熊猫在两个数据框之间进行比较，标记匹配的内容

Question

我必须要数据帧df和df1

df低于

Facility    Category ID   Part  Text
Centennial  History 11111   A   Drain
Centennial  History 11111   B   Read
Centennial  History 11111   C   EKG
Centennial  History 11111   D   Assistant 
Centennial  History 11111   E   Primary

df1在下面（仅包含一个小样本问题，实际上是50,000行）

Facility    Category  ID      Part   Text
Centennial  History  11111    D      Assistant 

基本上，我想比较数据框之间的行，如果行在两个数据框之间匹配，则在第一个数据框df创建另一个列，列标题为['MatchingFlag']

我的最终结果数据框如下所示，因为我担心那些不匹配的数据框。

Facility    Category  ID    Part    Text      MatchingFlag
Centennial  History  11111  A     Drain         No
Centennial  History  11111  B     Read          No
Centennial  History  11111  C     EKG           No
Centennial  History  11111  D     Assistant     Yes
Centennial  History  11111  E     Primary       No

有什么帮助吗？ 我试过合并df = pd.merge(df1, df, how='left', on=['Facility', 'Category', 'ID', 'Part', 'Text'])然后根据空白或NaN值创建一个标志，但这并没有达到我的期望。

Answer 1

在要匹配的列上设置索引，然后使用该索引来排序匹配的行可能是有意义的

columns = ['Facility', 'Category', 'ID', 'Part', 'Text']

# It's always a good idea to sort after creating a MultiIndex like this
df = df.set_index(columns).sortlevel()
df1 = df1.set_index(columns).sortlevel()

# You don't have to use Yes here, anything will do
# The boolean True might be more appropriate
df['MatchingFlag'] = "Yes"
df1['MatchingFlag'] = "Yes"

# Add them together, matching rows will have the value "YesYes"
# Non-matches will be nan
result = df + df1

# If you'd rather not have NaN's 
result.loc[:,'MatchingFlag'] = result.loc[:,'MatchingFlag'].replace('YesYes','Yes')
result.loc[:,'MatchingFlag'] = result['MatchingFlag'].fillna('No')