weighted-random | Select randomly from a list of weighted values
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kandi X-RAY | weighted-random Summary
Select randomly from a list of weighted values
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QUESTION
I want a circle with more circles inside it (doesn't strictly need to be inside it). The position of inner circles are determined randomly in a way that there should be most circles at the centre and less and less and it goes out towards the edge of the circle.
From this question, I gathered that numbers can be biased using f(x) instead of just x, x being the random number, of course. Here is the code:
...ANSWER
Answered 2021-Nov-28 at 10:27Actually you are generating random numbers in the range [0.0, 1.0] and assigning them to the range [_min
, _max
]. Therfore 0 is mapped to _min
. As a result, there are more points near _min
.
You have to generate random numbers in range [-1.0, 1.0] and map them to the range [_min
, _max
]. So 0 is in the middle of the range and most of the points are near (_min + _max) / 2
:
QUESTION
Let's say I have a list of objects (in Python) that looks something like this (contains an identifier and a ranking/weighting):
...ANSWER
Answered 2021-Oct-19 at 16:57If I'm not mistaken one approach could be to weighted sample without replacement:
QUESTION
I have a bunch of names from the web (first name, last name, of people in different countries). Some of the countries have statistics on how many people have each last name, as shown in some places like here.
Well, that Japanese surname list only lists the top 100. I have other lists like for Vietnamese listing the top 20, and other lists the top 50 or 1000 even in some places. But I have real name lists that are up to the 1000+ count. So I might have 2000 Japanese surnames, with only 100 that have listed the actual count of people with that surname.
What I would like to do is built a "faker" sort of library, that generates realistic names based on these statistics. I know how to pick a random element from a weighted array in JavaScript, so once the "weights" (number of people with that name) are included for each name, it is just a matter of plugging it into that algorithm.
My question is, how can I "complete the curve" on the names that don't have a weight on them? That is, say we have an exponential-like curve sort of, from the 20 or 100 names that have weights on them. I would then like to randomly pick names from the remaining unweighted list, and give them a value that places them somewhat realistically in the remaining tail of the curve. How can that be done?
For example, here is a list of Vietnamese names with weights:
...ANSWER
Answered 2021-Aug-04 at 09:34I'm no mathematician, so I've simply fitted the data to a y=A*x^B
equation using these equations, although Wolfram has some others that might fit your data better. Perhaps some papers around the distribution of (sur)names might hint at a better equation.
Nonetheless, the current prediction doesn't seem too bad:
QUESTION
Using randint()
how do I give lower values a higher weight (higher chance to be picked)?
I have the following code:
...ANSWER
Answered 2021-Apr-23 at 14:29The following method satisfies your requirements. It uses the rejection sampling approach: Generate an integer uniformly at random, and accept it with probability proportional to its weight. If the number isn't accepted, we reject it and try again (see also this answer of mine).
QUESTION
Is there a way to do a general, performant groupby-operation that does not rely on pd.groupby?
Input ...ANSWER
Answered 2020-Aug-07 at 18:46Before ditching groupby
I'd suggest first evaluating whether you are truly taking advantage of what groupby
has to offer.
lambda
in favor of built-in pd.DataFrameGroupBy
methods.
Many of the Series
and DataFrame
methods are implemented as pd.DataFrameGroupBy
methods. You should use those directly as opposed to calling them with a groupby
+ apply(lambda x: ...)
Further, for many calculations you can re-frame the problem as some vectorized operation on an entire DataFrame that then uses a groupby method implemented in cython. This will be fast.
A common example of this would be finding the proportion of 'Y'
answers within a group. A straight-forward approach would be to check the condition within each group then get the proportion:
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