CODEWITHSUNDEEP
pythonnumpymath
Rob
Rob

Reputation: 3459

Random Sample From Probability Distribution a certain number of times

Let's say I have a normal distribution, and I want to sample from it 1000 times and if the value is in the top 20th percentile, I want to add 1 to a counter. What's the optimal solution for this?

I'm trying to solve this using numpy and can't figure out the math behind it. Right now I have this but I feel like it could be done mathematically:

import numpy as np
s = np.random.normal(0, 1, 1000)
sum([val for val in s if val > np.percentile(s, 80)])

Upvotes: 1

Views: 97

Answers (2)

user2285236
user2285236

Reputation:

This is the generated array (you can play with mean and standard deviation, they are the default ones):

mu = 0
std = 1
arr = np.random.normal(mu, std, 1000)

This gives you the number of items in the top 20 percentile:

arr[arr > np.percentile(arr, 80)].size
Out[30]: 200

Edit: Your code is also nice. But you don't want to sum the values, you want to count them. So whenever val > np.percentile(s, 80) you want to sum 1's:

sum([1 for val in s if val > np.percentile(s, 80)])
Out[35]: 200

This will be slower than numpy's methods though.

Upvotes: 1

Brian
Brian

Reputation: 1667

If you have a normal distribution, I'll assume you know mu and sigma, otherwise this question requires some extra post-processing. You'll have a certain Z=(X-mu)/sigma for each element you take out. Your 20th percentile would be any X that makes Z>0.842.

You can do something like:

if Z_val(x, mu, sigma) > 0.842: counter+=1

Upvotes: 1

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