python - 根据另一列中的值，用字符串替换一列中的 NaN

简单地说，在 B 列 = 't3' 的地方，我想用一个新字符串替换 A 列中的 NaN 值。

我下面的尝试都失败了。

d = pd.DataFrame({"A":[np.nan, 't2', np.nan, 't3', np.nan], "B":['t1', 't2', 't3', 't4', 't3']})
print "Original Dataframe:\n", d

# Does not work
d[d.B == 't3'].A = 'new_val'

# Does not work
d[d.B == 't3'].A.replace(np.nan, 'new_val')


# Does not work
d[d.B == 't3'].A.replace(np.nan, 'new_val', inplace=True)

print "Final Dataframe:\n", d

这是输出:

Original Dataframe:
     A   B
0  NaN  t1
1   t2  t2
2  NaN  t3
3   t3  t4
4  NaN  t3

[5 rows x 2 columns]
Final Dataframe:
     A   B
0  NaN  t1
1   t2  t2
2  NaN  t3
3   t3  t4
4  NaN  t3

最佳答案

使用 loc 见 http://pandas.pydata.org/pandas-docs/stable/indexing.html#different-choices-for-indexing-loc-iloc-and-ix

In [5]:

d.loc[(d['A'].isnull()) & (d.B == 't3'), 'A']='new_val'

d

Out[5]:

         A   B
0      NaN  t1
1       t2  t2
2  new_val  t3
3       t3  t4
4  new_val  t3

[5 rows x 2 columns]

关于python - 根据另一列中的值，用字符串替换一列中的 NaN，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/21943688/