根據多個條件合並兩個數據幀

Question

我希望比較兩個數據幀（df-a 和 df-b）並搜索來自 1 個數據幀（df-b）的給定 ID 和日期位於該 ID 在另一個數據幀（df-a）中匹配的日期范圍內的位置）。 然后我想去除 df-a 中的所有列，並將它們連接到 df-b 匹配的地方。 例如

如果我有一個數據幀 df-a，格式如下 df-a：

    ID       Start_Date    End_Date     A   B   C   D   E 
0   cd2      2020-06-01    2020-06-24   'a' 'b' 'c' 10  20
1   cd2      2020-06-24    2020-07-21
2   cd56     2020-06-10    2020-07-03
3   cd915    2020-04-28    2020-07-21
4   cd103    2020-04-13    2020-04-24

和 df-b 在

    ID      Date
0   cd2     2020-05-12
1   cd2     2020-04-12
2   cd2     2020-06-10
3   cd15    2020-04-28
4   cd193   2020-04-13

我想要一個像這樣的輸出 df-c=

    ID      Date        Start_Date  End_Date    A   B   C   D   E 
0   cd2     2020-05-12      -           -       -   -   -   -   -
1   cd2     2020-04-12      -           -       -   -   -   -   -
2   cd2     2020-06-10 2020-06-01 2020-06-11    'a' 'b' 'c' 10  20
3   cd15    2020-04-28      -           -       -   -   -   -   -
4   cd193   2020-04-13      -           -       -   -   -   -   -

在上一篇文章中，我得到了一個很好的答案，它允許比較數據幀並在滿足此條件的任何地方刪除，但我正在努力弄清楚如何從 df-a 中適當地提取信息。 目前的嘗試如下！

df_c=df_b.copy()

ar=[]
for i in range(df_c.shape[0]):
    currentID = df_c.stafnum[i]
    currentDate = df_c.Date[i]
    df_a_entriesForCurrentID = df_a.loc[df_a.stafnum == currentID]

    for j in range(df_a_entriesForCurrentID.shape[0]):
        startDate = df_a_entriesForCurrentID.iloc[j,:].Leave_Start_Date
        endDate = df_a_entriesForCurrentID.iloc[j,:].Leave_End_Date

        if (startDate <= currentDate <= endDate):
            print(df_c.loc[i])
            print(df_a_entriesForCurrentID.iloc[j,:])
            
            #df_d=pd.concat([df_c.loc[i], df_a_entriesForCurrentID.iloc[j,:]], axis=0)
            
            #df_fin_2=df_fin.append(df_d, ignore_index=True)
            #ar.append(df_d)

Answer 1

所以你想進行一種“軟”匹配。 這是一個嘗試矢量化日期范圍匹配的解決方案。

# notice working with dates as strings, inequalities will only work if dates in format y-m-d
# otherwise it is safer to parse all date columns like `df_a.Date = pd.to_datetime(df_a)`

# create a groupby object once so we can efficiently filter df_b inside the loop
# good idea if df_b is considerably large and has many different IDs
gdf_b = df_b.groupby('ID')
b_IDs = gdf_b.indices # returns a dictionary with grouped rows {ID: arr(integer-indices)}

matched = [] # so we can collect matched rows from df_b
# iterate over rows with `.itertuples()`, more efficient than iterating range(len(df_a))
for i, ID, date in df_a.itertuples():
    if ID in b_IDs:
        gID = gdf_b.get_group(ID) # get the filtered df_b
        inrange = gID.Start_Date.le(date) & gID.End_Date.ge(date)
        if any(inrange):
            matched.append(
                gID.loc[inrange.idxmax()] # get the first row with date inrange
                .values[1:] # use the array without column indices and slice `ID` out
            )
        else:
            matched.append([np.nan] * (df_b.shape[1] - 1)) # no date inrange, fill with NaNs
    else:
        matched.append([np.nan] * (df_b.shape[1] - 1)) # no ID match, fill with NaNs
df_c = df_a.join(pd.DataFrame(matched, columns=df_b.columns[1:]))
print(df_c)

輸出

      ID        Date  Start_Date    End_Date    A    B    C     D     E
0    cd2  2020-05-12         NaN         NaN  NaN  NaN  NaN   NaN   NaN
1    cd2  2020-04-12         NaN         NaN  NaN  NaN  NaN   NaN   NaN
2    cd2  2020-06-10  2020-06-01  2020-06-24    a    b    c  10.0  20.0
3   cd15  2020-04-28         NaN         NaN  NaN  NaN  NaN   NaN   NaN
4  cd193  2020-04-13         NaN         NaN  NaN  NaN  NaN   NaN   NaN