由于其值而对键进行采样(Sampling keys due to their values)

由于其值而对键进行采样(Sampling keys due to their values)

我在python中有一个字典，key-> value为str->int 。 如果我必须根据它自己的值选择一个键，那么当值变大时，键的选择可能性就会降低。

例如，如果key1=2且key2->1 ，则key1的态度应为2:1 。

我怎样才能做到这一点？

I have a dictionary in python with key->value as str->int. If I have to chose a key based on it's own value, then as the value gets larger the key has a lower possibility of being chosen.

For example, if key1=2 and key2->1, then the attitude of key1 should be 2:1.

How can I do this?

最满意答案

1.构建类似CDF的列表，如下所示：

def build_cdf(distrib): cdf = [] val = 0 for key, freq in distrib.items(): val += freq cdf.append((val, key)) return (val, cdf)

此函数返回一个元组，第一个值是概率之和，第二个值是CDF。

2.像这样构造采样器：

import random def sample_from_cdf(val_and_cdf): (val, cdf) = val_and_cdf; rand = random.uniform(0, val) # use bisect.bisect_left to reduce search time from O(n) to O(log n). return [key for index, key in cdf if index > rand][0]

用法：

x = build_cdf({"a":0.2, "b":0.3, "c":0.5}); y = [sample_from_cdf(x) for i in range(0,100000)]; print (len([t for t in y if t == "a"])) # 19864 print (len([t for t in y if t == "b"])) # 29760 print (len([t for t in y if t == "c"])) # 50376

你可能想把它变成一个类。

1. Construct a CDF-like list like this:

def build_cdf(distrib): cdf = [] val = 0 for key, freq in distrib.items(): val += freq cdf.append((val, key)) return (val, cdf)

This function returns a tuple, the 1st value is the sum of probabilities, and 2nd value is the CDF.

2. Construct the sampler like this:

import random def sample_from_cdf(val_and_cdf): (val, cdf) = val_and_cdf; rand = random.uniform(0, val) # use bisect.bisect_left to reduce search time from O(n) to O(log n). return [key for index, key in cdf if index > rand][0]

Usage:

x = build_cdf({"a":0.2, "b":0.3, "c":0.5}); y = [sample_from_cdf(x) for i in range(0,100000)]; print (len([t for t in y if t == "a"])) # 19864 print (len([t for t in y if t == "b"])) # 29760 print (len([t for t in y if t == "c"])) # 50376

You may want to make this into a class.