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Kirill Smelkov
cpython
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7c7da38e
Commit
7c7da38e
authored
Nov 17, 2016
by
Raymond Hettinger
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Further refinements to the examples and recipes for the random module
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70be6747
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Doc/library/random.rst
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Doc/library/random.rst
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7c7da38e
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@@ -322,36 +322,49 @@ change across Python versions, but two aspects are guaranteed not to change:
Examples and Recipes
--------------------
Basic
usage
::
Basic
examples
::
>>> random
.random() # Random float x,
0.0 <= x < 1.0
>>> random
() # Random float:
0.0 <= x < 1.0
0.37444887175646646
>>>
random.uniform(1, 10) # Random float x, 1
.0 <= x < 10.0
1
.1800146073117523
>>>
uniform(2, 10) # Random float: 2
.0 <= x < 10.0
3
.1800146073117523
>>> random.randrange(10) # Integer from 0 to 9
>>> expovariate(1/5) # Interval between arrivals averaging 5 seconds
5.148957571865031
>>> randrange(10) # Integer from 0 to 9
7
>>> rand
om.randrange(0, 101, 2) # Even integer from 0 to 100
>>> rand
range(0, 101, 2) # Even integer from 0 to 100 inclusive
26
>>>
random.choice('abcdefghij') # Single random element
>>>
choice('abcdefghij') # Single random element from a sequence
'c'
>>> deck =
['jack', 'queen', 'king', 'ace']
>>> shuffle(deck)
>>> deck =
'ace two three four'.split()
>>> shuffle(deck)
# Shuffle a list
>>> deck
['king', 'queen', 'ace', 'jack']
['four', 'two', 'ace', 'three']
>>> sample([10, 20, 30, 40, 50], k=4) # Four samples without replacement
[40, 10, 50, 30]
>>> random.sample([1, 2, 3, 4, 5], k=3) # Three samples without replacement
[4, 1, 5]
Simulations::
>>> # Six weighted samples 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']
# Probability of getting 5 or more heads from 7 spins
# 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).
>>> deck = collections.Counter(tens=16, low_cards=36)
>>> seen = sample(list(deck.elements()), k=20)
>>> print(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]
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@@ -360,8 +373,8 @@ Basic usage::
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 s
mall
s
ample of s
ize five::
with replacement to estimate a confidence interval for the mean of a s
ample of
size five::
# http://statistics.about.com/od/Applications/a/Example-Of-Bootstrapping.htm
from statistics import mean
...
...
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