Pandas：將一列與空白的行重疊

Question

我有一個看起來像這樣的數據框：

                dcc3   manager1   manager2
party_num                                 
L21635789  SBAS01030  A22677981        NaN
L21635789  SBAS02030        NaN  A22810282
L21635789  SBAS03030        NaN  A21721880

我正在嘗試將來自 manager2 的一行（與哪一行無關）“疊加”到包含 manager1 的行中，該行為空白/NaN，如下所示：

                dcc3   manager1   manager2
party_num                                 
L21635789  SBAS01030  A22677981  A22810282
L21635789  SBAS02030        NaN        NaN
L21635789  SBAS03030        NaN        NaN

或者

                dcc3   manager1   manager2
party_num                                 
L21635789  SBAS01030  A22677981  A21721880
L21635789  SBAS02030        NaN        NaN
L21635789  SBAS03030        NaN        NaN

顯然我們需要在 DCC3 上重新索引，但是然后呢？ 它只需要覆蓋這 2 列（並且只有這些列存在）

我真的可以使用幫助，在此先感謝您。

編輯 1：

對不起，我沒有澄清，這是一個基本案例。 在某些情況下，這只是一個值（這不適用），或最多 5-6。 我以 3 行為例。

Answer 1

您可以使用np.where來完成此操作：

df['manager2'] = np.where(df['manager1'].notnull() & df['manager2'].isnull(),
                          df['manager2'].dropna().iloc[0], np.nan) # You could do df['manager2'].dropna().iloc[1] for the other value
df
Out[1]: 
                dcc3   manager1   manager2
party_num                                 
L21635789  SBAS01030  A22677981  A22810282
L21635789  SBAS02030        NaN        nan
L21635789  SBAS03030        NaN        nan

Answer 2

這兩行代碼應該可以為您解決問題。

df.manager2 = df.manager2.bfill().ffill()
df.loc[df.manager1.isnull(), 'manager2'] = np.NaN

下面是我嘗試過的幾個場景，代碼是一樣的。 看看這是不是你想要的。

import pandas as pd
import numpy as np
c=['party_num','dcc3','manager1','manager2']

場景一：

第 1 行：manager1 = NaN，manager2 = 值

結果：將 manager2 值分配給第 2 行

print ('\nScenario 1')
print ('row 1: manager 1: NaN, manager 2: value; pick row2 manager 1 value')
d  = [['L21635789','SBAS01030',np.NaN,'A22810282'],
     ['L21635789','SBAS02030','A22677981',np.NaN],
     ['L21635789','SBAS03030',np.NaN,'A21721880']]

df = pd.DataFrame(data=d,columns=c)
print (df)
df.manager2 = df.manager2.bfill().ffill()
df.loc[df.manager1.isnull(), 'manager2'] = np.NaN
print ()
print (df)

場景 1 的輸出：

Scenario 1
row 1: manager 1: NaN, manager 2: value; pick row2 manager 1 value
   party_num       dcc3   manager1   manager2
0  L21635789  SBAS01030        NaN  A22810282
1  L21635789  SBAS02030  A22677981        NaN
2  L21635789  SBAS03030        NaN  A21721880

   party_num       dcc3   manager1   manager2
0  L21635789  SBAS01030        NaN        NaN
1  L21635789  SBAS02030  A22677981  A21721880
2  L21635789  SBAS03030        NaN        NaN

場景2：

第 1 行：manager1 = 值，manager2 = NaN

結果：將 manager2 值分配給第 1 行

print ('\nScenario 2')
print ('row 1: manager 1: value, manager 2: NaN; pick row2 manager 2 value')

d = [['L21635789','SBAS01030','A22677981',np.NaN],
     ['L21635789','SBAS02030',np.NaN,'A22810282'],
     ['L21635789','SBAS03030',np.NaN,'A21721880']]

df = pd.DataFrame(data=d,columns=c)
print (df)
df.manager2 = df.manager2.bfill().ffill()
df.loc[df.manager1.isnull(), 'manager2'] = np.NaN
print ()
print (df)

場景 2 的輸出：

Scenario 2
row 1: manager 1: value, manager 2: NaN; pick row2 manager 2 value
   party_num       dcc3   manager1   manager2
0  L21635789  SBAS01030  A22677981        NaN
1  L21635789  SBAS02030        NaN  A22810282
2  L21635789  SBAS03030        NaN  A21721880

   party_num       dcc3   manager1   manager2
0  L21635789  SBAS01030  A22677981  A22810282
1  L21635789  SBAS02030        NaN        NaN
2  L21635789  SBAS03030        NaN        NaN

場景3：

第 1 行：manager1 = NaN，manager2 = NaN

第 2 行：manager1 = 值； manager2 = NaN; 第 3 行：manager2 = 值

結果：將 manager3 值分配給第 2 行

print ('\nScenario 3')
print ('row 1: manager 1: NaN, manager 2: NaN; pick row2 manager 1 & row 3 manager 2')

d = [['L21635789','SBAS01030',np.NaN,np.NaN],
     ['L21635789','SBAS02030','A22677981',np.NaN],
     ['L21635789','SBAS03030',np.NaN,'A21721880']]

df = pd.DataFrame(data=d,columns=c)
print (df)
df.manager2 = df.manager2.bfill().ffill()
df.loc[df.manager1.isnull(), 'manager2'] = np.NaN
print ()
print (df)

場景 3 的輸出：

Scenario 3
row 1: manager 1: NaN, manager 2: NaN; pick row2 manager 1 & row 3 manager 2
   party_num       dcc3   manager1   manager2
0  L21635789  SBAS01030        NaN        NaN
1  L21635789  SBAS02030  A22677981        NaN
2  L21635789  SBAS03030        NaN  A21721880

   party_num       dcc3   manager1   manager2
0  L21635789  SBAS01030        NaN        NaN
1  L21635789  SBAS02030  A22677981  A21721880
2  L21635789  SBAS03030        NaN        NaN

場景 4：

第 1 行：manager1 = 值，manager2 = NaN

第 3 行：經理 1 = 價值，經理 2 = 價值

結果：忽略第 1 行和第 2 行，因為第 3 行對 manager1 和 manager2 都有值

print ('\nScenario 4')
print ('row 1: manager 1: NaN, manager 2: value; row3 has both manager 1 & manager 2')

d = [['L21635789','SBAS01030',np.NaN,'A21721880'],
     ['L21635789','SBAS02030',np.NaN,np.NaN],
     ['L21635789','SBAS03030','A22677981','A21721882']]

df = pd.DataFrame(data=d,columns=c)
print (df)
df.manager2 = df.manager2.bfill().ffill()
df.loc[df.manager1.isnull(), 'manager2'] = np.NaN
print ()
print (df)

場景 4 的輸出：

Scenario 4
row 1: manager 1: NaN, manager 2: value; row3 has both manager 1 & manager 2
   party_num       dcc3   manager1   manager2
0  L21635789  SBAS01030        NaN  A21721880
1  L21635789  SBAS02030        NaN        NaN
2  L21635789  SBAS03030  A22677981  A21721882

   party_num       dcc3   manager1   manager2
0  L21635789  SBAS01030        NaN        NaN
1  L21635789  SBAS02030        NaN        NaN
2  L21635789  SBAS03030  A22677981  A21721882