I am having trouble controlling the color and linestyle of histogram plotted using Matplotlib's hist function with stacked=True. For a single non-stacked histogram, I have no trouble:
import pylab as P
mu, sigma = 200, 25
x0 = mu + sigma*P.randn(10000)
n, bins, patches = P.hist(
x0, 20,
histtype='stepfilled',
facecolor='lightblue'
)
However, when I introduce additional histograms,
import pylab as P
mu, sigma = 200, 25
x0 = mu + sigma*P.randn(10000)
x1 = mu + sigma*P.randn(7000)
x2 = mu + sigma*P.randn(3000)
n, bins, patches = P.hist(
[x0,x1,x2], 20,
histtype='stepfilled',
stacked=True,
facecolor=['lightblue','lightgreen','crimson']
)
it throws the following error:
ValueError: to_rgba: Invalid rgba arg "['lightblue', 'lightgreen', 'crimson']"
could not convert string to float: lightblue
Using the color=['lightblue', 'lightgreen', 'crimson'] option does work, but I would like to have direct control of the fill and line colors separately while being able to use the named Matplotlib colors. I am using version 1.2.1 of Matplotlib.
facecolor needs to be a single named color, not a list, but adding this
after your P.hist usage might get the job done for you:
for patch in patches[0]: patch.set_facecolor('lightblue')
for patch in patches[1]: patch.set_facecolor('lightgreen')
for patch in patches[2]: patch.set_facecolor('crimson')
Related
mlab.csd from matplotlib: http://matplotlib.org/api/mlab_api.html#matplotlib.mlab.csd can be used to get real valued cross spectral density. If I want to get the phase information from the spectral density, I need a csd calculation which returns complex values. Is there one ?
This is discussed e.g. in this answer: https://stackoverflow.com/a/29306730/3920342
If you use csd of the mlab library you will get complex values so you can calculate phase angles (and the real valued coherence). In the following code s1 and and s2 contain the two signals (in time domain) to be correlated.
from matplotlib import mlab
# First create power sectral densities for normalization
(ps1, f) = mlab.psd(s1, Fs=1./dt, scale_by_freq=False)
(ps2, f) = mlab.psd(s2, Fs=1./dt, scale_by_freq=False)
plt.plot(f, ps1)
plt.plot(f, ps2)
# Then calculate cross spectral density
(csd, f) = mlab.csd(s1, s2, NFFT=256, Fs=1./dt,sides='default', scale_by_freq=False)
fig = plt.figure()
ax1 = fig.add_subplot(1, 2, 1)
# Normalize cross spectral absolute values by auto power spectral density
ax1.plot(f, np.absolute(csd)**2 / (ps1 * ps2))
ax2 = fig.add_subplot(1, 2, 2)
angle = np.angle(csd, deg=True)
angle[angle<-90] += 360
ax2.plot(f, angle)
# zoom in on frequency with maximum coherence
ax1.set_xlim(9, 11)
ax1.set_ylim(0, 1e-0)
ax1.set_title("Cross spectral density: Coherence")
ax2.set_xlim(9, 11)
ax2.set_ylim(0, 90)
ax2.set_title("Cross spectral density: Phase angle")
Here the real and imaginary(!) part of the cross spectral density:
This code is taken from the question How to use the cross-spectral density to calculate the phase shift of two related signals to create two signals s1 and s2:
"""
Compute the coherence of two signals
"""
import numpy as np
import matplotlib.pyplot as plt
# make a little extra space between the subplots
plt.subplots_adjust(wspace=0.5)
nfft = 256
dt = 0.01
t = np.arange(0, 30, dt)
nse1 = np.random.randn(len(t)) # white noise 1
nse2 = np.random.randn(len(t)) # white noise 2
r = np.exp(-t/0.05)
cnse1 = np.convolve(nse1, r, mode='same')*dt # colored noise 1
cnse2 = np.convolve(nse2, r, mode='same')*dt # colored noise 2
# two signals with a coherent part and a random part
s1 = 0.01*np.sin(2*np.pi*10*t) + cnse1
s2 = 0.01*np.sin(2*np.pi*10*t) + cnse2
I can use np.polyfit to fit a line in my scatter plot as shown bellow
a = np.array([1.08,2.05,1.56,0.73,1.1,0.73,0.34,0.73,0.88,2.05])
b=np.array([4.72131259, 6.60937492, 6.41485738, 6.82386894, 6.20293278, 7.22670489, 6.15681295, 5.91595178, 6.43917035, 6.64453907])
m1, b1 = np.polyfit(a, b, 1)
corr1 =a1.plot(a, m1*a+b1, '-', color='black')
a1.scatter(a, b)
Is there any way to fit a line using polyfit this time taking the errors for my points as shown bellow?
ae = np.empty(10)
ae.fill(0.15)
be = ae
sca1=a1.errorbar(a, b, ae, be, capsize=0, ls='none', color='black', elinewidth=1)
If you want to compute the fit and plot the (fixed at the given value) error bars over the fit points, like this:
Then this code will do the job:
import numpy as np
import matplotlib.pyplot as mp
a = np.array([1.08,2.05,1.56,0.73,1.1,0.73,0.34,0.73,0.88,2.05])
b=np.array([4.72131259, 6.60937492, 6.41485738, 6.82386894, 6.20293278, 7.22670489, 6.15681295, 5.91595178, 6.43917035, 6.64453907])
ae = np.empty(10)
ae.fill(0.15)
be = ae
m1, b1 = np.polyfit(a, b, 1)
mp.figure()
corr1 =mp.errorbar(a,m1*a+b1,ae,be, '-', color='black')
mp.scatter(a, b)
mp.show()
If you want to get the covariance of the fit, and use the standard deviation to set the error bars, instead use the code
import numpy as np
import matplotlib.pyplot as mp
import math
a = np.array([1.08,2.05,1.56,0.73,1.1,0.73,0.34,0.73,0.88,2.05])
b=np.array([4.72131259, 6.60937492, 6.41485738, 6.82386894, 6.20293278, 7.22670489, 6.15681295, 5.91595178, 6.43917035, 6.64453907])
coeff,covar = np.polyfit(a, b, 1,cov=True)
m1= coeff[0]
b1= coeff[1]
xe = math.sqrt(covar[0][0])
ye = math.sqrt(covar[1][2])
mp.figure()
corr1 =mp.errorbar(a,m1*a+b1,xe,ye, '-', color='black')
mp.scatter(a, b)
mp.show()
which gives a plot like this:
If you want to do a weighted fit, you can supply a weight vector to polyfit with the syntax
m2, b2 = np.polyfit(a, b, 1,w=weightvector)
According to the documentation the weightvector should contain 1 over the standard deviation of the data points.
If you want to do a least squares fit weighted by errors in BOTH x and y, I don't think polyfit does this - it will accept a weight vector for one dimension.
To supply errors in both dimensions as weights you would have to use scipy.optimize.leastsq.
There is a documentation page at this link of the Scipy documentation about doing fits with scipy.optimize.leastsq. The example talks about a fit to power law, but clearly a straight line could be done as well.
For errors in one dimension (Y) here, an example using leastsq is:
import numpy as np
import matplotlib.pyplot as mp
import math
from scipy import optimize
a = np.array([1.08,2.05,1.56,0.73,1.1,0.73,0.34,0.73,0.88,2.05])
b=np.array([4.72131259, 6.60937492, 6.41485738, 6.82386894, 6.20293278, 7.22670489, 6.15681295, 5.91595178, 6.43917035, 6.64453907])
aerr = np.empty(10)
aerr.fill(0.15)
berr=aerr
# fit a straight line with scipy scipy.optimize.leastsq
# define our (line) fitting function
fitfunc = lambda p, x: p[0] + p[1] * x
errfunc = lambda p, x, y, err: (y - fitfunc(p, x)) / err
pinit = [1.0, -1.0]
out = optimize.leastsq(errfunc, pinit,args=(a, b, aerr), full_output=1)
coeff = out[0]
covar = out[1]
print 'coeff', coeff
print 'covar', covar
m1= coeff[1]
b1= coeff[0]
xe = math.sqrt(covar[0][0])
ye = math.sqrt(covar[1][1])
# plot results
mp.figure()
corr2 =mp.errorbar(a,m1*a+b1,xe,ye, '-', color='red')
mp.scatter(a, b)
mp.show()
To take into account errors in both X and Y, you would have to change the definition of errfunc to reflect the specific technique you are using to do that. If a lambda isn't convenient you can instead define a function that will do that. I can't comment further on this without knowing what technique is being used to weight by X and Y errors, there are several in the literature.
I am rather new to matplotlib (and this is also my first question here). I'm trying to represent the scalp surface potential as recorded by an EEG. So far I have a two-dimensional figure of a sphere projection, which I generated using contourf, and pretty much boils down to an ordinary heat map.
Is there any way this can be done on half a sphere?, i.e. generating a 3D sphere with surface colours given by a list of values? Something like this, http://embal.gforge.inria.fr/img/inverse.jpg, but I have more than enough with just half a sphere.
I have seen a few related questions (for example, Matplotlib 3d colour plot - is it possible?), but they either don't really address my question or remain unanswered to date.
I have also spent the morning looking through countless examples. In most of what I've found, the colour at one particular point of a surface is indicative of its Z value, but I don't want that... I want to draw the surface, then specify the colours with the data I have.
You can use plot_trisurf and assign a custom field to the underlying ScalarMappable through set_array method.
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import matplotlib.tri as mtri
(n, m) = (250, 250)
# Meshing a unit sphere according to n, m
theta = np.linspace(0, 2 * np.pi, num=n, endpoint=False)
phi = np.linspace(np.pi * (-0.5 + 1./(m+1)), np.pi*0.5, num=m, endpoint=False)
theta, phi = np.meshgrid(theta, phi)
theta, phi = theta.ravel(), phi.ravel()
theta = np.append(theta, [0.]) # Adding the north pole...
phi = np.append(phi, [np.pi*0.5])
mesh_x, mesh_y = ((np.pi*0.5 - phi)*np.cos(theta), (np.pi*0.5 - phi)*np.sin(theta))
triangles = mtri.Triangulation(mesh_x, mesh_y).triangles
x, y, z = np.cos(phi)*np.cos(theta), np.cos(phi)*np.sin(theta), np.sin(phi)
# Defining a custom color scalar field
vals = np.sin(6*phi) * np.sin(3*theta)
colors = np.mean(vals[triangles], axis=1)
# Plotting
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
cmap = plt.get_cmap('Blues')
triang = mtri.Triangulation(x, y, triangles)
collec = ax.plot_trisurf(triang, z, cmap=cmap, shade=False, linewidth=0.)
collec.set_array(colors)
collec.autoscale()
plt.show()
I have many pictures as below:
My objective is to identify those "beads", try to mark it with a circle, and count the detected numbers.
I tried to use image segmentation algorithms via Python and the source codes are as below:
from matplotlib import pyplot as plt
from skimage import data
from skimage.feature import blob_dog, blob_log, blob_doh
from math import sqrt
from skimage.color import rgb2gray
from scipy import misc # try
image = misc.imread('test.jpg')
image_gray = rgb2gray(image)
blobs_log = blob_log(image_gray, max_sigma=10, num_sigma=5, threshold=.1)
# Compute radii in the 3rd column.
blobs_log[:, 2] = blobs_log[:, 2] * sqrt(2)
blobs_dog = blob_dog(image_gray, max_sigma=2, threshold=.051)
blobs_dog[:, 2] = blobs_dog[:, 2] * sqrt(2)
blobs_doh = blob_doh(image_gray, max_sigma=2, threshold=.01)
blobs_list = [blobs_log, blobs_dog, blobs_doh]
colors = ['yellow', 'lime', 'red']
titles = ['Laplacian of Gaussian', 'Difference of Gaussian',
'Determinant of Hessian']
sequence = zip(blobs_list, colors, titles)
for blobs, color, title in sequence:
fig, ax = plt.subplots(1, 1)
ax.set_title(title)
ax.imshow(image, interpolation='nearest')
for blob in blobs:
y, x, r = blob
c = plt.Circle((x, y), r, color=color, linewidth=2, fill=False)
ax.add_patch(c)
plt.show()
The best results obtained so far are still unsatisfactory:
How can I improve it ?
You could use Gimp or Photoshop and test some filters and colors changes to differentiate the circles from the background. Brightness and Contrast adjustments may work. Then you can apply an Edge detector to detect the circles.
by converting this image to grayscale you have effectively thrown away the most powerful cue you have to segment the beads - their distinctive green color. try running the same code but replace
image_gray = rgb2gray(image)
with
image_gray = image[:,:,1]
I am using matplotlib.pyplot to create histograms. I'm not actually interested in the plots of these histograms, but interested in the frequencies and bins (I know I can write my own code to do this, but would prefer to use this package).
I know I can do the following,
import numpy as np
import matplotlib.pyplot as plt
x1 = np.random.normal(1.5,1.0)
x2 = np.random.normal(0,1.0)
freq, bins, patches = plt.hist([x1,x1],50,histtype='step')
to create a histogram. All I need is freq[0], freq[1], and bins[0]. The problem occurs when I try and use,
freq, bins, patches = plt.hist([x1,x1],50,histtype='step')
in a function. For example,
def func(x, y, Nbins):
freq, bins, patches = plt.hist([x,y],Nbins,histtype='step') # create histogram
bincenters = 0.5*(bins[1:] + bins[:-1]) # center bins
xf= [float(i) for i in freq[0]] # convert integers to float
xf = [float(i) for i in freq[1]]
p = [ (bincenters[j], (1.0 / (xf[j] + yf[j] )) for j in range(Nbins) if (xf[j] + yf[j]) != 0]
Xt = [j for i,j in p] # separate pairs formed in p
Yt = [i for i,j in p]
Y = np.array(Yt) # convert to arrays for later fitting
X = np.array(Xt)
return X, Y # return arrays X and Y
When I call func(x1,x2,Nbins) and plot or print X and Y, I do not get my expected curve/values. I suspect it something to do with plt.hist, since there is a partial histogram in my plot.
I don't know if I'm understanding your question very well, but here, you have an example of a very simple home-made histogram (in 1D or 2D), each one inside a function, and properly called:
import numpy as np
import matplotlib.pyplot as plt
def func2d(x, y, nbins):
histo, xedges, yedges = np.histogram2d(x,y,nbins)
plt.plot(x,y,'wo',alpha=0.3)
plt.imshow(histo.T,
extent=[xedges.min(),xedges.max(),yedges.min(),yedges.max()],
origin='lower',
interpolation='nearest',
cmap=plt.cm.hot)
plt.show()
def func1d(x, nbins):
histo, bin_edges = np.histogram(x,nbins)
bin_center = 0.5*(bin_edges[1:] + bin_edges[:-1])
plt.step(bin_center,histo,where='mid')
plt.show()
x = np.random.normal(1.5,1.0, (1000,1000))
func1d(x[0],40)
func2d(x[0],x[1],40)
Of course, you may check if the centering of the data is right, but I think that the example shows some useful things about this topic.
My recommendation: Try to avoid any loop in your code! They kill the performance. If you look, In my example there aren't loops. The best practice in numerical problems with python is avoiding loops! Numpy has a lot of C-implemented functions that do all the hard looping work.
You can use np.histogram2d (for 2D histogram) or np.histogram (for 1D histogram):
hst = np.histogram(A, bins)
hst2d = np.histogram2d(X,Y,bins)
Output form will be the same as plt.hist and plt.hist2d, the only difference is there is no plot.
No.
But you can bypass the pyplot:
import matplotlib.pyplot
fig = matplotlib.figure.Figure()
ax = matplotlib.axes.Axes(fig, (0,0,0,0))
numeric_results = ax.hist(data)
del ax, fig
It won't impact active axes and figures, so it would be ok to use it even in the middle of plotting something else.
This is because any usage of plt.draw_something() will put the plot in current axis - which is a global variable.
If you would like to simply compute the histogram (that is, count the number of points in a given bin) and not display it, the np.histogram() function is available