How to generate random sequence of 1, 2, 3 provided that 80 % of numbers will 1, 15 % will 2 and 5 % will 3?
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Possibly related to stackoverflow.com/questions/4276787/…, stackoverflow.com/questions/2073235/random-weighted-choice, stackoverflow.com/questions/4265988/…Jérôme– Jérôme2016-05-13 14:38:13 +00:00Commented May 13, 2016 at 14:38
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2 Answers
Use random to get a random number in [0,1) and map your outputs to that interval.
from random import random
result = []
# Return 100 results (for instance)
for i in range(100):
res = random()
if res < 0.8:
result.append(1)
elif res < 0.95:
result.append(2)
else:
result.append(3)
return result
This is a trivial solution. You may want to write a more elegant one that allows you to specify the probabilities for each number in a dedicated structure (list, dict,...) rather than in a if/else statement.
But then you might be better off using a dedicated library, as suggested in this answer. Here's an example with scipy's stats.rv_discrete
from scipy import stats
xk = np.arange(1, 4)
pk = (0.8, 0.15, 0.05)
custm = stats.rv_discrete(name='custm', values=(xk, pk))
# Return 100 results (for instance)
return custm.rvs(size=100)
4 Comments
kanayamalakar
I was going to post this answer only
Rory Daulton
Shouldn't there be a colon at the end of the
elif line?
Kenenbek Arzymatov
pk = (0.8, 0.15, 0.05) or your variant?
Jérôme
You're right. Fixed. The doc says "(xk, pk) where xk are integers with non-zero probabilities pk with sum(pk) = 1."
This should do the trick
import random
sequence = []
desired_length = n # how many numbers you want
for i in range(desired_length):
random_value = random.random()
if random_value < .8:
sequence.append(1)
elif random_value < .95:
sequence.append(2)
else:
sequence.append(3)
print sequence
