Python 在熊猫身上没有得到确切的结果

运行此代码后，我得到的结果如下

df=pandas.DataFrame(processed_data_format, columns=["file_name", "innings", "over","ball", "individual ball", "runs","batsman", "wicket_status","bowler_name","fielder_name"])      
a = {'runs':['sum'],'ball':['sum'],'file_name':['unique']}
t = df.groupby('batsman').agg(a)

['Younis Khan', array(['225245', '225247'], dtype=object), 113, 121]

但是我想得到这样的结果

df=pandas.DataFrame(processed_data_format, columns=["file_name", "innings", "over","ball", "individual ball", "runs","batsman", "wicket_status","bowler_name","fielder_name"])      
a = {'runs':['sum'],'ball':['sum'],'file_name':['unique']}
t = df.groupby('batsman').agg(a)

['Younis Khan', array(['225245', '225247'], dtype=object), 113, 121]

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我不需要数组（['225245'，'225247'，dtype=object），我需要得到文件名的计数

a={'runs'：'sum'，'ball'：'sum'，'file_name'：'unique'}

有效吗？@EdChum不，它会给出相同的结果answer@EdChumnunique将发挥作用，而不是独一无二