Question

我有一个数据集，其中包含几列作为项目列表。我在下面给出一个例子。我正在尝试查找列表中具有100％匹配项的条目。我想找到90％或更低的产品。

>>> df2 = pd.DataFrame({ 'ID':['1', '2', '3', '4', '5', '6', '7', '8'], 'Productdetailed': [['Phone', 'Watch', 'Pen'], ['Pencil', 'fork', 'Eraser'], ['Apple', 'Mango', 'Orange'], ['Something', 'Nothing', 'Everything'], ['Eraser', 'fork', 'Pencil'], ['Phone', 'Watch', 'Pen'],['Apple', 'Mango'], ['Pen', 'Phone', 'Watch']]})

>>> df2
ID                   Productdetailed
0  1               [Phone, Watch, Pen]
1  2            [Pencil, fork, Eraser]
2  3            [Apple, Mango, Orange]
3  4  [Something, Nothing, Everything]
4  5            [Eraser, fork, Pencil]
5  6               [Phone, Watch, Pen]
6  7                    [Apple, Mango]
7  8               [Pen, Phone, Watch]

如果您注意到df2中的索引0和索引7，则它们具有相同的项目集，但顺序不同。其中索引0和索引5具有相同顺序的相同项目。我想将两者都视为比赛。我尝试了groupby和series.isin()。我还尝试通过将数据集一分为二来尝试相交，但是由于类型错误而失败。

首先，我想计算精确匹配项的数量（匹配行数的数量）以及与其匹配的行索引号。但是，当某些项目仅部分匹配时，例如df2中的索引2和索引6。我想说的是已匹配项目的百分比以及与之对应的列号。

我提到了。我试图将特定列值的数据分为两部分。然后

applied df2['Intersection'] = 
     [list(set(a).intersection(set(b))) 
         for a, b in zip(df2_part1.Productdetailed, df2_part2.Productdetailed)
     ]

，其中a和b是Productdetailed和df2_part1的片段中的df2_part2列。

有没有办法做到这一点？请帮助

Answer 1

要知道确切的匹配项：

df2["Productdetailed"]=df2["Productdetailed"].sort_values()
# create new colum from the sorted list. More easy to work with pivot table
df2['Productdetailed_str'] = df2['Productdetailed'].apply(lambda x: ', '.join(x))
df2["hit"] = 1
df3 = (df2.pivot_table(index=["Productdetailed_str"],
                 values=["ID", "hit"],
                aggfunc={'ID': lambda x: ', '.join(x), 'hit': 'sum'}
               ))

命中率是出现的次数。结果df3：

                                  ID  hit
Productdetailed_str                      
Apple, Mango                       7    1
Apple, Mango, Orange               3    1
Eraser, fork, Pencil               5    1
Pen, Phone, Watch                  8    1
Pencil, fork, Eraser               2    1
Phone, Watch, Pen               1, 6    2
Something, Nothing, Everything     4    1

部分匹配更加困难，但是您可以开始拆分列表并使用数据透视表进行播放：

test = df2.apply(lambda x: pd.Series(x['Productdetailed']),axis=1).stack().reset_index(level=1, drop=True).to_frame(name='list').join(df2)

如果运行测试。您在“列表列”中的“产品详细列”列表中的单词。另外，您有ID ...因此，我认为使用数据透视表可以提取信息。

Answer 2

此解决方案解决了完全匹配任务（代码复杂度很高，不建议使用）：

NSPersistentStoreCoordinator

用于完全匹配和部分匹配（如果至少两个值匹配，则部分匹配，也可以更改）：

#First create a dummy column of Productdetailed which is sorted
df2['dummy'] = df2['Productdetailed'].apply(sorted)
#Create Matching column which stores index of first matched list
df2['Matching'] = np.nan

#Code for finding the exact matches and assigning indices in Matching column
for index1,lst1 in enumerate(df2['dummy']):
    for index2,lst2 in enumerate(df2['dummy']):
        if index1<index2:
            if (lst1 == lst2):
                if np.isnan(df2.loc[index2,'Matching']):
                    df2.loc[index1,'Matching'] = index1
                    df2.loc[index2,'Matching'] = index1

#Finding the sum of total exact matches
print(df2['Matching'].notnull().sum())
5

#Deleting the dummy column
del df2['dummy']

#Final Dataframe
print(df2)

  ID                   Productdetailed  Matching
0  1               [Phone, Watch, Pen]       0.0
1  2            [Pencil, fork, Eraser]       1.0
2  3            [Apple, Mango, Orange]       NaN
3  4  [Something, Nothing, Everything]       NaN
4  5            [Eraser, fork, Pencil]       1.0
5  6               [Phone, Watch, Pen]       0.0
6  7                    [Apple, Mango]       NaN
7  8               [Pen, Phone, Watch]       0.0

#First create a dummy column of Productdetailed which is sorted
df2['dummy'] = df2['Productdetailed'].apply(sorted)
#Create Matching column which stores index of first matched list
df2['Matching'] = np.nan
#Create Column Stating Status of Matching
df2['Status'] = 'No Match'

#Code for finding the exact matches and assigning indices in Matching column
for index1,lst1 in enumerate(df2['dummy']):
    for index2,lst2 in enumerate(df2['dummy']):
        if index1<index2:
            if (lst1 == lst2):
                if np.isnan(df2.loc[index2,'Matching']):
                    df2.loc[index1,'Matching'] = index1
                    df2.loc[index2,'Matching'] = index1
                    df2.loc[[index1,index2],'Status'] = 'Fully Matched'
            else:
                count = sum([1 for v1 in lst1 for v2 in lst2 if v1==v2])
                if count>=2:
                    if np.isnan(df2.loc[index2,'Matching']):
                        df2.loc[index1,'Matching'] = index1
                        df2.loc[index2,'Matching'] = index1
                        df2.loc[[index1,index2],'Status'] = 'Partially Matched'

#Finding the sum of total exact matches
print(df2['Matching'].notnull().sum())

7

#Deleting the dummy column
del df2['dummy']

#Final Dataframe
print(df2)

在包含项目列表的列中查找通用值

2 个答案: