python - 根据 pandas DataFrame 中的列值有条件地替换多列

Question

我想根据第一组列中的值(具体来说，第一列中的一个为空白)同时用其他列中的相应值替换多个列的值。这是我正在尝试做的一个例子:

import pandas as pd

df = pd.DataFrame({'a1':['m', 'n', 'o', 'p'],
                   'a2':['q', 'r', 's', 't'],
                   'b1':['',  '',  'a', '' ],
                   'b2':['',  '',  'b',  '']})

df

#   a1 a2 b1 b2
# 0  m  q
# 1  n  r
# 2  o  s  a  b
# 3  p  t

我想用 a1 和 a2 中的相应值替换 b1 和 b2 中的 '' 值，其中 b1 为空:

#   a1 a2 b1 b2
# 0  m  q  m  q
# 1  n  r  n  r
# 2  o  s  a  b
# 3  p  t  p  t

这是我的思考过程(我对 pandas 比较陌生，所以我在这里说话可能带有很重的 R 口音):

missing = (df.b1 == '')

# First thought:
df[missing, ['b1', 'b2']] = df[missing, ['a1', 'a2']]
# TypeError: 'Series' objects are mutable, thus they cannot be hashed

# Fair enough  
df[tuple(missing), ('b1', 'b2')] = df[tuple(missing), ('a1', 'a2')]
# KeyError: ((True, True, False, True), ('a1', 'a2'))

# Obviously I'm going about this wrong.  Maybe I need to use indexing?
df[['b1', 'b2']].ix[missing,:]
#   b1 b2
# 0      
# 1      
# 3      

# That looks right
df[['b1', 'b2']][missing, :] = df[['a1', 'a2']].ix[missing, :]
# TypeError: 'Series' objects are mutable, thus they cannot be hashed
# Deja vu

df[['b1', 'b2']].ix[tuple(missing), :] = df[['a1', 'a2']].ix[tuple(missing), :]
# ValueError: could not convert string to float:
# Uhh...

我可以逐列进行:

df['b1'].ix[missing] = df['a1'].ix[missing]
df['b2'].ix[missing] = df['a2'].ix[missing]

...但我怀疑有更惯用的方法来做到这一点。想法？

更新:为了澄清，我特别想知道是否可以同时更新所有列。例如，对 Primer 答案的假设修改(这不起作用并导致 NaN，尽管我不确定为什么):

df.loc[missing, ['b1', 'b2']] = f.loc[missing, ['a1', 'a2']]

#   a1 a2   b1   b2
# 0  m  q  NaN  NaN
# 1  n  r  NaN  NaN
# 2  o  s    a    b
# 3  p  t  NaN  NaN

Answer 1

怎么样

df[['b1', 'b2']] = df[['b1', 'b2']].where(df[['b1', 'b2']] != '', df[['a1', 'a2']].values)

返回

  a1 a2 b1 b2
0  m  q  m  q
1  n  r  n  r
2  o  s  a  b
3  p  t  p  t