On 2018-12-28 17:31, Abdur-Rahmaan Janhangeer wrote:
> do you have something like
>
> choice(balls)
>
> >>> red
Don't specify the "k=" (which defaults to 1 if you omit it) and use
the first element of the results:
>>> from random import choices
>>> distribution = {"green":2, "red": 2, "blue": 1}
>>> data, weights = zip(*distribution.items())
>>> choices(data, weights)[0]
'red'
> and subsequent repetitions for long enough yield approximately 2/5
> times r 2/5 times g and 1/5 b
You can sample it yourself:
>>> from collections import defaultdict
>>> a = defaultdict(int)
>>> for i in range(10000):
... a[choices(data, weights=weights)[0]] += 1
...
>>> dict(a)
{'green': 3979, 'red': 4008, 'blue': 2013}
though if you plan to, then it might be better/faster to use
cum_weights instead, calculating it once and then reusing it rather
than having choices() re-calculate the cumulative-weights on every
call.
> like one without random choice over list/tuples
Not sure what you mean by this.
-tkc
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