比较熊猫中的多条记录

Question

尝试比较“Cntr No”的 df1 和 df2，并且 df2 [“人工成本”、“材料成本”、“估计货币金额”] 的任何一列中的值必须与 df1 的总计相匹配。

例如，df1 OOLU 3868088 与 df2 OOLU 3868088 匹配，并且 df1 “28”的总值与 df2 的“劳动力成本”值“28”匹配。

df:

df1 = pd.DataFrame({'Cntr No': ['OOLU 3868088','OOLU 3868088','OOLU 3868088','TRIU 0625840','TRIU 0625840','TRIU 0625840','TRIU 1234567','OOLU 6232016','OOLU 0981231','OOLU 1212444'], 
               'Total': [12,28,48,119,82.5,11.0,18.0,11.0,13.0,10.0]})

df2 = pd.DataFrame({'Cntr No': ['OOLU 3868088','OOLU 3868088','OOLU 3868088','TRIU 0625840','TRIU 0625840','TRIU 0625840','TRIU 1234567'],  
                  'Labour Cost': [0.0,0.0,28.0,0.0,54.0,0.0,0.0], 
                  'Material Cost':[0.00,12.0,58.91,82.5,54.0,0.0,16.0],
                  'Amount in Estimate Currency':[48.00,12.00,87.81,82.5,119.0,12.0,16.0]})

预期输出：

    Cntr No        Total    Tally_with_df2
0   OOLU 3868088    12.0    Yes
1   OOLU 3868088    28.0    Yes
2   OOLU 3868088    48.0    Yes
3   TRIU 0625840    119.0   Yes
4   TRIU 0625840    82.5    Yes
5   TRIU 0625840    11.0    No
6   TRIU 1234567    18.0    No

使用的代码：这是我尝试过但无法达到我的要求的以下代码

cols = ['Labour Cost', 'Material Cost', 'Amount in Estimate Currency']

 d = {k: set(v.values()) for k, v in \
    df_co.set_index('Cntr No')[cols].to_dict(orient='index').items()}

df['Tally'] = [j in d.get(i, set()) for i, j in zip(df['Cntr No'], df['Total'])]
df['Tally'] = df['Tally'].map({True: 'Yes', False: 'No'})

df1:

Cntr No                       object
Serviced By                   object
Location                      object
WO No                         object
WASH - CHEMICAL              float64
PTI - CHILL                  float64
WASHING CONTAINER AGENT      float64
WASH - CHEMICAL AGENT        float64
WASHING CONTAINER -AGENT     float64
BUNDLING/UNBUNDLING OF FR    float64
PTI - AUTO                   float64
PTI                          float64
Struct Repair - Labour       float64
Struct Repair - Material     float64
Machy Repair - Labour        float64
Total                        float64
Vendor                        object
Sz                            object
Ty                            object
CO                            object
WO Date                       object
WO ID                         object

df2:

 Cntr No                            object
Equipment Size/type Group Code     object
Labour Cost                       float64
Material Cost                     float64
Amount in Estimate Currency       float64
Remarks                            object

Answer 1

IIUC，我们可以为每个唯一的 Cntr 编号从 df2 创建一个groupby 数据。

## this is grouped data
to_remove = df2.select_dtypes(['object']).columns.values.tolist()

df3 = (df2
.groupby('Cntr No')
.apply(lambda df: set(np.concatenate(df.loc[:, df.columns.difference(to_remove)].values))))

## df3 looks like this - using set for faster speed
print(df3)

Cntr No
OOLU 3868088    {0.0, 12.0, 48.0, 87.81, 58.91, 28.0}
TRIU 0625840           {0.0, 12.0, 82.5, 54.0, 119.0}
TRIU 1234567                              {16.0, 0.0}


## this function ensures all cases are handles
def get_value(x, data):
    if x['Cntr No'] not in data.index:
        return 'Not Found'
    else:
        if x['Total'] in data[x['Cntr No']]:
            return 'Yes'
        else:
            return 'No'

## next we do a simple look-up
df1['Tally_with_df2'] = df1.apply(lambda x: get_value(x, df3), axis=1)

print(df1)

        Cntr No  Total Tally_with_df2
0  OOLU 3868088   12.0            Yes
1  OOLU 3868088   28.0            Yes
2  OOLU 3868088   48.0            Yes
3  TRIU 0625840  119.0            Yes
4  TRIU 0625840   82.5            Yes
5  TRIU 0625840   11.0             No
6  TRIU 1234567   18.0             No