(k)=-1; end if y(k,i)<.5*SideLength sy(k)=1; else sy(k)=-1; end end % %Calculate Acceleration w/ random "nudge" and repulive force Ax(:,i)=ConservAccel*Ax(:,i-1)+RandAccel*(rand(v,1)-.5*ones(v,1))+sx*G./rx.^2; Ay(:,i)=ConservAccel*Ay(:,i-1)+RandAccel*(rand(v,1)-.5*ones(v,1))+sy*G./ry.^2; % %Ad hoc method of trying to slow down particles from jumping outside of %feasible region for h=1:v if abs(vx(h,i-1)+Ax(h,i))<speedlimit vx(h,i)=vx(h,i-1)+Ax(h,i); elseif (vx(h,i-1)+Ax(h,i))<-speedlimit vx(h,i)=-speedlimit; else vx(h,i)=speedlimit; end end for h=1:v if abs(vy(h,i-1)+Ay(h,i))<speedlimit vy(h,i)=vy(h,i-1)+Ay(h,i); elseif (vy(h,i-1)+Ay(h,i))<-speedlimit vy(h,i)=-speedlimit; else vy(h,i)=speedlimit; end end % %Update position x(:,i)=x(:,i-1)+(vx(:,i-1)+vx(:,i))/2; y(:,i)=y(:,i-1)+(vy(:,i-1)+vy(:,1))/2; % end %Plot position clf; hold on; axis([-100,SideLength+100,-100,SideLength+100]); cc=hsv(v); for j=1:v plot(x(j,1),y(j,1),'ko') plot(x(j,:),y(j,:),'color',cc(j,:)) end hold off; % end My original plan was to place particles within a square, and move them around by allowing their acceleration in the x and y direction to be governed by a uniformly distributed random variable. To keep the particles within the square, I tried to create a repulsive force that would push the particles away from the boundaries of the square. In practice, the particles tend to leave the desired "feasible" region after a relatively small number of time steps (say, 1000)." I'd love to hear your suggestions on either modifying my existing code or considering the problem from another perspective. When reading the code, please don't feel the need to get hung up on any of the ad hoc parameters at the very beginning of the script. They seem to help, but I don't believe any beside the "G" constant should truly be necessary to make this system work. Here is an example of the current output: Many of the vehicles have found their way outside of the desired square region, [0,400] X [0,400].

    Read the article

  • iPhone Audio Queue Service sample units

    - by pion
    I am looking at Audio Queue Services document specifically on the following code: // Writing an audio queue buffer to disk AudioFileWritePackets ( // 1 pAqData->mAudioFile, // 2 false, // 3 inBuffer->mAudioDataByteSize, // 4 inPacketDesc, // 5 pAqData->mCurrentPacket, // 6 &inNumPackets, // 7 inBuffer->mAudioData // 8 ); inBuffer-mAudioDataByteSize is the number of bytes of audio data being written. inBuffer-mAudioData is the new audio data to write to the audio file. Assuming the sample rate is 44100. AudioStreamBasicDescription mDataFormat; mDataFormat.mSampleRate = 44100.0f; mDataFormat.mBitsPerChannel = 16; ... NSInteger numberSamples = inBuffer->mAudioDataByteSize / 2; SInt16 *audioSample = (SInt16 *)inBuffer->mAudioData; I use core-plot to plot the above where x axis is number of sample [1 .. numberSamples] and the y axis is audioSample[0] .. audioSample[numberSamples]. I can see the chart in "real-time" where the y axis goes up and down depending the loudness of my voice. Beginner questions: What does the audioSample represent? What am I looking at here? What is the unit of audioSample? What do I need to do if I just want to plot the range between 50 - 100 Hz? Thanks in advance for your help.

    Read the article

  • jqGrid - edit function never getting called

    - by dcp
    I'm having a problem with JQGrid using ASP.NET MVC. I'm trying to follow this example (after you go to that link, over on the left side of the page please click Live Data Manipulation, then Edit Row), but my edit function is never getting called (i.e. it's never getting into the $("#bedata").click(function(). Does anyone know what could be the problem? <script type="text/javascript"> var lastsel2; jQuery(document).ready(function() { jQuery("#editgrid").jqGrid({ url: '/Home/GetMovieData/', datatype: 'json', mtype: 'GET', colNames: ['id', 'Movie Name', 'Directed By', 'Release Date', 'IMDB Rating', 'Plot', 'ImageURL'], colModel: [ { name: 'id', index: 'Id', width: 55, sortable: false, hidden: true, editable: false, editoptions: { readonly: true, size: 10} }, { name: 'Movie Name', index: 'Name', width: 250, editable: true, editoptions: { size: 10} }, { name: 'Directed By', index: 'Director', width: 250, align: 'right', editable: true, editoptions: { size: 10} }, { name: 'Release Date', index: 'ReleaseDate', width: 100, align: 'right', editable: true, editoptions: { size: 10} }, { name: 'IMDB Rating', index: 'IMDBUserRating', width: 100, align: 'right', editable: true, editoptions: { size: 10} }, { name: 'Plot', index: 'Plot', width: 150, hidden: false, editable: true, editoptions: { size: 30} }, { name: 'ImageURL', index: 'ImageURL', width: 55, hidden: true, editable: false, editoptions: { readonly: true, size: 10} } ], pager: jQuery('#pager'), rowNum: 5, rowList: [5, 10, 20], sortname: 'id', sortorder: "desc", height: '100%', width: '100%', viewrecords: true, imgpath: '/Content/jqGridCss/redmond/images', caption: 'Movies from 2008', editurl: '/Home/EditMovieData/', caption: 'Movie List' }); }); $("#bedata").click(function() { var gr = jQuery("#editgrid").jqGrid('getGridParam', 'selrow'); if (gr != null) jQuery("#editgrid").jqGrid('editGridRow', gr, { height: 280, reloadAfterSubmit: false }); else alert("Hey dork, please select a row"); }); </script> The relevant HTML is here: <table id="editgrid"> </table> <div id="pager" style="text-align: center;"> </div> <input type="button" id="bedata" value="Edit Selected" />

    Read the article

  • Linq to find pair of points with longest length?

    - by Chris
    I have the following code: foreach (Tuple<Point, Point> pair in pointsCollection) { var points = new List<Point>() { pair.Value1, pair.Value2 }; } Within this foreach, I would like to be able to determine which pair of points has the most significant length between the coordinates for each point within the pair. So, let's say that points are made up of the following pairs: (1) var points = new List<Point>() { new Point(0,100), new Point(100,100) }; (2) var points = new List<Point>() { new Point(150,100), new Point(200,100) }; So I have two sets of pairs, mentioned above. They both will plot a horizontal line. I am interested in knowing what the best approach would be to find the pair of points that have the greatest distance between, them, whether it is vertically or horizontally. In the two examples above, the first pair of points has a difference of 100 between the X coordinate, so that would be the point with the most significant difference. But if I have a collection of pairs of points, where some points will plot a vertical line, some points will plot a horizontal line, what would be the best approach for retrieving the pair from the set of points whose difference, again vertically or horizontally, is the greatest among all of the points in the collection? Thanks! Chris

    Read the article

  • Using GNU Octave FFT functions

    - by CFP
    Hello world! I'm playing with octave's fft functions, and I can't really figure out how to scale their output: I use the following (very short) code to approximate a function: function y = f(x) y = x .^ 2; endfunction; X=[-4096:4095]/64; Y = f(X); # plot(X, Y); F = fft(Y); S = [0:2047]/2048; function points = approximate(input, count) size = size(input)(2); fourier = [fft(input)(1:count) zeros(1, size-count)]; points = ifft(fourier); endfunction; Y = f(X); plot(X, Y, X, approximate(Y, 10)); Basically, what it does is take a function, compute the image of an interval, fft-it, then keep a few harmonics, and ifft the result. Yet I get a plot that is vertically compressed (the vertical scale of the output is wrong). Any ideas? Thanks!

    Read the article

< Previous Page | 10 11 12 13 14 15 16 17 18 19 20 21  | Next Page >