I'm new to python.
Again, uh..I apologize for the noobnes.
so, I have the following script
Basically I'm trying to learn by making projects, for example a .py that changes your wallpaper, plays a creepy sound and shows a 'message box'. and it repeats for 3 times.
import ctypes
from ctypes import wintypes as w
from playsound import playsound
count = 0
while (count < 3):
count = count + 1
SPI_SETDESKWALLPAPER = 20
ctypes.windll.user32.SystemParametersInfoA(SPI_SETDESKWALLPAPER, 0, "horror.jpg" , 0)
user32 = ctypes.WinDLL('user32')
MessageBox = user32.MessageBoxW
MessageBox.argtypes = w.HWND,w.LPCWSTR,w.LPCWSTR,w.UINT
MessageBox.restype = ctypes.c_int
MessageBox(None, u'works 3 times', u'test:', 0)
playsound('cursed.mp3')
my question would be : why is it working so slow or what am I doing wrong ?
For example, it firsts changes the wallpaper and shows the notification...after I click ok it plays the sound :/
Then it repeats...but it takes like 2 seconds in-between-commands...if that makes sense...I apologise for the extreme stupid question, but I would like some clarification...Have I skipped the 'syntax' class ?
Windows 10
Python 3.7
Anaconda 1.9.7
Spyder 3.3.3
PsychoPy for Python 2.7
I am coding an experiment that needs to present images in a random order for the participant to respond to. I am able to get the images in an array, but to present them one at a time I am using a while loop with a variable that increases by 1 every time it goes through the loop. It is not recognizing the variable as a number and therefore the array cannot call anything.
I've tried not randomizing the variable to see if that is the issue, but it just seems to be that my variable i is not being read as a number
#import packages
import random, os
from psychopy import core, visual, event
from PIL import Image
#setup screen with specs and draw
win = visual.Window([400, 300], monitor="testMonitor")
message = visual.TextStim(win, text="")
message.draw()
win.flip()
core.wait(3.0)
#set image size and populate array with images
stim_size = (0.8, 0.8)
image = [i for i in os.listdir('C:/Users/*/psychopy-tests')
if i.endswith('.bmp')]
#randomize image order
images = random.shuffle(image)
this is where my issue seems to be
i = 0
while i != 29: #there are only 28 images
new = images[i] #this is where the issue is
image_stim = Image.open(new)
stim = visual.ImageStim(win, image_stim, size = (stim_size))
stim.draw()
win.update()
output = []
if event.getKeys(keyList=['space']):
output[i] = 1
if event.getKeys(['escape']):
win.close()
core.quit()
if event.getKeys(keyList=None):
output[i] = 0
core.wait(5.0)
i = i + 1
The random.shuffle shuffles in place and doesn’t return anything i.e., It returns None.
Therefor images is None and not subscriptable.
source
I am new to python. I tried the example given in here http://docs.bokeh.org/en/latest/docs/gallery/color_scatter.html with my own dataset, which looks like this
Unnamed: 0 fans id stars
0 0 69 18kPq7GPye-YQ3LyKyAZPw 4.14
1 1 1345 rpOyqD_893cqmDAtJLbdog 3.67
2 2 105 4U9kSBLuBDU391x6bxU-YA 3.68
3 3 2 fHtTaujcyKvXglE33Z5yIw 4.64
4 4 5 SIBCL7HBkrP4llolm4SC2A 3.80
here's my code:
import pandas as pd
from bokeh.plotting import figure, show, output_file
op = pd.read_csv('FansStars.csv')
x = op.stars
y = op.fans
radii = 1.5
colors = ["#%02x%02x%02x" % (int(r), int(g), 150) for r, g in zip(50+2*x, 30+2*y)]
TOOLS="hover,crosshair,pan,wheel_zoom,zoom_in,zoom_out,box_zoom,undo,redo,reset,tap,save,box_select,poly_select,lasso_select,"
p = figure(tools=TOOLS)
p.scatter(x, y, radius=radii,
fill_color=colors, fill_alpha=0.6,
line_color=None)
output_file("color_scatter.html", title="color_scatter.py example")
show(p)
However, when I run this code, I get no error and a webpage is opened, but BLANK. On reloading several times, I can finally see the tools, but that's all.
Can anyone tell me where am I going wrong?
Thanks!
I cant replicate this on Python 3.4 with Bokeh 0.12.3. So in that way, your code seems fine. I tried it both in the notebook (output_notebook) and to a file like you do and both seem to work fine.
The radius of 1.5 which you specify is taken in data units (x apparently), this makes the circles extremely big, covering the entire screen at first render. But using the wheelzoom to zoom out a bit reveals all circles as expected. Here is what your code looks like in Firefox for me (after zooming out):
I am trying to build a model for the likelihood function of a particular outcome of a Langevin equation (Brownian particle in a harmonic potential):
Here is my model in pymc2 that seems to work:
https://github.com/hstrey/BayesianAnalysis/blob/master/Langevin%20simulation.ipynb
#define the model/function to be fitted.
def model(x):
t = pm.Uniform('t', 0.1, 20, value=2.0)
A = pm.Uniform('A', 0.1, 10, value=1.0)
#pm.deterministic(plot=False)
def S(t=t):
return 1-np.exp(-4*delta_t/t)
#pm.deterministic(plot=False)
def s(t=t):
return np.exp(-2*delta_t/t)
path = np.empty(N, dtype=object)
path[0]=pm.Normal('path_0',mu=0, tau=1/A, value=x[0], observed=True)
for i in range(1,N):
path[i] = pm.Normal('path_%i' % i,
mu=path[i-1]*s,
tau=1/A/S,
value=x[i],
observed=True)
return locals()
mcmc = pm.MCMC( model(x) )
mcmc.sample( 20000, 2000, 10 )
The basic idea is that each point depends on the previous point in the chain (Markov chain). Btw, x is an array of data, N is its length, delta_t is the time step =0.01. Any idea how to implement this in pymc3? I tried:
# define the model/function for diffusion in a harmonic potential
DHP_model = pm.Model()
with DHP_model:
t = pm.Uniform('t', 0.1, 20)
A = pm.Uniform('A', 0.1, 10)
S=1-pm.exp(-4*delta_t/t)
s=pm.exp(-2*delta_t/t)
path = np.empty(N, dtype=object)
path[0]=pm.Normal('path_0',mu=0, tau=1/A, observed=x[0])
for i in range(1,N):
path[i] = pm.Normal('path_%i' % i,
mu=path[i-1]*s,
tau=1/A/S,
observed=x[i])
Unfortunately the model crashes as soon as I try to run it. I tried some pymc3 examples (tutorial) on my machine and this is working.
Thanks in advance. I am really hoping that the new samplers in pymc3 will help me with this model. I am trying to apply Bayesian methods to single-molecule experiments.
Rather than creating many individual normally-distributed 1-D variables in a loop, you can make a custom distribution (by extending Continuous) that knows the formula for computing the log likelihood of your entire path. You can bootstrap this likelihood formula off of the Normal likelihood formula that pymc3 already knows. See the built-in AR1 class for an example.
Since your particle follows the Markov property, your likelihood looks like
import theano.tensor as T
def logp(path):
now = path[1:]
prev = path[:-1]
loglik_first = pm.Normal.dist(mu=0., tau=1./A).logp(path[0])
loglik_rest = T.sum(pm.Normal.dist(mu=prev*ss, tau=1./A/S).logp(now))
loglik_final = loglik_first + loglik_rest
return loglik_final
I'm guessing that you want to draw a value for ss at every time step, in which case you should make sure to specify ss = pm.exp(..., shape=len(x)-1), so that prev*ss in the block above gets interpreted as element-wise multiplication.
Then you can just specify your observations with
path = MyLangevin('path', ..., observed=x)
This should run much faster.
Since I did not see an answer to my question, let me answer it myself. I came up with the following solution:
# now lets model this data using pymc
# define the model/function for diffusion in a harmonic potential
DHP_model = pm.Model()
with DHP_model:
D = pm.Gamma('D',mu=mu_D,sd=sd_D)
A = pm.Gamma('A',mu=mu_A,sd=sd_A)
S=1.0-pm.exp(-2.0*delta_t*D/A)
ss=pm.exp(-delta_t*D/A)
path=pm.Normal('path_0',mu=0.0, tau=1/A, observed=x[0])
for i in range(1,N):
path = pm.Normal('path_%i' % i,
mu=path*ss,
tau=1.0/A/S,
observed=x[i])
start = pm.find_MAP()
print(start)
trace = pm.sample(100000, start=start)
unfortunately, this code takes at N=50 anywhere between 6hours to 2 days to compile. I am running on a pretty fast PC (24Gb RAM) running Ubuntu. I tried to using the GPU but that runs slightly slower. I suspect memory problems since it uses 99.8% of the memory when running. I tried the same calculation with Stan and it only takes 2min to run.
As my title suggests, I'm trying to fit a Gaussian to some data and I'm just getting a straight line. I've been looking at these other discussion Gaussian fit for Python and Fitting a gaussian to a curve in Python which seem to suggest basically the same thing. I can make the code in those discussions work fine for the data they provide, but it won't do it for my data.
My code looks like this:
import pylab as plb
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
from scipy import asarray as ar,exp
y = y - y[0] # to make it go to zero on both sides
x = range(len(y))
max_y = max(y)
n = len(y)
mean = sum(x*y)/n
sigma = np.sqrt(sum(y*(x-mean)**2)/n)
# Someone on a previous post seemed to think this needed to have the sqrt.
# Tried it without as well, made no difference.
def gaus(x,a,x0,sigma):
return a*exp(-(x-x0)**2/(2*sigma**2))
popt,pcov = curve_fit(gaus,x,y,p0=[max_y,mean,sigma])
# It was suggested in one of the other posts I looked at to make the
# first element of p0 be the maximum value of y.
# I also tried it as 1, but that did not work either
plt.plot(x,y,'b:',label='data')
plt.plot(x,gaus(x,*popt),'r:',label='fit')
plt.legend()
plt.title('Fig. 3 - Fit for Time Constant')
plt.xlabel('Time (s)')
plt.ylabel('Voltage (V)')
plt.show()
The data I am trying to fit is as follows:
y = array([ 6.95301373e+12, 9.62971320e+12, 1.32501876e+13,
1.81150568e+13, 2.46111132e+13, 3.32321345e+13,
4.45978682e+13, 5.94819771e+13, 7.88394616e+13,
1.03837779e+14, 1.35888594e+14, 1.76677210e+14,
2.28196006e+14, 2.92781632e+14, 3.73133045e+14,
4.72340762e+14, 5.93892782e+14, 7.41632194e+14,
9.19750269e+14, 1.13278296e+15, 1.38551838e+15,
1.68291212e+15, 2.02996957e+15, 2.43161742e+15,
2.89259207e+15, 3.41725793e+15, 4.00937676e+15,
4.67187762e+15, 5.40667931e+15, 6.21440313e+15,
7.09421973e+15, 8.04366842e+15, 9.05855930e+15,
1.01328502e+16, 1.12585509e+16, 1.24257598e+16,
1.36226443e+16, 1.48356404e+16, 1.60496345e+16,
1.72482199e+16, 1.84140400e+16, 1.95291969e+16,
2.05757166e+16, 2.15360187e+16, 2.23933053e+16,
2.31320228e+16, 2.37385276e+16, 2.42009864e+16,
2.45114362e+16, 2.46427484e+16, 2.45114362e+16,
2.42009864e+16, 2.37385276e+16, 2.31320228e+16,
2.23933053e+16, 2.15360187e+16, 2.05757166e+16,
1.95291969e+16, 1.84140400e+16, 1.72482199e+16,
1.60496345e+16, 1.48356404e+16, 1.36226443e+16,
1.24257598e+16, 1.12585509e+16, 1.01328502e+16,
9.05855930e+15, 8.04366842e+15, 7.09421973e+15,
6.21440313e+15, 5.40667931e+15, 4.67187762e+15,
4.00937676e+15, 3.41725793e+15, 2.89259207e+15,
2.43161742e+15, 2.02996957e+15, 1.68291212e+15,
1.38551838e+15, 1.13278296e+15, 9.19750269e+14,
7.41632194e+14, 5.93892782e+14, 4.72340762e+14,
3.73133045e+14, 2.92781632e+14, 2.28196006e+14,
1.76677210e+14, 1.35888594e+14, 1.03837779e+14,
7.88394616e+13, 5.94819771e+13, 4.45978682e+13,
3.32321345e+13, 2.46111132e+13, 1.81150568e+13,
1.32501876e+13, 9.62971320e+12, 6.95301373e+12,
4.98705540e+12])
I would show you what it looks like, but apparently I don't have enough reputation points...
Anyone got any idea why it's not fitting properly?
Thanks for your help :)
The importance of the initial guess, p0 in curve_fit's default argument list, cannot be stressed enough.
Notice that the docstring mentions that
[p0] If None, then the initial values will all be 1
So if you do not supply it, it will use an initial guess of 1 for all parameters you're trying to optimize for.
The choice of p0 affects the speed at which the underlying algorithm changes the guess vector p0 (ref. the documentation of least_squares).
When you look at the data that you have, you'll notice that the maximum and the mean, mu_0, of the Gaussian-like dataset y, are
2.4e16 and 49 respectively. With the peak value so large, the algorithm, would need to make drastic changes to its initial guess to reach that large value.
When you supply a good initial guess to the curve fitting algorithm, convergence is more likely to occur.
Using your data, you can supply a good initial guess for the peak_value, the mean and sigma, by writing them like this:
y = np.array([...]) # starting from the original dataset
x = np.arange(len(y))
peak_value = y.max()
mean = x[y.argmax()] # observation of the data shows that the peak is close to the center of the interval of the x-data
sigma = mean - np.where(y > peak_value * np.exp(-.5))[0][0] # when x is sigma in the gaussian model, the function evaluates to a*exp(-.5)
popt,pcov = curve_fit(gaus, x, y, p0=[peak_value, mean, sigma])
print(popt) # prints: [ 2.44402560e+16 4.90000000e+01 1.20588976e+01]
Note that in your code, for the mean you take sum(x*y)/n , which is strange, because this would modulate the gaussian by a polynome of degree 1 (it multiplies a gaussian with a monotonously increasing line of constant slope) before taking the mean. That will offset the mean value of y (in this case to the right). A similar remark can be made for your calculation of sigma.
Final remark: the histogram of y will not resemble a Gaussian, as y is already a Gaussian. The histogram will merely bin (count) values into different categories (answering the question "how many datapoints in y reach a value between [a, b]?").