如何从Pandas嵌套列表列取最小值？为何np.min()失效而np.mean()可用？

我在处理Pandas中嵌套列表类型的列时，发现np.mean()能正常计算邻居值的均值，但替换为np.min()时会抛出错误：ValueError: zero-size array to reduction operation minimum which has no identity。需要明确报错原因，以及实现取最小值的预期效果。

可正常运行的原代码（使用np.mean()） #

import pandas as pd
import numpy as np

def transformation(custom_df):
    dic = dict(zip(custom_df['customers'], custom_df['values']))
    custom_df['values'] = np.where(custom_df['values'].isna() & (custom_df['valid_neighbors'] >= 1),
                                   custom_df['neighbors'].apply(
                                       lambda row: np.mean([dic[v] for v in row if dic.get(v)])),
                                   custom_df['values'])
    return custom_df

customers = [1, 2, 3, 4, 5, 6]
values = [np.nan, np.nan, 10, np.nan, 11, 12]
neighbors = [[6], [3], [], [3, 5], [6], [5]]
vn = [1, 1, 0, 2, 1, 1]
df2 = pd.DataFrame({'customers': customers, 'values': values, 'neighbors': neighbors, 'valid_neighbors': vn})

print("原数据：")
print(df2)
df2 = transformation(df2)
print("\n处理后结果：")
print(df2)

原代码运行结果 #

原数据：
   customers  values neighbors  valid_neighbors
0          1     NaN       [6]                1
1          2     NaN       [3]                1
2          3    10.0        []                0
3          4     NaN    [3, 5]                2
4          5    11.0       [6]                1
5          6    12.0       [5]                1

处理后结果：
   customers  values neighbors  valid_neighbors
0          1    12.0       [6]                1
1          2    10.0       [3]                1
2          3    10.0        []                0
3          4    10.5    [3, 5]                2
4          5    11.0       [6]                1
5          6    12.0       [5]                1

报错原因 #

• np.mean()对空数组会返回np.nan，不会触发错误；但np.min()没有预设的空数组默认返回值，当传入空数组时，直接抛出ValueError。
• 即使代码中判断了valid_neighbors >=1，如果邻居对应的values全为NaN，[dic[v] for v in row if dic.get(v)]会生成空列表，此时调用np.min()就会报错。

解决方法及修改后代码 #

可以自定义一个处理函数，先检查列表是否为空，空则返回np.nan（或其他默认值），非空再计算最小值；也可以用pd.Series.min()，它对空序列返回np.nan，不会报错。

修改后代码（使用自定义函数） #

import pandas as pd
import numpy as np

def get_min(arr):
    if not arr:
        return np.nan
    return np.min(arr)

def transformation(custom_df):
    dic = dict(zip(custom_df['customers'], custom_df['values']))
    custom_df['values'] = np.where(custom_df['values'].isna() & (custom_df['valid_neighbors'] >= 1),
                                   custom_df['neighbors'].apply(
                                       lambda row: get_min([dic[v] for v in row if dic.get(v)])),
                                   custom_df['values'])
    return custom_df

customers = [1, 2, 3, 4, 5, 6]
values = [np.nan, np.nan, 10, np.nan, 11, 12]
neighbors = [[6], [3], [], [3, 5], [6], [5]]
vn = [1, 1, 0, 2, 1, 1]
df2 = pd.DataFrame({'customers': customers, 'values': values, 'neighbors': neighbors, 'valid_neighbors': vn})

df2 = transformation(df2)
print(df2)

修改后运行结果（符合预期） #

customers  values neighbors  valid_neighbors
0          1    12.0       [6]                1
1          2    10.0       [3]                1
2          3    10.0        []                0
3          4    10.0    [3, 5]                2
4          5    11.0       [6]                1
5          6    12.0       [5]                1

内容的提问来源于stack exchange，提问作者enriicoo