python – 根据pandas DataFrame中的列值有条件地替换多个列

我想基于第一组列中的值(具体地,第一列中的一列为空白),同时将多列的值替换为其他列中的对应值.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

我想将b1和b2中的”值替换为a1和a2中的相应值,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(我对熊猫来说相对较新,所以我可能会说这里有一个很重的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...

我可以逐列完成：

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

…但我怀疑这是一种更惯用的方法.思考？

更新：为了澄清,我特别想知道是否所有列都可以同时更新.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

解决方法:

怎么样

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

这回来了

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