根據列值從熊貓數據框中排除記錄

Question

我嘗試根據兩個條件（if,else）篩選我的熊貓數據框。只有if聲明有效。 if語句基於2個條件（logic1和logic2）保留（標記）我的DataFrame中的所有記錄。參見第4節

在else發言，我想排除已標記（logic1和logic2）所有的ID，而無需創建一個額外的列表或遍歷每個記錄。有沒有辦法將所有這些記錄過濾掉而不將ID存儲在額外的列表中？

我只想使用過濾器功能，如果可能的話。目前我從第3部分獲得輸出。它是錯誤的，因爲id = 2已被標記，但仍包含在輸出中。我需要的輸出部4中顯示。

代碼

logic1 = (potatoes['Desc'] == 'Bla2') & (potatoes['Value'] == True) & (potatoes['Enabled'] == True) 
logic2 = (potatoes['Desc'].isin(['Bla8', 'Bla9'])) & (potatoes['Active'] == True) & (potatoes['Enabled'] == True) 

if flagged: 
    potatoes_flagged = potatoes[logic1 | logic2] 
    return potatoes_flagged 
else: 
    potatoes_not_flagged = potatoes[~logic1 & ~logic2] 
    return potatoes_not_flagged 

1.輸入（馬鈴薯）

id | Desc | Active | Enabled | Value | [A LOT OF OTHER COLUMNS] 
1 | Bla1 | 1  | 0  | 1  | [A LOT OF OTHER COLUMNS] 
2 | Bla2 | 1  | 1  | 1  | [A LOT OF OTHER COLUMNS] 
2 | Bla3 | 1  | 1  | 0  | [A LOT OF OTHER COLUMNS] 
2 | Bla4 | 0  | 0  | 0  | [A LOT OF OTHER COLUMNS] 
2 | Bla5 | 0  | 0  | 0  | [A LOT OF OTHER COLUMNS] 
3 | Bla6 | 1  | 1  | 0  | [A LOT OF OTHER COLUMNS] 
4 | Bla7 | 0  | 0  | 1  | [A LOT OF OTHER COLUMNS] 

2.輸出爲標記（如果）（CORRECT ）

id | Desc | Active | Enabled | Value | [A LOT OF OTHER COLUMNS] 
2 | Bla2 | 1  | 1  | 1  | [A LOT OF OTHER COLUMNS] 

3.輸出不標記（否則）（錯）

id | Desc | Active | Enabled | Value | [A LOT OF OTHER COLUMNS] 
1 | Bla1 | 1  | 0  | 1  | [A LOT OF OTHER COLUMNS] 
2 | Bla3 | 1  | 1  | 0  | [A LOT OF OTHER COLUMNS] 
2 | Bla4 | 0  | 0  | 0  | [A LOT OF OTHER COLUMNS] 
2 | Bla5 | 0  | 0  | 0  | [A LOT OF OTHER COLUMNS] 
3 | Bla6 | 1  | 1  | 0  | [A LOT OF OTHER COLUMNS] 
4 | Bla7 | 0  | 0  | 1  | [A LOT OF OTHER COLUMNS] 

4.輸出需要未標記（正確）

id | Desc | Active | Enabled | Value | [A LOT OF OTHER COLUMNS] 
1 | Bla1 | 1  | 0  | 1  | [A LOT OF OTHER COLUMNS] 
3 | Bla6 | 1  | 1  | 0  | [A LOT OF OTHER COLUMNS] 
4 | Bla7 | 0  | 0  | 1  | [A LOT OF OTHER COLUMNS] 

Answer 1

它看起來像你想找到所有id s不會由potatoes[logic1 | logic2]返回。你可以使用一個倒置的isin調用來做到這一點。

idx_flagged = potatoes.loc[logic1 | logic2, 'id'].values 
potatoes[~potatoes.id.isin(idx_flagged)] 

    id Desc Active Enabled Value 
0 1 Bla1  1  0  1 
5 3 Bla6  1  1  0 
6 4 Bla7  0  0  1