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Good day to everyone. I was wondering if there is any way to extract a mass map and a mass density map for a scatter plot of mass distributions.
Developing the code for the mass distributions:
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.cm as cm
from scipy.ndimage.filters import gaussian_filter
from numpy.random import rand
# Finds nran number of random points in two dimensions
def randomizer(nran):
arr = rand(nran, 2)
return arr
# Calculates a sort of 'density' plot. Using this from a previous StackOverflow Question: https://stackoverflow.com/questions/2369492/generate-a-heatmap-in-matplotlib-using-a-scatter-data-set
def myplot(x, y, s, bins = 1000):
plot, xedges, yedges = np.histogram2d(x, y, bins = bins)
plot = gaussian_filter(plot, sigma = s)
extent = [xedges[0], xedges[-1], yedges[0], yedges[-1]]
return plot.T, extent
Trying out an example:
arr = randomizer(1000)
plot, extent = myplot(arr[:, 0], arr[:, 1], 20)
fig, ax = plt.subplots(1, 2, figsize = (15, 5))
ax[0].scatter(arr[:, 0], arr[:, 1])
ax[0].set_aspect('equal')
ax[0].set_xlabel('x')
ax[0].set_ylabel('y')
ax[0].set_title('Scatter Plot')
img = ax[1].imshow(plot)
ax[1].set_title('Density Plot?')
ax[1].set_aspect('equal')
ax[1].set_xlabel('x')
ax[1].set_ylabel('y')
plt.colorbar(img)
This yields a scatter plot and what I think kind of represents a density plot (please correct if wrong). Now, suppose that each dot has a mass of 50 kg. Does the "density plot" represent a map of the total mass distribution (if that makes sense?)since the colorbar has a max value much less than 50. Then, using this, how can I compute a mass density for this mass distribution? I would really appreciate if someone could help. Thank you.
Edit: Added the website from where I got the heatmap function.
Okay, I think I've got the solution. I've been meaning to upload this for quite an amount of time. Here it goes:
# Importing packages
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.cm as cm
from numpy.random import random
from scipy.stats import binned_statistic_2d
# Finds nran number of random points in two dimensions
def randomizer(nran):
arr_x = []
arr_y = []
for i in range(nran):
arr_x += [10 * random()] # Since random() only produces floats in (0, 1), I multiply by 10 (for illustrative purposes)
arr_y += [10 *random()] # Since random() only produces floats in (0, 1), I multiply by 10 (for illustrative purposes)
return arr_x, arr_y
# Computing weight array
def weights_array(weight, length):
weights = np.array([weight] * length)
return weights
# Computes a weighted histogram and divides it by the total grid area to get the density
def histogramizer(x_array, y_array, weights, num_pixels, Dimension):
Range = [0, Dimension] # Assumes the weights are distributed in a square area
grid, _, _, _ = binned_statistic_2d(x_array, y_array, weights, 'sum', bins=num_pixels, range=[Range,Range])
area = int(np.max(x_array)) * int(np.max(y_array))
density = grid/area
return density
Then, actually implementing this, one finds:
arr_x, arr_y = randomizer(1000000)
weights = []
for i in range(len(arr_x)):
weights += [50]
density = histogramizer(arr_x, arr_y, weights, [400,400], np.max(arr_x))
fig, ax = plt.subplots(figsize = (15, 5))
plt.imshow(density, extent = [0, int(np.max(arr_x)), 0, int(np.max(arr_x))]);
plt.colorbar(label = '$kg m^-2$');
The result I got for this was the following plot (I know it's generally not recommended to add a photo, but I wanted to add it for sake of showing my code's output):
I have some data as numpy arrays x, y, v as shown in the code below.
This is actually dummy data for velocity (v) of dust particles in a x-y plane.
I have binned my data into 4 bins and for each bin I have calculated mean of entries in each bin and made a heat map.
Now what I want to do is make a histogram/distribution of v in each bin with 0 as the centre of the histogram.
I do not want to plot the mean anymore, just want to divide my data into the same bins as this code and for each bin I want to generate a histogram of the values in the bins.
How should I do it?
I think this is a way to model the spectrum of an emission line from the gas particles. Any help is appreciated! Thanks!
from scipy import stats
import numpy as np
import matplotlib.pyplot as plt
x = np.array([-10,-2,4,12,3,6,8,14,3])
y = np.array([5,5,-6,8,-20,10,2,2,8])
v = np.array([4,-6,-10,40,22,-14,20,8,-10])
x_bins = np.linspace(-20, 20, 3)
y_bins = np.linspace(-20, 20, 3)
H, xedges, yedges = np.histogram2d(x, y, bins = [x_bins, y_bins], weights = v)
pstat = stats.binned_statistic_2d(x, y, v, statistic='mean', bins = [x_bins, y_bins])
plt.xlabel("x")
plt.ylabel("y")
plt.imshow(pstat.statistic.T, origin='lower', cmap='RdBu',
extent=[xedges[0], xedges[-1], yedges[0], yedges[-1]])
plt.colorbar().set_label('mean', rotation=270)
EDIT: Please note that my original data is huge. My arrays for x,y, v are very large and I am using 30x30 grid, that is, not just 4quadrants but 900 bins. I might also need to increase the bin number. So, we want to find a way to automatically divide the 'v' data into the regularly spaced bins and then be able to plot the histograms of the 'v' data in each bin.
I would iterate over the zipped x and y, then flag if v is inside the quadrant and append them to a quadrant list. after, you can plot whatever you'd like:
x = np.array([-10,-2,4,12,3,6,8,14,3])
y = np.array([5,5,-6,8,-20,10,2,2,8])
v = np.array([4,-6,-10,40,22,-14,20,8,-10])
q1 = []
q2 = []
q3 = []
q4 = []
for i, (x1,y1) in enumerate(zip(x,y)):
if x1<0 and y1>=0:
q1.append(v[i])
elif x1>=0 and y1>=0:
q2.append(v[i])
elif x1>=0 and y1<0:
q3.append(v[i])
elif x1<0 and y1<0:
q4.append(v[i])
print(q1)
print(q2)
print(q3)
print(q4)
#[4, -6]
#[40, -14, 20, 8, -10]
#[-10, 22]
#[]
plt.hist(q1, density=True)
plt.hist(q2, density=True)
plt.hist(q3, density=True)
#q4 is empty
I have a rectilinear (not regular) grid of data (x,y,V) where V is the value at the position (x,y). I would like to use this data source to interpolate my results so that I can fill in the gaps and plot the interpolated values (inside the range) in the future. (Also I need functionality of griddata to check arbitrary values inside the range).
I looked at the documentation at SciPy and here.
Here is what I tried, and the result:
It clearly doesn't match the data.
# INTERPOLATION ATTEMPT?
from scipy.interpolate import Rbf
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.cm as cm
edges = np.linspace(-0.05, 0.05, 100)
centers = edges[:-1] + np.diff(edges[:2])[0] / 2.
XI, YI = np.meshgrid(centers, centers)
# use RBF
rbf = Rbf(x, y, z, epsilon=2)
ZI = rbf(XI, YI)
# plot the result
plt.subplots(1,figsize=(12,8))
X_edges, Y_edges = np.meshgrid(edges, edges)
lims = dict(cmap='viridis')
plt.pcolormesh(X_edges, Y_edges, ZI, shading='flat', **lims)
plt.scatter(x, y, 200, z, edgecolor='w', lw=0.1, **lims)
#decoration
plt.title('RBF interpolation?')
plt.xlim(-0.05, 0.05)
plt.ylim(-0.05, 0.05)
plt.colorbar()
plt.show()
For reference, here is my data (extracted), it has a circular pattern that I need interpolation to recognize.
#DATA
experiment1raw = np.array([
[0,40,1,11.08,8.53,78.10,2.29],
[24,-32,2,16.52,11.09,69.03,3.37],
[8,-32,4,14.27,10.68,71.86,3.19],
[-8,-32,6,10.86,9.74,76.69,2.72],
[-24,-32,8,6.72,12.74,77.08,3.45],
[32,-24,9,18.49,13.67,64.32,3.52],
[-32,-24,17,6.72,12.74,77.08,3.45],
[16,-16,20,13.41,21.33,59.92,5.34],
[0,-16,22,12.16,14.67,69.04,4.12],
[-16,-16,24,9.07,13.37,74.20,3.36],
[32,-8,27,19.35,17.88,57.86,4.91],
[-32,-8,35,6.72,12.74,77.08,3.45],
[40,0,36,19.25,20.36,54.97,5.42],
[16,0,39,13.41,21.33,59.952,5.34],
[0,0,41,10.81,19.55,64.37,5.27],
[-16,0,43,8.21,17.83,69.34,4.62],
[-40,0,46,5.76,13.43,77.23,3.59],
[32,8,47,15.95,23.61,54.34,6.10],
[-32,8,55,5.97,19.09,70.19,4.75],
[16,16,58,11.27,26.03,56.36,6.34],
[0,16,60,9.19,24.94,60.06,5.79],
[-16,16,62,7.10,22.75,64.57,5.58],
[32,24,65,12.39,29.19,51.17,7.26],
[-32,24,73,5.40,24.55,64.33,5.72],
[24,32,74,10.03,31.28,50.96,7.73],
[8,32,76,8.68,30.06,54.34,6.92],
[-8,32,78,6.88,28.78,57.84,6.49],
[-24,32,80,5.83,26.70,61.00,6.46],
[0,-40,81,7.03,31.55,54.40,7.01],
])
#Atomic Percentages are set here
Cr1 = experiment1raw[:,3]
Mn1 = experiment1raw[:,4]
Fe1 = experiment1raw[:,5]
Co1 = experiment1raw[:,6]
#COORDINATE VALUES IN PRE-T
x_pret = experiment1raw[:,0]/1000
y_pret = experiment1raw[:,1]/1000
#important translation
x = -y_pret
y = -x_pret
z = Cr1
You used a larger epsilon in RBF. Best bet is to set it as default and let scipy calculate an appropriate value. See the implementation here.
So setting default epsilon:
rbf = Rbf(x, y, z)
I got a pretty good interpolation for your data (subjective opinion).
I am doing a Kernel Density Estimation in Python and getting the contours and paths as shown below. (here is my sample data: https://pastebin.com/193PUhQf).
from numpy import *
from math import *
import numpy as np
import matplotlib.pyplot as plt
from scipy import stats
x_2d = []
y_2d = []
data = {}
data['nodes'] = []
# here is the sample data:
# https://pastebin.com/193PUhQf
X = [.....]
for Picker in xrange(0, len(X)):
x_2d.append(X[Picker][0])
y_2d.append(X[Picker][1])
# convert to arrays
m1 = np.array([x_2d])
m2 = np.array([y_2d])
x_min = m1.min() - 30
x_max = m1.max() + 30
y_min = m2.min() - 30
y_max = m2.max() + 30
x, y = np.mgrid[x_min:x_max:200j, y_min:y_max:200j]
positions = np.vstack([x.ravel(), y.ravel()])
values = np.vstack([m1, m2])
kde = stats.gaussian_kde(values)
z = np.reshape(kde(positions).T, x.shape)
fig = plt.figure(2, dpi=200)
ax = fig.add_subplot(111)
pc = ax.pcolor(x, y, z)
cb = plt.colorbar(pc)
cb.ax.set_ylabel('Probability density')
c_s = plt.contour(x, y, z, 20, linewidths=1, colors='k')
ax.plot(m1, m2, 'o', mfc='w', mec='k')
ax.set_title("My Title", fontsize='medium')
plt.savefig("kde.png", dpi=200)
plt.show()
There is a similar way to get the contours using R, which is described here:
http://bl.ocks.org/diegovalle/5166482
Question: how can I achieve the same output using my python script or as a start point?
the desired output should be like contours_tj.json which can be used by leaflet.js lib.
UPDATE:
My input data structure is composed of three columns, comma separated:
first one is the X value
second one is the Y value
third one is the ID of my data, it has no numerical value, it is simply an identifier of the data point.
Update 2:
Question, if simply put, is that I want the same output as in the above link using my input file which is in numpy array format.
update 3:
my input data structure is of list type:
print type(X)
<type 'list'>
and here are the first few lines:
print X[0:5]
[[10.800584, 11.446064, 4478597], [10.576840,11.020229, 4644503], [11.434276,10.790881, 5570870], [11.156718,11.034633, 6500333], [11.054956,11.100243, 6513301]]
geojsoncontour is a python library to convert matplotlib contours to geojson
geojsoncontour.contour_to_geojson requires a contour_levels argument. The levels in pyplot.contour are chosen automatically, but you can access them with c_s._levels
So, for your example you could do:
import geojsoncontour
# your code here
c_s = plt.contour(x, y, z, 20, linewidths=1, colors='k')
# Convert matplotlib contour to geojson
geojsoncontour.contour_to_geojson(
contour=c_s,
geojson_filepath='out.geojson',
contour_levels=c_s._levels,
ndigits=3,
unit='m'
)
Suppose I create a histogram using scipy/numpy, so I have two arrays: one for the bin counts, and one for the bin edges. If I use the histogram to represent a probability distribution function, how can I efficiently generate random numbers from that distribution?
It's probably what np.random.choice does in #Ophion's answer, but you can construct a normalized cumulative density function, then choose based on a uniform random number:
from __future__ import division
import numpy as np
import matplotlib.pyplot as plt
data = np.random.normal(size=1000)
hist, bins = np.histogram(data, bins=50)
bin_midpoints = bins[:-1] + np.diff(bins)/2
cdf = np.cumsum(hist)
cdf = cdf / cdf[-1]
values = np.random.rand(10000)
value_bins = np.searchsorted(cdf, values)
random_from_cdf = bin_midpoints[value_bins]
plt.subplot(121)
plt.hist(data, 50)
plt.subplot(122)
plt.hist(random_from_cdf, 50)
plt.show()
A 2D case can be done as follows:
data = np.column_stack((np.random.normal(scale=10, size=1000),
np.random.normal(scale=20, size=1000)))
x, y = data.T
hist, x_bins, y_bins = np.histogram2d(x, y, bins=(50, 50))
x_bin_midpoints = x_bins[:-1] + np.diff(x_bins)/2
y_bin_midpoints = y_bins[:-1] + np.diff(y_bins)/2
cdf = np.cumsum(hist.ravel())
cdf = cdf / cdf[-1]
values = np.random.rand(10000)
value_bins = np.searchsorted(cdf, values)
x_idx, y_idx = np.unravel_index(value_bins,
(len(x_bin_midpoints),
len(y_bin_midpoints)))
random_from_cdf = np.column_stack((x_bin_midpoints[x_idx],
y_bin_midpoints[y_idx]))
new_x, new_y = random_from_cdf.T
plt.subplot(121, aspect='equal')
plt.hist2d(x, y, bins=(50, 50))
plt.subplot(122, aspect='equal')
plt.hist2d(new_x, new_y, bins=(50, 50))
plt.show()
#Jaime solution is great, but you should consider using the kde (kernel density estimation) of the histogram. A great explanation why it's problematic to do statistics over histogram, and why you should use kde instead can be found here
I edited #Jaime's code to show how to use kde from scipy. It looks almost the same, but captures better the histogram generator.
from __future__ import division
import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import gaussian_kde
def run():
data = np.random.normal(size=1000)
hist, bins = np.histogram(data, bins=50)
x_grid = np.linspace(min(data), max(data), 1000)
kdepdf = kde(data, x_grid, bandwidth=0.1)
random_from_kde = generate_rand_from_pdf(kdepdf, x_grid)
bin_midpoints = bins[:-1] + np.diff(bins) / 2
random_from_cdf = generate_rand_from_pdf(hist, bin_midpoints)
plt.subplot(121)
plt.hist(data, 50, normed=True, alpha=0.5, label='hist')
plt.plot(x_grid, kdepdf, color='r', alpha=0.5, lw=3, label='kde')
plt.legend()
plt.subplot(122)
plt.hist(random_from_cdf, 50, alpha=0.5, label='from hist')
plt.hist(random_from_kde, 50, alpha=0.5, label='from kde')
plt.legend()
plt.show()
def kde(x, x_grid, bandwidth=0.2, **kwargs):
"""Kernel Density Estimation with Scipy"""
kde = gaussian_kde(x, bw_method=bandwidth / x.std(ddof=1), **kwargs)
return kde.evaluate(x_grid)
def generate_rand_from_pdf(pdf, x_grid):
cdf = np.cumsum(pdf)
cdf = cdf / cdf[-1]
values = np.random.rand(1000)
value_bins = np.searchsorted(cdf, values)
random_from_cdf = x_grid[value_bins]
return random_from_cdf
Perhaps something like this. Uses the count of the histogram as a weight and chooses values of indices based on this weight.
import numpy as np
initial=np.random.rand(1000)
values,indices=np.histogram(initial,bins=20)
values=values.astype(np.float32)
weights=values/np.sum(values)
#Below, 5 is the dimension of the returned array.
new_random=np.random.choice(indices[1:],5,p=weights)
print new_random
#[ 0.55141614 0.30226256 0.25243184 0.90023117 0.55141614]
I had the same problem as the OP and I would like to share my approach to this problem.
Following Jaime answer and Noam Peled answer I've built a solution for a 2D problem using a Kernel Density Estimation (KDE).
Frist, let's generate some random data and then calculate its Probability Density Function (PDF) from the KDE. I will use the example available in SciPy for that.
import numpy as np
import matplotlib.pyplot as plt
from scipy import stats
def measure(n):
"Measurement model, return two coupled measurements."
m1 = np.random.normal(size=n)
m2 = np.random.normal(scale=0.5, size=n)
return m1+m2, m1-m2
m1, m2 = measure(2000)
xmin = m1.min()
xmax = m1.max()
ymin = m2.min()
ymax = m2.max()
X, Y = np.mgrid[xmin:xmax:100j, ymin:ymax:100j]
positions = np.vstack([X.ravel(), Y.ravel()])
values = np.vstack([m1, m2])
kernel = stats.gaussian_kde(values)
Z = np.reshape(kernel(positions).T, X.shape)
fig, ax = plt.subplots()
ax.imshow(np.rot90(Z), cmap=plt.cm.gist_earth_r,
extent=[xmin, xmax, ymin, ymax])
ax.plot(m1, m2, 'k.', markersize=2)
ax.set_xlim([xmin, xmax])
ax.set_ylim([ymin, ymax])
And the plot is:
Now, we obtain random data from the PDF obtained from the KDE, which is the variable Z.
# Generate the bins for each axis
x_bins = np.linspace(xmin, xmax, Z.shape[0]+1)
y_bins = np.linspace(ymin, ymax, Z.shape[1]+1)
# Find the middle point for each bin
x_bin_midpoints = x_bins[:-1] + np.diff(x_bins)/2
y_bin_midpoints = y_bins[:-1] + np.diff(y_bins)/2
# Calculate the Cumulative Distribution Function(CDF)from the PDF
cdf = np.cumsum(Z.ravel())
cdf = cdf / cdf[-1] # Normalização
# Create random data
values = np.random.rand(10000)
# Find the data position
value_bins = np.searchsorted(cdf, values)
x_idx, y_idx = np.unravel_index(value_bins,
(len(x_bin_midpoints),
len(y_bin_midpoints)))
# Create the new data
new_data = np.column_stack((x_bin_midpoints[x_idx],
y_bin_midpoints[y_idx]))
new_x, new_y = new_data.T
And we can calculate the KDE from this new data and the plot it.
kernel = stats.gaussian_kde(new_data.T)
new_Z = np.reshape(kernel(positions).T, X.shape)
fig, ax = plt.subplots()
ax.imshow(np.rot90(new_Z), cmap=plt.cm.gist_earth_r,
extent=[xmin, xmax, ymin, ymax])
ax.plot(new_x, new_y, 'k.', markersize=2)
ax.set_xlim([xmin, xmax])
ax.set_ylim([ymin, ymax])
Here is a solution, that returns datapoints that are uniformly distributed within each bin instead of the bin center:
def draw_from_hist(hist, bins, nsamples = 100000):
cumsum = [0] + list(I.np.cumsum(hist))
rand = I.np.random.rand(nsamples)*max(cumsum)
return [I.np.interp(x, cumsum, bins) for x in rand]
A few things do not work well for the solutions suggested by #daniel, #arco-bast, et al
Taking the last example
def draw_from_hist(hist, bins, nsamples = 100000):
cumsum = [0] + list(I.np.cumsum(hist))
rand = I.np.random.rand(nsamples)*max(cumsum)
return [I.np.interp(x, cumsum, bins) for x in rand]
This assumes that at least the first bin has zero content, which may or may not be true. Secondly, this assumes that the value of the PDF is at the upper bound of the bins, which it isn't - it's mostly in the centre of the bin.
Here's another solution done in two parts
def init_cdf(hist,bins):
"""Initialize CDF from histogram
Parameters
----------
hist : array-like, float of size N
Histogram height
bins : array-like, float of size N+1
Histogram bin boundaries
Returns:
--------
cdf : array-like, float of size N+1
"""
from numpy import concatenate, diff,cumsum
# Calculate half bin sizes
steps = diff(bins) / 2 # Half bin size
# Calculate slope between bin centres
slopes = diff(hist) / (steps[:-1]+steps[1:])
# Find height of end points by linear interpolation
# - First part is linear interpolation from second over first
# point to lowest bin edge
# - Second part is linear interpolation left neighbor to
# right neighbor up to but not including last point
# - Third part is linear interpolation from second to last point
# over last point to highest bin edge
# Can probably be done more elegant
ends = concatenate(([hist[0] - steps[0] * slopes[0]],
hist[:-1] + steps[:-1] * slopes,
[hist[-1] + steps[-1] * slopes[-1]]))
# Calculate cumulative sum
sum = cumsum(ends)
# Subtract off lower bound and scale by upper bound
sum -= sum[0]
sum /= sum[-1]
# Return the CDF
return sum
def sample_cdf(cdf,bins,size):
"""Sample a CDF defined at specific points.
Linear interpolation between defined points
Parameters
----------
cdf : array-like, float, size N
CDF evaluated at all points of bins. First and
last point of bins are assumed to define the domain
over which the CDF is normalized.
bins : array-like, float, size N
Points where the CDF is evaluated. First and last points
are assumed to define the end-points of the CDF's domain
size : integer, non-zero
Number of samples to draw
Returns
-------
sample : array-like, float, of size ``size``
Random sample
"""
from numpy import interp
from numpy.random import random
return interp(random(size), cdf, bins)
# Begin example code
import numpy as np
import matplotlib.pyplot as plt
# initial histogram, coarse binning
hist,bins = np.histogram(np.random.normal(size=1000),np.linspace(-2,2,21))
# Calculate CDF, make sample, and new histogram w/finer binning
cdf = init_cdf(hist,bins)
sample = sample_cdf(cdf,bins,1000)
hist2,bins2 = np.histogram(sample,np.linspace(-3,3,61))
# Calculate bin centres and widths
mx = (bins[1:]+bins[:-1])/2
dx = np.diff(bins)
mx2 = (bins2[1:]+bins2[:-1])/2
dx2 = np.diff(bins2)
# Plot, taking care to show uncertainties and so on
plt.errorbar(mx,hist/dx,np.sqrt(hist)/dx,dx/2,'.',label='original')
plt.errorbar(mx2,hist2/dx2,np.sqrt(hist2)/dx2,dx2/2,'.',label='new')
plt.legend()
Sorry, I don't know how to get this to show up in StackOverflow, so copy'n'paste and run to see the point.
I stumbled upon this question when I was looking for a way to generate a random array based on a distribution of another array. If this would be in numpy, I would call it random_like() function.
Then I realized, I have written a package Redistributor which might do this for me even though the package was created with a bit different motivation (Sklearn transformer capable of transforming data from an arbitrary distribution to an arbitrary known distribution for machine learning purposes). Of course I understand unnecessary dependencies are not desired, but at least knowing this package might be useful to you someday. The thing OP asked about is basically done under the hood here.
WARNING: under the hood, everything is done in 1D. The package also implements multidimensional wrapper, but I have not written this example using it as I find it to be too niche.
Installation:
pip install git+https://gitlab.com/paloha/redistributor
Implementation:
import numpy as np
import matplotlib.pyplot as plt
def random_like(source, bins=0, seed=None):
from redistributor import Redistributor
np.random.seed(seed)
noise = np.random.uniform(source.min(), source.max(), size=source.shape)
s = Redistributor(bins=bins, bbox=[source.min(), source.max()]).fit(source.ravel())
s.cdf, s.ppf = s.source_cdf, s.source_ppf
r = Redistributor(target=s, bbox=[noise.min(), noise.max()]).fit(noise.ravel())
return r.transform(noise.ravel()).reshape(noise.shape)
source = np.random.normal(loc=0, scale=1, size=(100,100))
t = random_like(source, bins=80) # More bins more precision (0 = automatic)
# Plotting
plt.figure(figsize=(12,4))
plt.subplot(121); plt.title(f'Distribution of source data, shape: {source.shape}')
plt.hist(source.ravel(), bins=100)
plt.subplot(122); plt.title(f'Distribution of generated data, shape: {t.shape}')
plt.hist(t.ravel(), bins=100); plt.show()
Explanation:
import numpy as np
import matplotlib.pyplot as plt
from redistributor import Redistributor
from sklearn.metrics import mean_squared_error
# We have some source array with "some unknown" distribution (e.g. an image)
# For the sake of example we just generate a random gaussian matrix
source = np.random.normal(loc=0, scale=1, size=(100,100))
plt.figure(figsize=(12,4))
plt.subplot(121); plt.title('Source data'); plt.imshow(source, origin='lower')
plt.subplot(122); plt.title('Source data hist'); plt.hist(source.ravel(), bins=100); plt.show()
# We want to generate a random matrix from the distribution of the source
# So we create a random uniformly distributed array called noise
noise = np.random.uniform(source.min(), source.max(), size=(100,100))
plt.figure(figsize=(12,4))
plt.subplot(121); plt.title('Uniform noise'); plt.imshow(noise, origin='lower')
plt.subplot(122); plt.title('Uniform noise hist'); plt.hist(noise.ravel(), bins=100); plt.show()
# Then we fit (approximate) the source distribution using Redistributor
# This step internally approximates the cdf and ppf functions.
s = Redistributor(bins=200, bbox=[source.min(), source.max()]).fit(source.ravel())
# A little naming workaround to make obj s work as a target distribution
s.cdf = s.source_cdf
s.ppf = s.source_ppf
# Here we create another Redistributor but now we use the fitted Redistributor s as a target
r = Redistributor(target=s, bbox=[noise.min(), noise.max()])
# Here we fit the Redistributor r to the noise array's distribution
r.fit(noise.ravel())
# And finally, we transform the noise into the source's distribution
t = r.transform(noise.ravel()).reshape(noise.shape)
plt.figure(figsize=(12,4))
plt.subplot(121); plt.title('Transformed noise'); plt.imshow(t, origin='lower')
plt.subplot(122); plt.title('Transformed noise hist'); plt.hist(t.ravel(), bins=100); plt.show()
# Computing the difference between the two arrays
print('Mean Squared Error between source and transformed: ', mean_squared_error(source, t))
Mean Squared Error between source and transformed: 2.0574123162302143