我有一个函数,它接受一个数据框,进行一些转换,然后将数字和分类列名作为列表返回。cat_cols, num_cols = Data_Type_And_Transformation(df_data_sample, 'MEAN')cat_cols = ['var1_m2_Transform', 'var2_m2_Transform', 'var2_m3_Transform', 'var3_m3_Transform', 'var5_m3_Transform', 'var8_m3_Transform', 'var9_m3_Transform']num_cols = ['ttl_change_3m', 'ttl_change_6m', 'base_rev_3m', 'csc_ttl_6m']然后,我试图创建一个字典,其键将是列名,值将是数据类型-NUM或CAT,如下所示:attribute_df_benford_cat = pd.DataFrame()attribute_df_benford_num = pd.DataFrame()attribute_df_cat['Attribute'] = cat_colsattribute_df_cat['Type'] = 'CAT'attribute_df_num['Attribute'] = num_colsattribute_df_num['Type'] = 'NUM'attribute_df = attribute_df_cat.append(attribute_df_num)attribute_df.set_index('Attribute',inplace = True)attribute_dict = OrderedDict(attribute_df.to_dict('index'))但这给了我以下形式的格言:Key Type Size Valuettl_change_3m dict 1 {'Type': 'NUM'}ttl_change_6m dict 1 {'Type': 'NUM'}base_rev_3m dict 1 {'Type': 'NUM'}csc_ttl_6m dict 1 {'Type': 'NUM'}var1_m2_Transform dict 1 {'Type': 'CAT'}var2_m2_Transform dict 1 {'Type': 'CAT'}var2_m3_Transform dict 1 {'Type': 'CAT'}var3_m3_Transform dict 1 {'Type': 'CAT'}var5_m3_Transform dict 1 {'Type': 'CAT'}var9_m3_Transform dict 1 {'Type': 'CAT'}var8_m3_Transform dict 1 {'Type': 'CAT'}而我想要以下格式:Key Type Size Valuettl_change_3m str 1 NUMttl_change_6m str 1 NUMbase_rev_3m str 1 NUMcsc_ttl_6m str 1 NUMvar1_m2_Transform str 1 CATvar2_m2_Transform str 1 CATvar2_m3_Transform str 1 CATvar3_m3_Transform str 1 CATvar5_m3_Transform str 1 CATvar9_m3_Transform str 1 CATvar8_m3_Transform str 1 CAT另外,我认为我正在做太多的步骤来获得结果,并且可能会有较短/有效的代码版本来做到这一点。有人可以帮我吗?
1 回答

慕森卡
TA贡献1806条经验 获得超8个赞
我认为您需要np.where,
import numpy as np
import pandas as pd
df=pd.DataFrame({'Key':pd.Series(num_cols+cat_cols)})
df['Value']=np.where(df['Key'].isin(cat_cols), 'CAT','NUM')
#print(df)
Key Value
# ttl_change_3m NUM
# ttl_change_6m NUM
# base_rev_3m NUM
# csc_ttl_6m NUM
# var1_m2_Transform CAT
# var2_m2_Transform CAT
# var2_m3_Transform CAT
# var3_m3_Transform CAT
# var5_m3_Transform CAT
# var8_m3_Transform CAT
# var9_m3_Transform CAT
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