CODEWITHSUNDEEP
pythonlistnumpyrandomscipy
Subhankar Ghosh
Subhankar Ghosh

Reputation: 469

Weighted random numbers in Python from a list of values

I am trying to create a list of 10,000 random numbers between 1 and 1000. But I want 80-85% of the numbers to be the same category( I mean some 100 numbers out of these should appear 80% of the times in the list of random numbers) and the rest appear around 15-20% of the times. Any idea if this can be done in Python/NumPy/SciPy. Thanks.

Upvotes: 1

Views: 1311

Answers (2)

Divakar
Divakar

Reputation: 221574

Here's an approach -

a = np.arange(1,1001) # Input array to extract numbers from

# Select 100 random unique numbers from input array and get also store leftovers
p1 = np.random.choice(a,size=100,replace=0)
p2 = np.setdiff1d(a,p1)

# Get random indices for indexing into p1 and p2
p1_idx = np.random.randint(0,p1.size,(8000))
p2_idx = np.random.randint(0,p2.size,(2000))

# Index and concatenate and randomize their positions
out = np.random.permutation(np.hstack((p1[p1_idx], p2[p2_idx])))

Let's verify after run -

In [78]: np.in1d(out, p1).sum()
Out[78]: 8000

In [79]: np.in1d(out, p2).sum()
Out[79]: 2000

Upvotes: 1

martianwars
martianwars

Reputation: 6500

This can be easily done using 1 call to random.randint() to select a list and another call to random.choice() on the correct list. I'll assume list frequent contain 100 elements which are to be chose 80 percent times and rare contains 900 elements to be chose 20 percent times.

import random
a = random.randint(1,5)
if a == 1:
    # Case for rare numbers
    choice = random.choice(rare)
else:
    # case for frequent numbers
    choice = random.choice(frequent)

Upvotes: 2

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