如何根据平均值、中位数、第 1 和第 9 十分位数值生成数据集？

我有以下描述数据集的值：

Number of Samples: 5388
Mean: 4173
Median: 4072
1st Decile: 2720
9th Decile: 5676

我需要生成适合这些值的任何数据集。 我发现的所有例子都要求你有标准差，但我没有。 如何做到这一点？ 谢谢！

有趣的问题！ 根据斯科特的建议，我快速尝试了一下。

Inputs:

import random
import pandas as pd
import numpy as np

# fixing the random seed
random.seed(a=1, version=2)
# formating floats
pd.options.display.float_format = '{:.1f}'.format

# given inputs
count = 5388
mean = 4173
median = 4072

lower_percentile = 10
lower_percentile_value = 2720

upper_percentile = 90
upper_percentile_value = 5676

max_value = 6325
min_value = 2101

功能：

def generate_dataset(count, mean, median, lower_percentile, upper_percentile,
    lower_percentile_value, upper_percentile_value,
    min_value, max_value
    ):
        
    # Calculate the number of values that fall within each percentile
    p_1_size = int(float(lower_percentile) * float(count) / 100)
    p_4_size = int(count - (float(upper_percentile) * float(count) / 100))
    p_2_size = int((count / 2) - p_1_size)
    p_3_size = int((count / 2) - p_4_size)
    
    # can be used to adjust the mean
    mean_adjuster = 5790

    # randomly pick values of right size from a range 
    p_1 = random.choices(range(min_value, lower_percentile_value), k=p_1_size)
    p_2 = random.choices(range(lower_percentile_value, median), k=p_2_size)
    p_3 = random.choices(range(median, mean_adjuster), k=p_3_size)
    p_4 = random.choices(range(upper_percentile_value, max_value), k=p_4_size)
    
    return p_1 + p_2 + p_3 + p_4
    
dataset = generate_dataset(
    count, mean, median, lower_percentile, upper_percentile,
    lower_percentile_value, upper_percentile_value, min_value, max_value
    )

比较：

# converting into DataFrame
df = pd.DataFrame({"x": dataset})

new_count = len(df)
new_mean = np.mean(df.x)
new_median = np.quantile(df.x, 0.5)
new_lower_percentile = np.quantile(df.x, lower_percentile/100)
new_upper_percentile = np.quantile(df.x, upper_percentile/100)

compare = pd.DataFrame(
    {
        "value": ["count", "mean", "median", "low_p", "high_p"],
        "original": [count, mean, median, lower_percentile_value, upper_percentile_value],
        "new":[new_count, new_mean, new_median, new_lower_percentile, new_upper_percentile]
    }
)

print(compare)

Output:

   value  original    new
0   count      5388 5388.0
1    mean      4173 4173.4
2  median      4072 4072.5
3   low_p      2720 2720.4
4  high_p      5676 5743.0

当所有值都是整数而不是浮点数时，使值完全相等有点棘手。

您可以添加另一个变量来控制两个数字的平均值，或者更改随机种子，看看是否可以获得更接近的值。或者，您可以编写一个函数来更改种子，直到值相等。 （可能需要几分钟或几个世纪：）

Cheers!