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我有以下列表记录随机对象的计数频率：

counter_obj= [('oranges', 66), ('apple', 13), ('banana', 13), ('pear', 12), ('strawberry', 10), ('watermelon', 10), ('avocado', 8) ... ('blueberry',1),('pineapple',1)]

我试图通过从每个等级 quartile中随机选择两个对象来选择八个元素。

我为第一个（25％）四分位数尝试了以下方法：

from collections import Counter
dct = {('oranges', 66), ('apple', 13), ('banana', 13), ('pear', 12), ('strawberry', 10), ('watermelon', 10), ('avocado', 8) ... ('blueberry',1),('pineapple',1)}
[tup for tup in Counter(dct).most_common(len(dct)//4)] # 25th percentile by frequency count

知道我在 1 处有很多值（它们只出现一次），我该如何处理剩下的 2 个四分位数 50% 和
75 %

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1 回答 1

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我会使用熊猫来解决这个问题：

import pandas as pd

dct = {('oranges', 66), ('apple', 13), ('banana', 13), ('pear', 12), ('strawberry', 10), ('watermelon', 10), ('avocado', 8) , ('blueberry',1),('pineapple',1)}

df = pd.DataFrame(dct, columns = ['Fruit','Count'])  # convert to DataFrame


select = []

for quant in [.25,.5,.75,1]:
  curr_q = df['Count'].quantile(quant)  # this calculates the quantile value
  curr_choice = df[df['Count']<=curr_q].sample(2)  # this selects all rows of your dataframe within current quantile, then samples two of these rows
  select.append(curr_choice)


select = pd.concat(select).reset_index(drop=True)  # concatenates the selected rows to get a nice dataframe, resets the indices.
于 2020-04-14T09:02:19.767 回答