updated populationvector project

projects/Makefile

@@ -1,4 +1,4 @@
-all: projects evalutation
+all: projects evaluation
 
 evaluation: evaluation.pdf
 evaluation.pdf: evaluation.tex

projects/evaluation.tex

@@ -15,7 +15,7 @@
 \begin{document}
 
 \sffamily
-\section*{Scientific computing WS16/17}
+\section*{Scientific computing WS17/18}
 
 \begin{tabular}{|p{0.15\textwidth}|p{0.07\textwidth}|p{0.07\textwidth}|p{0.07\textwidth}|p{0.07\textwidth}|p{0.07\textwidth}|p{0.07\textwidth}|p{0.07\textwidth}|p{0.07\textwidth}|p{0.07\textwidth}|}
 \hline

projects/project_noiseficurves/noiseficurves.tex

@@ -11,7 +11,7 @@
 
 
 %%%%%%%%%%%%%% Questions %%%%%%%%%%%%%%%%%%%%%%%%%
-%\section{REPLACE BY SUBTHRESHOLD RESONANCE PROJECT!}
+\section{REPLACE BY SUBTHRESHOLD RESONANCE PROJECT!}
 \begin{questions}
   \question You are recording the activity of a neuron in response to
   constant stimuli of intensity $I$ (think of that, for example,

projects/project_populationvector/code/generatedata.py

@@ -69,15 +69,15 @@ plt.show()
 
 # tuning curves:
 nunits = 6
-unitphases = np.linspace(0.0, 1.0, nunits) + 0.05*np.random.randn(nunits)/float(nunits)
+unitphases = np.linspace(0.04, 0.96, nunits) + 0.05*np.random.randn(nunits)/float(nunits)
 unitgains = 15.0 + 5.0*(2.0*np.random.rand(nunits)-1.0)
 nangles = 12
 angles = 180.0*np.arange(nangles)/nangles
 for unit, (phase, gain) in enumerate(zip(unitphases, unitgains)):
-    print '%.1f %.0f' % (gain, phase*180.0)
+    print 'gain=%.1f phase=%.0f' % (gain, phase*180.0)
     allspikes = []
     for k, angle in enumerate(angles):
-        spikes = lifadaptspikes(0.5*(1.0-np.cos(2.0*np.pi*(angle/180.0-phase))), gain)
+        spikes = lifadaptspikes(0.5*(1.0+np.cos(2.0*np.pi*(angle/180.0-phase))), gain)
         allspikes.append(spikes)
     spikesobj = np.zeros((len(allspikes), len(allspikes[0])), dtype=np.object)
     for k in range(len(allspikes)):
@@ -89,10 +89,10 @@ for unit, (phase, gain) in enumerate(zip(unitphases, unitgains)):
 nangles = 50
 angles = 180.0*np.random.rand(nangles)
 for k, angle in enumerate(angles):
-    print '%.0f' % angle
+    print 'angle = %.0f' % angle
     allspikes = []
     for unit, (phase, gain) in enumerate(zip(unitphases, unitgains)):
-        spikes = lifadaptspikes(0.5*(1.0-np.cos(2.0*np.pi*(angle/180.0-phase))), gain)
+        spikes = lifadaptspikes(0.5*(1.0+np.cos(2.0*np.pi*(angle/180.0-phase))), gain)
         allspikes.append(spikes)
     spikesobj = np.zeros((len(allspikes), len(allspikes[0])), dtype=np.object)
     for i in range(len(allspikes)):

projects/project_populationvector/populationvector.tex

@@ -35,8 +35,6 @@
   \texttt{spikes} variables of the \texttt{population*.mat} files.
   The \texttt{angle} variable holds the angle of the presented bar.
 
-%NOTE: the orientation is angle plus 90 degree!!!!!!
-
 \continue
   \begin{parts}
     \part Illustrate the spiking activity of the V1 cells in response
@@ -50,13 +48,11 @@
     of the neurons.
 
     \part Fit the function \[ r(\varphi) =
-    g(1-\cos(\varphi-\varphi_0))/2 \] to the measured tuning curves in
+    g(1+\cos(\varphi-\varphi_0))/2 \] to the measured tuning curves in
     order to estimated the orientation angle at which the neurons
     respond strongest. In this function $\varphi_0$ is the position of
-    the peak (really? How exactly is $\varphi_0$ related to the
+    the peak and $g$ is a gain factor that sets the maximum firing
-    position of the peak? Do you find a better function where
+    rate.
-    $\varphi_0$ is identical with the peak position?) and $g$ is a
-    gain factor that sets the maximum firing rate.
 
     \part How can the orientation angle of the presented bar be read
     out from one trial of the population activity of the 6 neurons?
@@ -70,8 +66,8 @@
     An alternative read out is maximum likelihood (see script).
 
     Load one of the \texttt{population*.mat} files, illustrate the
-    data, and estimate the orientation angle of the bar by the two
+    data, and estimate the orientation angle of the bar from single
-    different methods.
+    trial data by the two different methods.
 
     \part Compare, illustrate and discuss the performance of your two
     decoding methods by using all of the recorded responses (all
@@ -81,16 +77,16 @@
   \end{parts}
 \end{questions}
 
-%NOTE: change data generation such that the phase variable is indeed
-%the maximum response and not the minumum!
-
 \end{document}
 
+
 gains and angles of the 6 neurons:
 
-14.6 0
+gain=10.7 phase=5
-17.1 36
+gain=18.0 phase=38
-17.6 72
+gain=11.3 phase=71
-14.1 107
+gain=14.1 phase=108
-10.7 144
+gain=19.0 phase=138
-11.4 181
+gain=16.4 phase=174
+
+

projects/project_populationvector/solution/populationvector.m

@@ -1,7 +1,9 @@
 close all
-files = dir('../unit*.mat');
+datapath = '../';
+% datapath = '../code/';
+files = dir(strcat(datapath, 'unit*.mat'));
 for file = files'
-	a = load(strcat('../', file.name));
+	a = load(strcat(datapath, file.name));
 	spikes = a.spikes;
 	angles = a.angles;
 	figure()
@@ -14,13 +16,13 @@ end
 
 %% tuning curves:
 close all
-cosine = @(p,xdata)0.5*p(1).*(1.0-cos(2.0*pi*(xdata/180.0-p(2))));
+cosine = @(p,xdata)0.5*p(1).*(1.0+cos(2.0*pi*(xdata/180.0-p(2))));
-files = dir('../unit*.mat');
+files = dir(strcat(datapath, 'unit*.mat'));
 phases = zeros(length(files), 1);
 figure()
 for j = 1:length(files)
 	file = files(j);
-	a = load(strcat('../', file.name));
+	a = load(strcat(datapath, file.name));
 	spikes = a.spikes;
 	angles = a.angles;
 	rates = zeros(size(spikes, 1), 1);
@@ -32,10 +34,13 @@ for j = 1:length(files)
 	p0 = [mr, angles(maxi)/180.0-0.5];
 	%p = p0;
 	p = lsqcurvefit(cosine, p0, angles, rates');
-	phase = p(2)*180.0 + 90.0;
+	phase = p(2)*180.0;
     if phase > 180.0
         phase = phase - 180.0;
     end
+    if phase < 0.0
+        phase = phase + 180.0;
+    end
     phases(j) = phase;
 	subplot(2, 3, j);
 	plot(angles, rates, 'b');
@@ -49,40 +54,45 @@ end
 
 
 %% read out:
-a = load('../population04.mat');
+a = load(strcat(datapath, 'population04.mat'));
 spikes = a.spikes;
 angle = a.angle;
-unitphases = a.phases*180.0 + 90.0;
+unitphases = a.phases*180.0;
 unitphases(unitphases>180.0) = unitphases(unitphases>180.0) - 180.0;
 figure();
 subplot(1, 3, 1);
 angleestimates1 = zeros(size(spikes, 2), 1);
 angleestimates2 = zeros(size(spikes, 2), 1);
+[x, inx] = sort(phases);
+% loop over trials:
 for j = 1:size(spikes, 2)
     rates = zeros(size(spikes, 1), 1);
     for k = 1:size(spikes, 1)
         r = firingrate(spikes(k, j), 0.0, 0.2);
         rates(k) = r;
     end
-    [x, inx] = sort(phases);
     plot(phases(inx), rates(inx), '-o');
     hold on;
     angleestimates1(j) = popvecangle(phases, rates);
     [m, i] = max(rates);
     angleestimates2(j) = phases(i);
 end
+xlabel('preferred angle')
+ylabel('firing rate')
 hold off;
 subplot(1, 3, 2);
 hist(angleestimates1);
+xlabel('stimulus angle')
 subplot(1, 3, 3);
 hist(angleestimates2);
+xlabel('stimulus angle')
 angle
 mean(angleestimates1)
 mean(angleestimates2)
 
 
 %% read out robustness:
-files = dir('../population*.mat');
+files = dir(strcat(datapath, 'population*.mat'));
 angles = zeros(length(files), 1);
 e1m = zeros(length(files), 1);
 e1s = zeros(length(files), 1);
@@ -90,7 +100,7 @@ e2m = zeros(length(files), 1);
 e2s = zeros(length(files), 1);
 for i = 1:length(files)
 	file = files(i);
-	a = load(strcat('../', file.name));
+	a = load(strcat(datapath, file.name));
     spikes = a.spikes;
     angle = a.angle;
     angleestimates1 = zeros(size(spikes, 2), 1);
@@ -114,5 +124,9 @@ end
 figure();
 subplot(1, 2, 1);
 scatter(angles, e1m);
+xlabel('stimuluis angle')
+ylabel('estimated angle')
 subplot(1, 2, 2);
 scatter(angles, e2m);
+xlabel('stimuluis angle')
+ylabel('estimated angle')