I am new in it, and cant fully understand the manual. I am running test code, but I want to make blank map, without this gradient colors. I think it's not hard for those who know. Help me please.Here what i got
import numpy as np
import healpy as hp
import pylab as pl
import matplotlib as plt
NSIDE = 32
m = np.arange(hp.nside2npix(NSIDE))
hp.mollview(m)
pl.show()
The numpy function np.arange gives out an array increasing from zero to 12288, which is why you have a gradient on your map.
You can replace that line of code to
m = np.zeros(hp.nside2npix(NSIDE))
if you want to treat blanks in your map as zeros, or to
m = np.full(hp.nside2npix(NSIDE), np.nan)
if you want to use NaN (not a number) value for the blank values on the map.
Related
I want to randomly select numbers with the probability acording to the sum of 2 normal/gaussian distributions and make a histogram.
basicaly
import numpy as np
u, vth = 0,1 # mean and standard deviation
v= np.random.normal(u, vth, 1000)+np.random.normal(-u, vth, 1000)
v.histogram()
However I get the error numpy.ndarray' object has no attribute 'histogram'. Another problem is that this isn't normalized so my results shouldn't be right...
I think what you are looking for is to make a histogram plot of a normal distribution. Have you tried using matplotlib library's histogram function? It can be done easily like so:
import numpy as np
u, vth = 0,1 # mean and standard deviation
v= np.random.normal(u, vth, 1000)+np.random.normal(-u, vth, 1000)
import matplotlib.pyplot as plt
plt.hist(v)
I'm working with the Mnist data set, in order to learn about Machine learning, and as for now I'm trying to display the first digit in the Mnist data set as an image, and I have encountered a problem.
I have a matrix with the dimensions 784x10000, where each column is a digit in the data set. I have created the matrix myself, because the Mnist data set came in the form of a text file, which in itself caused me quite a lot of problems, but that's a question for itself.
The MN_train matrix below, is my large 784x10000 matrix. So what I'm trying to do below, is to fill up a 28x28 matrix, in order to display my image.
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.image as mpimg
grey = np.zeros(shape=(28,28))
k = 0
for l in range(28):
for p in range(28):
grey[p,l]=MN_train[k,0]
k = k + 1
print grey
plt.show(grey)
But when I try to display the image, I get the following error:
The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
Followed by a image plot that does not look like the number five, as I would expect.
Is there something I have overlooked, or does this tell me that my manipulation of the text file, in order to construct the MN_train matrix, has resulted in an error?
The error you get is because you supply the array to show. show accepts only a single boolean argument hold=True or False.
In order to create an image plot, you need to use imshow.
plt.imshow(grey)
plt.show() # <- no argument here
Also note that the loop is rather inefficient. You may just reshape the input column array.
The complete code would then look like
import numpy as np
import matplotlib.pyplot as plt
MN_train = np.loadtxt( ... )
grey = MN_train[:,0].reshape((28,28))
plt.imshow(grey)
plt.show()
Here is my first steps within the NumPy world.
As a matter of fact the target is plotting below 2-D function as a 3-D mesh:
N = \frac{n}{2\sigma\sqrt{\pi}}\exp^{-\frac{n^{2}x^{2}}{4\sigma^{2}}}
That could been done as a piece a cake in Matlab with below snippet:
[x,n] = meshgrid(0:0.1:20, 1:1:100);
mu = 0;
sigma = sqrt(2)./n;
f = normcdf(x,mu,sigma);
mesh(x,n,f);
But the bloody result is ugly enough to drive me trying Python capabilities to generate scientific plots.
I searched something and found that the primary steps to hit above mark in Pyhton might be acquired by below snippet:
from matplotlib.patches import Polygon
import numpy as np
from scipy.integrate import quad
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
sigma = 1
def integrand(x,n):
return (n/(2*sigma*np.sqrt(np.pi)))*np.exp(-(n**2*x**2)/(4*sigma**2))
t = np.linespace(0, 20, 0.01)
n = np.linespace(1, 100, 1)
lower_bound = -100000000000000000000 #-inf
upper_bound = t
tt, nn = np.meshgrid(t,n)
real_integral = quad(integrand(tt,nn), lower_bound, upper_bound)
Axes3D.plot_trisurf(real_integral, tt,nn)
Edit: With due attention to more investigations on Greg's advices, above code is the most updated snippet.
Here is the generated exception:
RuntimeError: infinity comparisons don't work for you
It is seemingly referring to the quad call...
Would you please helping me to handle this integrating-plotting problem?!...
Best
Just a few hints to get you in the right direction.
numpy.meshgrid can do the same as MatLABs function:
http://docs.scipy.org/doc/numpy/reference/generated/numpy.meshgrid.html
When you have x and n you can do math just like in matlab:
sigma = numpy.sqrt(2)/n
(in python multiplication/division is default index by index - no dot needed)
scipy has a lot more advanced functions, see for example How to calculate cumulative normal distribution in Python for a 1D case.
For plotting you can use matplotlibs pcolormesh:
import matplotlib.pyplot as plt
plt.pcolormesh(x,n,real_integral)
Hope this helps until someone can give you a more detailed answer.
I intend for part of a program I'm writing to automatically generate Gaussian distributions of various statistics over multiple raw text sources, however I'm having some issues generating the graphs as per the guide at:
python pylab plot normal distribution
The general gist of the plot code is as follows.
import numpy as np
import matplotlib.mlab as mlab
import matplotlib.pyplot as pyplot
meanAverage = 222.89219487179491 # typical value calculated beforehand
standardDeviation = 3.8857889432054091 # typical value calculated beforehand
x = np.linspace(-3,3,100)
pyplot.plot(x,mlab.normpdf(x,meanAverage,standardDeviation))
pyplot.show()
All it does is produce a rather flat looking and useless y = 0 line!
Can anyone see what the problem is here?
Cheers.
If you read documentation of matplotlib.mlab.normpdf, this function is deprycated and you should use scipy.stats.norm.pdf instead.
Deprecated since version 2.2: scipy.stats.norm.pdf
And because your distribution mean is about 222, you should use np.linspace(200, 220, 100).
So your code will look like:
import numpy as np
from scipy.stats import norm
import matplotlib.pyplot as pyplot
meanAverage = 222.89219487179491 # typical value calculated beforehand
standardDeviation = 3.8857889432054091 # typical value calculated beforehand
x = np.linspace(200, 220, 100)
pyplot.plot(x, norm.pdf(x, meanAverage, standardDeviation))
pyplot.show()
It looks like you made a few small but significant errors. You either are choosing your x vector wrong or you swapped your stddev and mean. Since your mean is at 222, you probably want your x vector in this area, maybe something like 150 to 300. This way you get all the good stuff, right now you are looking at -3 to 3 which is at the tail of the distribution. Hope that helps.
I see that, for the *args which are sending meanAverage, standardDeviation, the correct thing to be sent is:
mu : a numdims array of means of a
sigma : a numdims array of atandard deviation of a
Does this help?
I am working in image processing right now in python using numpy and scipy all the time. I have one piece of code that can enlarge an image, but not sure how this works.
So please some expert in scipy/numpy in python can explain to me line by line. I am always eager to learn.
import numpy as N
import os.path
import scipy.signal
import scipy.interpolate
import matplotlib.pyplot as plt
import matplotlib.cm as cm
def enlarge(img, rowscale, colscale, method='linear'):
x, y = N.meshgrid(N.arange(img.shape[1]), N.arange(img.shape[0]))
pts = N.column_stack((x.ravel(), y.ravel()))
xx, yy = N.mgrid[0.:float(img.shape[1]):1/float(colscale),
0.:float(img.shape[0]):1/float(rowscale)]
large = scipy.interpolate.griddata(pts, img.flatten(), (xx, yy), method).T
large[-1,:] = large[-2,:]
large[:,-1] = large[:,-2]
return large
Thanks a lot.
First, a grid of empty points is created with point per pixel.
x, y = N.meshgrid(N.arange(img.shape[1]), N.arange(img.shape[0]))
The actual image pixels are placed into the variable pts which will be needed later.
pts = N.column_stack((x.ravel(), y.ravel()))
After that, it creates a mesh grid with one point per pixel for the enlarged image; if the original image was 200x400, the colscale set to 4 and rowscale set to 2, the mesh grid would have (200*4)x(400*2) or 800x800 points.
xx, yy = N.mgrid[0.:float(img.shape[1]):1/float(colscale),
0.:float(img.shape[0]):1/float(rowscale)]
Using scipy, the points in pts variable are interpolated into the larger grid. Interpolation is the manner in which missing points are filled or estimated usually when going from a smaller set of points to a larger set of points.
large = scipy.interpolate.griddata(pts, img.flatten(), (xx, yy), method).T
I am not 100% certain what the last two lines do without going back and looking at what the griddata method returns. It appears to be throwing out some additional data that isn't needed for the image or performing a translation.
large[-1,:] = large[-2,:]
large[:,-1] = large[:,-2]
return large