基于条件将Pandas DataFrame列从String转换为Int

Question

我有一个看起来像的数据框

DF

viz  a1_count  a1_mean     a1_std
n         3        2   0.816497
y         0      NaN        NaN 
n         2       51  50.000000

我想基于条件将"viz"列转换为0和1.我试过了:

df['viz'] = 0 if df['viz'] == "n" else 1

但我得到:

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

Answer 1

你试图将一个标量与ValueError你所看到的整个系列进行比较.一个简单的方法是将布尔系列转换为int:

In [84]:
df['viz'] = (df['viz'] !='n').astype(int)
df

Out[84]:
   viz  a1_count  a1_mean     a1_std
0    0         3        2   0.816497
1    1         0      NaN        NaN
2    0         2       51  50.000000

您还可以使用np.where:

In [86]:
df['viz'] = np.where(df['viz'] == 'n', 0, 1)
df

Out[86]:
   viz  a1_count  a1_mean     a1_std
0    0         3        2   0.816497
1    1         0      NaN        NaN
2    0         2       51  50.000000

布尔比较的输出:

In [89]:
df['viz'] !='n'

Out[89]:
0    False
1     True
2    False
Name: viz, dtype: bool

然后投射到int:

In [90]:
(df['viz'] !='n').astype(int)

Out[90]:
0    0
1    1
2    0
Name: viz, dtype: int32