pandas 从熊猫中的单个字符串列创建新的二进制列
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Creating new binary columns from single string column in pandas
提问by user1610719
I've seen this before and simply can't remember the function.
我以前见过这个，只是不记得这个功能。
Say I have a column "Speed" and each row has 1 of these values:
假设我有一列“速度”，每一行都有以下值之一：
'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:
如何使用除“速度”列之外的所有行创建新数据框，该列现在有 3 列：“慢”、“正常”和“快”，其中我的所有行在旧的“速度”列中都标有 1 " 列了。所以如果我有：
print df['Speed'].ix[0]
> 'Normal'
I would not expect this:
我不希望这样：
print df['Normal'].ix[0]
>1
print df['Slow'].ix[0]
>0
回答by joris
You can do this easily with pd.get_dummies(docs):
您可以使用pd.get_dummies( docs)轻松完成此操作：
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
回答by aha
Here is one solution:
这是一种解决方案：
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)
回答by sun
This has another method：
这还有一个方法：
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
