从pandas中的单个字符串列创建新的二进制列

Question

I've seen this before and simply can't remember the function.

Say I have a column "Speed" and each row has 1 of these values:

'Slow', 'Normal', 'Fast'

How do I create a new dataframe with all my rows except the column "Speed" which is now 3 columns: "Slow" "Normal" and "Fast" which has all of my rows labeled with a 1 in whichever column the old "Speed" column was. So if I had:

print df['Speed'].ix[0]
> 'Normal'

I would not expect this:

print df['Normal'].ix[0]
>1

print df['Slow'].ix[0]
>0

Answer 1

You can do this easily with pd.get_dummies ( docs ):

In [37]: df = pd.DataFrame(['Slow', 'Normal', 'Fast', 'Slow'], columns=['Speed'])

In [38]: df
Out[38]:
    Speed
0    Slow
1  Normal
2    Fast
3    Slow

In [39]: pd.get_dummies(df['Speed'])
Out[39]:
   Fast  Normal  Slow
0     0       0     1
1     0       1     0
2     1       0     0
3     0       0     1

Answer 2

Here is one solution:

df['Normal'] = df.Speed.apply(lambda x: 1 if x == "Normal" else 0)
df['Slow'] = df.Speed.apply(lambda x: 1 if x == "Slow" else 0)
df['Fast'] = df.Speed.apply(lambda x: 1 if x == "Fast" else 0)

Answer 3

This has another method：

df           = pd.DataFrame(['Slow','Fast','Normal','Normal'],columns=['Speed'])
df['Normal'] = np.where(df['Speed'] == 'Normal', 1 ,0)
df['Fast']   = np.where(df['Speed'] == 'Fast', 1 ,0)
df['Slow']   = np.where(df['Speed'] == 'Slow', 1 ,0)

df 
     Speed  Normal  Fast  Slow
0    Slow       0     0     1
1    Fast       0     1     0
2  Normal       1     0     0
3  Normal       1     0     1