IPython Scientific Graphs

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     "input": "%pylab inline\nimport matplotlib.pyplot as plt # so we don't have to look at mpl's backend\nimport matplotlib.gridspec as gridspec # for subplots\nimport matplotlib.cm as cm # for fun-with-colors\nimport numpy as np",
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": "\nWelcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\nFor more information, type 'help(pylab)'.\n"
      }
     ],
     "prompt_number": 56
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    {
     "cell_type": "code",
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     "input": "from matplotlylib import fig_to_plotly\nusername = 'IPython.Demo'\napi_key = '1fw3zw2o13'",
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 59
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     "cell_type": "code",
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     "input": "fig1 = plt.figure()\nfrom random import gauss\ngrands = []\nfor i in range(100):\n    grands.append(gauss(0,1))\nplot(grands)\nfig_to_plotly(fig1, username, api_key, notebook= True)",
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     "cell_type": "code",
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     "input": "fig2 = plt.figure()\nplot(rand(100))\nfig_to_plotly(fig2, username, api_key, notebook= True)",
     "language": "python",
     "metadata": {},
     "outputs": [
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     "cell_type": "code",
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     "input": "fig3 = plt.figure()\nnpts = 5000\nxs = 2*rand(npts)-1\nys = 2*rand(npts)-1\nr = xs**2+ys**2\nninside = (r<1).sum()\nfigsize(6,6) # make the figure square\ntitle(\"Approximation to pi = %f\" % (4*ninside/float(npts)))\nplot(xs[r<1],ys[r<1],'b.')\nplot(xs[r>1],ys[r>1],'r.')\nfigsize(8,6) # change the figsize back to 4x3 for the rest of the notebook\nfig_to_plotly(fig3, username, api_key, notebook= True)",
     "language": "python",
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     "outputs": [
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     "cell_type": "code",
     "collapsed": false,
     "input": "fig4 = plt.figure()\nplot (rand(100))\nfig_to_plotly(fig4, username, api_key, notebook = True)",
     "language": "python",
     "metadata": {},
     "outputs": [
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     "cell_type": "code",
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     "input": "fig5 = plt.figure()\nfrom scipy.fftpack import fft,fftfreq\n\nnpts = 4000\nnplot = npts/10\nt = linspace(0,120,npts)\ndef acc(t): return 10*sin(2*pi*2.0*t) + 5*sin(2*pi*8.0*t) + 2*rand(npts)\n\nsignal = acc(t)\n\nFFT = abs(fft(signal))\nfreqs = fftfreq(npts, t[1]-t[0])\n\nsubplot(211)\nplot(t[:nplot], signal[:nplot])\nsubplot(212)\nplot(freqs,20*log10(FFT),',')\nshow()\nfig_to_plotly(fig5, username, api_key, notebook = True)",
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     "outputs": [
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     "input": "class Schrod1d:\n    \"\"\"\\\n    Schrod1d: Solver for the one-dimensional Schrodinger equation.\n    \"\"\"\n    def __init__(self,V,start=0,end=1,npts=50,**kwargs):\n        m = kwargs.get('m',1.0)\n        self.x = linspace(start,end,npts)\n        self.Vx = V(self.x)\n        self.H = (-0.5/m)*self.laplacian() + diag(self.Vx)\n        return\n    \n    def plot(self,*args,**kwargs):\n        titlestring = kwargs.get('titlestring',\"Eigenfunctions of the 1d Potential\")\n        xstring = kwargs.get('xstring',\"Displacement (bohr)\")\n        ystring = kwargs.get('ystring',\"Energy (hartree)\")\n        if not args:\n            args = [3]\n        x = self.x\n        E,U = eigh(self.H)\n        h = x[1]-x[0]\n\n        # Plot the Potential\n        plot(x,self.Vx,color='k')\n\n        for i in range(*args):\n            # For each of the first few solutions, plot the energy level:\n            axhline(y=E[i],color='k',ls=\":\")\n            # as well as the eigenfunction, displaced by the energy level so they don't\n            # all pile up on each other:\n            plot(x,U[:,i]/sqrt(h)+E[i])\n        title(titlestring)\n        xlabel(xstring)\n        ylabel(ystring) \n        return\n        \n    def laplacian(self):\n        x = self.x\n        h = x[1]-x[0] # assume uniformly spaced points\n        n = len(x)\n        M = -2*identity(n,'d')\n        for i in range(1,n):\n            M[i,i-1] = M[i-1,i] = 1\n        return M/h**2",
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     "metadata": {},
     "outputs": [],
     "prompt_number": 44
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    {
     "cell_type": "code",
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     "input": "fig6 = plt.figure()\nsquare_well = Schrod1d(lambda x: 0*x,m=10)\nsquare_well.plot(4,titlestring=\"Square Well Potential\")\nfig_to_plotly(fig6, username, api_key, notebook = True)",
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     "cell_type": "code",
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     "input": "fig7 = plt.figure()\nho = Schrod1d(lambda x: x**2,start=-3,end=3)\nho.plot(6,titlestring=\"Harmonic Oscillator\")\nfig_to_plotly(fig7, username, api_key, notebook = True)",
     "language": "python",
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     "input": "fig8 = plt.figure()\ndef finite_well(x,V_left=1,V_well=0,V_right=1,d_left=10,d_well=10,d_right=10):\n    V = zeros(x.size,'d')\n    for i in range(x.size):\n        if x[i] < d_left: \n            V[i] = V_left\n        elif x[i] > (d_left+d_well):\n            V[i] = V_right\n        else:\n            V[i] = V_well\n    return V\n        \nfw = Schrod1d(finite_well,start=0,end=30,npts=100)\nfw.plot()\nfig_to_plotly(fig8, username, api_key, notebook = True)",
     "language": "python",
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     "cell_type": "code",
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     "input": "fig9 = plt.figure()\ndef triangular(x,F=30): return F*x\n\ntw = Schrod1d(triangular,m=10)\ntw.plot()\nfig_to_plotly(fig9, username, api_key, notebook = True)",
     "language": "python",
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     "cell_type": "code",
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     "input": "fig10 = plt.figure()\ndef tri_finite(x): return finite_well(x)+triangular(x,F=0.025)\n\ntfw = Schrod1d(tri_finite,start=0,end=30,npts=100)\ntfw.plot()\nfig_to_plotly(fig10, username, api_key, notebook = True)",
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