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Neaten-up and extend the examples in the random module docs.

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Raymond Hettinger 2016-12-04 11:00:34 -08:00
parent 223813111e
commit 71c62e14aa
1 changed files with 14 additions and 10 deletions
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24
Doc/library/random.rst
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@ -364,25 +364,29 @@ Basic examples::
Simulations::
# Six roulette wheel spins (weighted sampling with replacement)
>>> # Six roulette wheel spins (weighted sampling with replacement)
>>> choices(['red', 'black', 'green'], [18, 18, 2], k=6)
['red', 'green', 'black', 'black', 'red', 'black']
# Deal 20 cards without replacement from a deck of 52 playing cards
# and determine the proportion of cards with a ten-value (i.e. a ten,
# jack, queen, or king).
>>> # Deal 20 cards without replacement from a deck of 52 playing cards
>>> # and determine the proportion of cards with a ten-value
>>> # (a ten, jack, queen, or king).
>>> deck = collections.Counter(tens=16, low_cards=36)
>>> seen = sample(list(deck.elements()), k=20)
>>> print(seen.count('tens') / 20)
>>> seen.count('tens') / 20
0.15
# Estimate the probability of getting 5 or more heads from 7 spins
# of a biased coin that settles on heads 60% of the time.
>>> n = 10000
>>> cw = [0.60, 1.00]
>>> sum(choices('HT', cum_weights=cw, k=7).count('H') >= 5 for i in range(n)) / n
>>> # Estimate the probability of getting 5 or more heads from 7 spins
>>> # of a biased coin that settles on heads 60% of the time.
>>> trial = lambda: choices('HT', cum_weights=(0.60, 1.00), k=7).count('H') >= 5
>>> sum(trial() for i in range(10000)) / 10000
0.4169
>>> # Probability of the median of 5 samples being in middle two quartiles
>>> trial = lambda : 2500 <= sorted(choices(range(10000), k=5))[2] < 7500
>>> sum(trial() for i in range(10000)) / 10000
0.7958
Example of `statistical bootstrapping
<https://en.wikipedia.org/wiki/Bootstrapping_(statistics)>`_ using resampling
with replacement to estimate a confidence interval for the mean of a sample of