I have some problems about this question. Generate 1000 independent random variables with normal distribution N(9,10) and plot the result in a histogram.
For the histogram use the intervals with length 1,namely[-1,0),[0,1),[1,2)...etc
import numpy as np
import matplotlib.pyplot as plt
def demo1():
mu ,sigma = 9, 10
sampleNo = 1000
s = np.random.normal(mu, sigma, sampleNo)
plt.hist(s, bins=100, density=True)
plt.show()
demo1()
I wonder how to choose bins number and I don't know how to adjust intervals with length 1.
You can feed a sequence to the bins argument. For example, if you give bins = np.arange(100) the histogram will have 100 bins of length 1 going as [0,1),[1,2)...
Another example, you can write plt.hist(s, bins=np.linspace(-50,50,100), density=True) which would plot the histogram of range (-50,50) with bin length 1
Related
I have a matplotlib histogram that works fine.
hist_bin_width = 4
on_hist = plt.hist(my_data,bins=range(-100, 200,hist_bin_width),alpha=.3,color='#6e9bd1',label='on')
All I want to do is to rescale by a factor of, say, 2. I don't want to change the bin width, or to change the y axis labels. I want to take the counts in all the bins (e.g. bin 1 has 17 counts) and multiply by 2 so that bin 1 now has 34 counts in it.
Is this possible?
Thank you.
As it's just a simple rescaling of the y-axis, this must be possible. The complication arises because Matplotlib's hist computes and draws the histogram, making it difficult to intervene. However, as the documentation also notes, you can use the weights parameter to "draw a histogram of data that has already been binned". You can bin the data in a first step with Numpy's histogram function. Applying the scaling factor is then straightforward:
from matplotlib import pyplot
import numpy
numpy.random.seed(0)
data = numpy.random.normal(50, 20, 10000)
(counts, bins) = numpy.histogram(data, bins=range(101))
factor = 2
pyplot.hist(bins[:-1], bins, weights=factor*counts)
pyplot.show()
pyplot.hist's weights argument can be used to weight each data point with a factor like
import matplotlib.pyplot as plt
import numpy as np; np.random.seed(42)
data = np.random.normal(50, 20, 10000)
factor = 2
hist_bin_width = 40
plt.hist(data, bins=range(-100, 200, hist_bin_width),
weights=factor*np.ones_like(data))
plt.show()
I have two arrays of corresponding data (x and y) that I plot as above on a log-log plot. The data is currently too granular and I would like to bin them to get a smoother relationship. Could I get some guidance on how I can bin along the x-axis, in exponential bin sizes, so that it appears linear on the log-log scale?
For example, if the first bin is of range x = 10^0 to 10^1, I want to collect all y-values with corresponding x in that range and average them into one value for that bin. I don't think np.hist or plt.hist quite does the trick, since they do binning by counting occurrences.
Edit: For context, if it helps, the above plot is an assortativity plot that plots the in vs out degree of a certain network.
You may use scipy.stats.binned_statistic to get the mean of the data in each bin. The bins would best be created via numpy.logspace. You may then plot those means e.g. as horiziontal lines spanning the bin width or as scatter at the mean position.
import numpy as np; np.random.seed(42)
from scipy.stats import binned_statistic
import matplotlib.pyplot as plt
x = np.logspace(0,5,300)
y = np.logspace(0,5,300)+np.random.rand(300)*1.e3
fig, ax = plt.subplots()
ax.scatter(x,y, s=9)
s, edges, _ = binned_statistic(x,y, statistic='mean', bins=np.logspace(0,5,6))
ys = np.repeat(s,2)
xs = np.repeat(edges,2)[1:-1]
ax.hlines(s,edges[:-1],edges[1:], color="crimson", )
for e in edges:
ax.axvline(e, color="grey", linestyle="--")
ax.scatter(edges[:-1]+np.diff(edges)/2, s, c="limegreen", zorder=3)
ax.set_xscale("log")
ax.set_yscale("log")
plt.show()
You can achieve this with pandas. The idea is to assign each X value to an interval using np.digitize. Since you are using a log scale, it makes sense to use np.logspace to choose intervals of exponentially changing lengths. Finally, you can group X values in each interval and compute mean Y values.
import pandas as pd
import numpy as np
x_max = 10
xs = np.exp(x_max * np.random.rand(1000))
ys = np.exp(np.random.rand(1000))
df = pd.DataFrame({
'X': xs,
'Y': ys,
})
df['Xbins'] = np.digitize(df.X, np.logspace(0, x_max, 30, base=np.exp(1)))
df['Ymean'] = df.groupby('Xbins').Y.transform('mean')
df.plot(kind='scatter', x='X', y='Ymean')
I'm trying to generate random samples from a lognormal distribution in Python, the application is for simulating network traffic. I'd like to generate samples such that:
The modal sample result is 320 (~10^2.5)
80% of the samples lie within the range 100 to 1000 (10^2 to 10^3)
My strategy is to use the inverse CDF (or Smirnov transform I believe):
Use the PDF for a normal distribution centred around 2.5 to calculate the PDF for 10^x where x ~ N(2.5,sigma).
Calculate the CDF for the above distribution.
Generate random uniform data along the interval 0 to 1.
Use the inverse CDF to transform the random uniform data into the required range.
The problem is, when I calculate the 10 and 90th percentile at the end, I have completely the wrong numbers.
Here is my code:
%matplotlib inline
import matplotlib
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import scipy.stats
from scipy.stats import norm
# find value of mu and sigma so that 80% of data lies within range 2 to 3
mu=2.505
sigma = 1/2.505
norm.ppf(0.1, loc=mu,scale=sigma),norm.ppf(0.9, loc=mu,scale=sigma)
# output: (1.9934025, 3.01659743)
# Generate normal distribution PDF
x = np.arange(16,128000, 16) # linearly spaced here, with extra range so that CDF is correctly scaled
x_log = np.log10(x)
mu=2.505
sigma = 1/2.505
y = norm.pdf(x_log,loc=mu,scale=sigma)
fig, ax = plt.subplots()
ax.plot(x_log, y, 'r-', lw=5, alpha=0.6, label='norm pdf')
x2 = (10**x_log) # x2 should be linearly spaced, so that cumsum works (later)
fig, ax = plt.subplots()
ax.plot(x2, y, 'r-', lw=5, alpha=0.6, label='norm pdf')
ax.set_xlim(0,2000)
# Calculate CDF
y_CDF = np.cumsum(y) / np.cumsum(y).max()
fig, ax = plt.subplots()
ax.plot(x2, y_CDF, 'r-', lw=2, alpha=0.6, label='norm pdf')
ax.set_xlim(0,8000)
# Generate random uniform data
input = np.random.uniform(size=10000)
# Use CDF as lookup table
traffic = x2[np.abs(np.subtract.outer(y_CDF, input)).argmin(0)]
# Discard highs and lows
traffic = traffic[(traffic >= 32) & (traffic <= 8000)]
# Check percentiles
np.percentile(traffic,10),np.percentile(traffic,90)
Which produces the output:
(223.99999999999997, 2480.0000000000009)
... and not the (100, 1000) that I would like to see. Any advice appreciated!
First, I'm not sure about Use the PDF for a normal distribution centred around 2.5. After all, log-normal is about base e logarithm (aka natural log), which means 320 = 102.5 = e5.77.
Second, I would approach problem in a different way. You need m and s to sample from Log-Normal.
If you look at wiki article above, you could see that it is two-parametric distribution. And you have exactly two conditions:
Mode = exp(m - s*s) = 320
80% samples in [100,1000] => CDF(1000,m,s) - CDF(100,m,s) = 0.8
where CDF is expressed via error function (which is pretty much common function found in any library)
So two non-linear equations for two parameters. Solve them, find m and s and put it into any standard log-normal sampling
Severin's approach is much leaner than my original attempt using the Smirnov transform. This is the code that worked for me (using fsolve to find s, although its quite trivial to do it manually):
# Find lognormal distribution, with mode at 320 and 80% of probability mass between 100 and 1000
# Use fsolve to find the roots of the non-linear equation
%matplotlib inline
import matplotlib
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import fsolve
from scipy.stats import lognorm
import math
target_modal_value = 320
# Define function to find roots of
def equation(s):
# From Wikipedia: Mode = exp(m - s*s) = 320
m = math.log(target_modal_value) + s**2
# Get probability mass from CDF at 100 and 1000, should equal to 0.8.
# Rearange equation so that =0, to find root (value of s)
return (lognorm.cdf(1000,s=s, scale=math.exp(m)) - lognorm.cdf(100,s=s, scale=math.exp(m)) -0.8)
# Solve non-linear equation to find s
s_initial_guess = 1
s = fsolve(equation, s_initial_guess)
# From s, find m
m = math.log(target_modal_value) + s**2
print('m='+str(m)+', s='+str(s)) #(m,s))
# Plot
x = np.arange(0,2000,1)
y = lognorm.pdf(x,s=s, scale=math.exp(m))
fig, ax = plt.subplots()
ax.plot(x, y, 'r-', lw=5, alpha=0.6, label='norm pdf')
plt.plot((100,100), (0,1), 'k--')
plt.plot((320,320), (0,1), 'k-.')
plt.plot((1000,1000), (0,1), 'k--')
plt.ylim(0,0.0014)
plt.savefig('lognormal_100_320_1000.png')
Here is the histogram
To generate this plot, I did:
bins = np.array([0.03, 0.3, 2, 100])
plt.hist(m, bins = bins, weights=np.zeros_like(m) + 1. / m.size)
However, as you noticed, I want to plot the histogram of the relative frequency of each data point with only 3 bins that have different sizes:
bin1 = 0.03 -> 0.3
bin2 = 0.3 -> 2
bin3 = 2 -> 100
The histogram looks ugly since the size of the last bin is extremely large relative to the other bins. How can I fix the histogram? I want to change the width of the bins but I do not want to change the range of each bin.
As #cel pointed out, this is no longer a histogram, but you can do what you are asking using plt.bar and np.histogram. You then just need to set the xticklabels to a string describing the bin edges. For example:
import numpy as np
import matplotlib.pyplot as plt
bins = [0.03,0.3,2,100] # your bins
data = [0.04,0.07,0.1,0.2,0.2,0.8,1,1.5,4,5,7,8,43,45,54,56,99] # random data
hist, bin_edges = np.histogram(data,bins) # make the histogram
fig,ax = plt.subplots()
# Plot the histogram heights against integers on the x axis
ax.bar(range(len(hist)),hist,width=1)
# Set the ticks to the middle of the bars
ax.set_xticks([0.5+i for i,j in enumerate(hist)])
# Set the xticklabels to a string that tells us what the bin edges were
ax.set_xticklabels(['{} - {}'.format(bins[i],bins[i+1]) for i,j in enumerate(hist)])
plt.show()
EDIT
If you update to matplotlib v1.5.0, you will find that bar now takes a kwarg tick_label, which can make this plotting even easier (see here):
hist, bin_edges = np.histogram(data,bins)
ax.bar(range(len(hist)),hist,width=1,align='center',tick_label=
['{} - {}'.format(bins[i],bins[i+1]) for i,j in enumerate(hist)])
If your actual values of the bins are not important but you want to have a histogram of values of completely different orders of magnitude, you can use a logarithmic scaling along the x axis. This here gives you bars with equal widths
import numpy as np
import matplotlib.pyplot as plt
data = [0.04,0.07,0.1,0.2,0.2,0.8,1,1.5,4,5,7,8,43,45,54,56,99]
plt.hist(data,bins=10**np.linspace(-2,2,5))
plt.xscale('log')
plt.show()
When you have to use your bin values you can do
import numpy as np
import matplotlib.pyplot as plt
data = [0.04,0.07,0.1,0.2,0.2,0.8,1,1.5,4,5,7,8,43,45,54,56,99]
bins = [0.03,0.3,2,100]
plt.hist(data,bins=bins)
plt.xscale('log')
plt.show()
However, in this case the widths are not perfectly equal but still readable. If the widths must be equal and you have to use your bins I recommend #tom's solution.
I have plotted a histogram in python, using matplotlib and I need the y-axis to be the probability, I cannot find how to do this. For example i want it to look similar to this http://www.mathamazement.com/images/Pre-Calculus/10_Sequences-Series-and-Summation-Notation/10_07_Probability/10-coin-toss-histogram.JPG
Here is my code, I will attached my plot aswell if needed
plt.figure(figsize=(10,10))
mu = np.mean(a) #mean of distribution
sigma = np.std(a) # standard deviation of distribution
n, bins,patches=plt.hist(a,bin, normed=True, facecolor='white')
y = mlab.normpdf(bins, mu, sigma)
plt.plot(bins,y,'r--')
print np.sum(n*np.diff(bins))# proved the intergal over bars is unity
plt.show()
Just divide all your sample counts by the total number of samples. This gives the probability rather than the count.
As #SteveBarnes points out, divide the sample counts by the total number of samples to get the probabilities for each bin. To get a plot like the one you linked to, your "bins" should just be the integers from 0 to 10. A simple way to compute the histogram for a sample from a discrete distribution is np.bincount.
Here's a snippet that creates a plot like the one you linked to:
import numpy as np
import matplotlib.pyplot as plt
n = 10
num_samples = 10000
# Generate a random sample.
a = np.random.binomial(n, 0.5, size=num_samples)
# Count the occurrences in the sample.
b = np.bincount(a, minlength=n+1)
# p is the array of probabilities.
p = b / float(b.sum())
plt.bar(np.arange(len(b)) - 0.5, p, width=1, facecolor='white')
plt.xlim(-0.5, n + 0.5)
plt.xlabel("Number of heads (k)")
plt.ylabel("P(k)")
plt.show()