场景1：

Question

我有一个看起来像这样的数据框：

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

我正在尝试将存在的manager2的一行（无论哪个）“覆盖”到具有空白/ NaN的manager1的行中，如下所示：

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

OR

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

显然，我们需要在DCC3上重新索引，但是那又如何呢？它只需要覆盖这两列（并且只要存在其他列就可以覆盖这些列）

我真的可以使用帮助，在此先谢谢您。

编辑1：

对不起，我没有弄清楚，这是一个基本情况。在某些情况下，这可能只是一个值（不适用于此值），或最多5-6。我以3行为例。

Answer 1

您可以使用np.where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

这两行代码应该可以为您解决问题。

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

以下是我尝试过的几种情况，代码是相同的。看看这是否是您想要的。

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

场景1：

行1：manager1 = NaN，manager2 =值

结果：将manager2的值分配给第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)

方案1的输出

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

方案2：

行1：manager1 =值，manager2 = NaN

结果：将manager2的值分配给第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)

方案2的输出

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

场景3：

行1：manager1 = NaN，manager2 = NaN

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

结果：将manager3的值分配给第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)

方案3的输出

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

场景4：

行1：manager1 =值，manager2 = NaN

第3行：manager1 =值，manager2 =值

结果：忽略第1行和第2行，因为第3行同时具有manager1和manager2的值

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)

方案4的输出：

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

熊猫：将一列覆盖为空白

编辑1：

2 个答案:

场景1：

方案2：

场景3：

场景4：