您要做的第一件事是将字符串转换为列表，然后将其与groupby一起用作agg：

df['3LETTER_SUBSTRINGS'] = df['3LETTER_SUBSTRINGS'].str.split(',')

df.groupby('ETHNICITY_RECAT').agg({'TOTAL_LENGTH':['min','max','mean'],'3LETTER_SUBSTRINGS':'sum'})

输出：

                TOTAL_LENGTH                             3LETTER_SUBSTRINGS
                         min max  mean                                  sum
ETHNICITY_RECAT                                                            
en                        16  18  17.0                           [ann,tom]
fr                        14  16  15.0  [jos,ose,sep,eph,tom,omm,mmy]
ir                        14  19  16.5                      [ann,nne,ann]

我认为您的大多数代码都还不错，您只是误解了该错误：它与字符串转换无关。在3LETTER_SUBSTRING列的每个单元格中都有列表/元组。使用lambda x:list(x)函数时，将创建一个元组列表。因此，没有像split(",")那样要做的事情，也不会强制转换为字符串并返回表...

相反，在创建新列表时，您只需要取消嵌套表格即可。因此，这是一个可重复的小代码：（请注意，我专注于您的元组/聚合问题，因为我相信您会很快找到其余的代码）

import pandas as pd
# Create some data
names = [("joseph","fr"),("ann","en"),("anne","ir"),("tom",("tommy","fr")]
df = pd.DataFrame(names,columns=["NAMES","ethnicity"])
df["3LETTER_SUBSTRING"] = df["NAMES"].apply(lambda name: [name[i:i+3] for i in range(len(name) - 2)])
print(df)
# Aggregate the 3LETTER per ethnicity,and unnest the result in a new table for each ethnicity:
df.groupby('ethnicity').agg({
    "3LETTER_SUBSTRING": lambda x:[z for y in x for z in y]
})

使用您指定的计数器，我得到了

dfg = df.groupby('ethnicity',as_index=False).agg({
    "3LETTER_SUBSTRING": lambda x:[z for y in x for z in y]
})
from collections import Counter
print(Counter(dfg[dfg["ethnicity"] == "en"]["3LETTER_SUBSTRING"][0]))
# Counter({'ann': 1,'tom': 1})

要以元组列表的形式获取它，只需使用内置的字典功能，例如dict.items()。

更新：如问题中那样使用预格式化的字符串列表：

import pandas as pd
# Create some data
names = [("joseph","fr","jos,eph"),"en","ann"),"ir","ann,nne"),"tom"),"tom,mmy"),"ann")]
df = pd.DataFrame(names,"ethnicity","3LETTER_SUBSTRING"])
def transform_3_letter_to_table(x):
    """
    Update this function with regard to your data format
    """
    return x.split(",")
df["3LETTER_SUBSTRING"] = df["3LETTER_SUBSTRING"].apply(transform_3_letter_to_table)
print(df)
# Applying aggregation
dfg = df.groupby('ethnicity',as_index=False).agg({
    "3LETTER_SUBSTRING": lambda x:[z for y in x for z in y]
})
print(dfg)
# test on some data
from collections import Counter
c = Counter(dfg[dfg["ethnicity"] == "en"]["3LETTER_SUBSTRING"][0])
print(c)
print(list(c.items()))