将 Pandas DataFrame 列拆分为 OneHot/Binary 列

Question

我有一个我正在为 SciKit Learn PCA 格式化的 DataFrame 看起来像这样：

datetime |  mood |  activities |  notes

8/27/2017 |  "good" | ["friends", "party", "gaming"] | NaN

8/28/2017 |  "meh" |  ["work", "friends", "good food"] | "Stuff stuff"

8/29/2017 |  "bad" |  ["work", "travel"] |  "Fell off my bike"

...等等

我想把它改成这个，我认为这对机器学习工作会更好：

datetime |  mood |  friends | party | gaming | work | good food | travel |  notes

8/27/2017 |  "good" | True | True | True | False | False | False | NaN

8/28/2017 |  "meh" |  True | False | False | True | True | False | "Stuff stuff"

8/29.2017 | "bad" | False | False | False | False | True | False | True | "Fell off my bike"

---

我已经尝试过这里概述的方法，它只是为我提供了所有活动的左对齐矩阵。列没有任何意义。如果我尝试传递columns给DataFrame构造函数，则会收到错误消息“传递了 26 列，传递的数据有 9 列。我相信这是因为即使我有 26 个离散事件，我在同一天所做的最多也是 9 个。如果在该特定行中找不到该列，有没有办法让它用 0/False 填充？谢谢。

Answer 1

你可以简单地使用 get_dummies

让我们假设这个数据框：

df = pd.DataFrame({'datetime':pd.date_range('2017-08-27', '2017-08-29'),
              'mood':['good','meh','bad'],'activities':[['friends','party','gaming'],
                                                        ["work", "friends", "good food"],
                                                        ["work", "travel"]],
              'notes':[np.nan, 'stuff stuff','fell off my bike']})
df.set_index(['datetime'], inplace=True)

            mood      activities                notes
datetime            
2017-08-27  good    [friends, party, gaming]    NaN
2017-08-28  meh     [work, friends, good food]  stuff stuff
2017-08-29  bad     [work, travel]              fell off my bike

只是concat和get_dummies：

df2 = pd.concat([df[['mood','notes']], pd.get_dummies(df['activities'].apply(pd.Series),
                                                      prefix='activity')], axis=1)


            mood    notes   activity_friends    activity_work   activity_friends    activity_party  activity_travel activity_gaming activity_good food
datetime                                    
2017-08-27  good    NaN             1               0                 0                 1                   0                   1                   0
2017-08-28  meh     stuff stuff     0               1                 1                 0                   0                   0                   1
2017-08-29  bad    fell off my bike 0               1                 0                 0                   1                   0                   0

如果你想使用，你可以将它们更改为布尔值loc：

df2.loc[:,df2.columns[2:]] = df2.loc[:,df2.columns[2:]].astype(bool)