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Generating non-random normally distributed values between two points

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0

I've stumbled across this code in an answer to a question and I'd like to automate the process of getting the distribution to fit neatly between two bounds.

import numpy as np
from scipy import stats

bounds = [0, 100]
n = np.mean(bounds)
# your distribution:
distribution = stats.norm(loc=n, scale=20)

# percentile point, the range for the inverse cumulative distribution function:
bounds_for_range = distribution.cdf(bounds)

# Linspace for the inverse cdf:
pp = np.linspace(*bounds_for_range, num=1000)

x = distribution.ppf(pp)

# And just to check that it makes sense you can try:
from matplotlib import pyplot as plt
plt.hist(x)
plt.show()

Let's say I have the values [720, 965], or any other bounds, that I would like to fit my distribution across. Is there a way to soft-code the adjustment of scale in stats.norm to fit this distribution across my bounds without any unreasonable gaps? Or are there any functions that have this type of functionality?

A scale of ~20 works well for the example code, but I have to adjust it to ~50 for the example of [720, 965]

python numpy scipy

share|improve this question

asked Mar 24 at 20:42

EstifEstif

245

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Is scale=(bounds[1] - bounds[0]) * 0.2 good enough?
  
  – Elias Strehle
  Mar 24 at 21:25
  
  

  
  
  
  

add a comment | 

0

I've stumbled across this code in an answer to a question and I'd like to automate the process of getting the distribution to fit neatly between two bounds.

import numpy as np
from scipy import stats

bounds = [0, 100]
n = np.mean(bounds)
# your distribution:
distribution = stats.norm(loc=n, scale=20)

# percentile point, the range for the inverse cumulative distribution function:
bounds_for_range = distribution.cdf(bounds)

# Linspace for the inverse cdf:
pp = np.linspace(*bounds_for_range, num=1000)

x = distribution.ppf(pp)

# And just to check that it makes sense you can try:
from matplotlib import pyplot as plt
plt.hist(x)
plt.show()

Let's say I have the values [720, 965], or any other bounds, that I would like to fit my distribution across. Is there a way to soft-code the adjustment of scale in stats.norm to fit this distribution across my bounds without any unreasonable gaps? Or are there any functions that have this type of functionality?

A scale of ~20 works well for the example code, but I have to adjust it to ~50 for the example of [720, 965]

python numpy scipy

share|improve this question

asked Mar 24 at 20:42

EstifEstif

245

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Is scale=(bounds[1] - bounds[0]) * 0.2 good enough?
  
  – Elias Strehle
  Mar 24 at 21:25
  
  

  
  
  
  

add a comment | 

I've stumbled across this code in an answer to a question and I'd like to automate the process of getting the distribution to fit neatly between two bounds.

import numpy as np
from scipy import stats

bounds = [0, 100]
n = np.mean(bounds)
# your distribution:
distribution = stats.norm(loc=n, scale=20)

# percentile point, the range for the inverse cumulative distribution function:
bounds_for_range = distribution.cdf(bounds)

# Linspace for the inverse cdf:
pp = np.linspace(*bounds_for_range, num=1000)

x = distribution.ppf(pp)

# And just to check that it makes sense you can try:
from matplotlib import pyplot as plt
plt.hist(x)
plt.show()

Let's say I have the values [720, 965], or any other bounds, that I would like to fit my distribution across. Is there a way to soft-code the adjustment of scale in stats.norm to fit this distribution across my bounds without any unreasonable gaps? Or are there any functions that have this type of functionality?

A scale of ~20 works well for the example code, but I have to adjust it to ~50 for the example of [720, 965]

python numpy scipy

share|improve this question

asked Mar 24 at 20:42

EstifEstif

245

import numpy as np
from scipy import stats

bounds = [0, 100]
n = np.mean(bounds)
# your distribution:
distribution = stats.norm(loc=n, scale=20)

# percentile point, the range for the inverse cumulative distribution function:
bounds_for_range = distribution.cdf(bounds)

# Linspace for the inverse cdf:
pp = np.linspace(*bounds_for_range, num=1000)

x = distribution.ppf(pp)

# And just to check that it makes sense you can try:
from matplotlib import pyplot as plt
plt.hist(x)
plt.show()

share|improve this question

asked Mar 24 at 20:42

EstifEstif

245

share|improve this question

asked Mar 24 at 20:42

EstifEstif

245

asked Mar 24 at 20:42

EstifEstif

245

asked Mar 24 at 20:42

EstifEstif

245

1 Answer
1

active

oldest

votes

0

I am not sure, but truncated normal distribution should be exactly what you are looking for.

from scipy.stats import truncnorm
distr_ab = truncnorm(a, b) # truncated normal distribution in the interval [a, b]
distr_ab.rvs(size=100) # get 100 samples from the distribution
# distr_ab.cdf, distr_ab.ppf etc... all accessible

share|improve this answer

answered Mar 25 at 1:48

bubblebubble

1,180713

add a comment | 

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0

I am not sure, but truncated normal distribution should be exactly what you are looking for.

from scipy.stats import truncnorm
distr_ab = truncnorm(a, b) # truncated normal distribution in the interval [a, b]
distr_ab.rvs(size=100) # get 100 samples from the distribution
# distr_ab.cdf, distr_ab.ppf etc... all accessible

share|improve this answer

answered Mar 25 at 1:48

bubblebubble

1,180713

add a comment | 

0

I am not sure, but truncated normal distribution should be exactly what you are looking for.

from scipy.stats import truncnorm
distr_ab = truncnorm(a, b) # truncated normal distribution in the interval [a, b]
distr_ab.rvs(size=100) # get 100 samples from the distribution
# distr_ab.cdf, distr_ab.ppf etc... all accessible

share|improve this answer

answered Mar 25 at 1:48

bubblebubble

1,180713

add a comment | 

I am not sure, but truncated normal distribution should be exactly what you are looking for.

from scipy.stats import truncnorm
distr_ab = truncnorm(a, b) # truncated normal distribution in the interval [a, b]
distr_ab.rvs(size=100) # get 100 samples from the distribution
# distr_ab.cdf, distr_ab.ppf etc... all accessible

share|improve this answer

answered Mar 25 at 1:48

bubblebubble

1,180713

from scipy.stats import truncnorm
distr_ab = truncnorm(a, b) # truncated normal distribution in the interval [a, b]
distr_ab.rvs(size=100) # get 100 samples from the distribution
# distr_ab.cdf, distr_ab.ppf etc... all accessible

share|improve this answer

answered Mar 25 at 1:48

bubblebubble

1,180713

answered Mar 25 at 1:48

bubblebubble

1,180713

answered Mar 25 at 1:48

bubblebubble

1,180713