反转 MultiLabelBinarizer 以在列中创建列表

Question

在 Python3 中，我有一个多标签二进制数据格式的起始数据框：

df1:

"a" "b" "c" "d" "e"

 1   1   0   0   1
 0   0   1   0   1
 1   0   0   0   0
 0   1   1   0   1

我需要实现的是：

df2:

"a" "b" "c" "d" "e" "labels"

 1   1   0   0   1   ["a", "b", "e"]
 0   0   1   0   1   ["c", "e"]
 1   0   0   0   0   ["a"]
 0   1   1   0   1   ["b", "c", "e"]

首先，我尝试使用来自 sklearn 的 MultiLabelBinarizer 的 inverse_transform() 函数，该函数基于之前的堆栈 question。

from sklearn.preprocessing import MultiLabelBinarizer
mlb = MultiLabelBinarizer()
mlb.fit(df1.columns)
mlb.inverse_transform(df1.values)

ValueError: Expected indicator for 15 classes, but got 5

我尝试遵循来自 sklearn 的确切文档，但我不确定我哪里出错了。我尝试调整了一些参数，但我不明白问题是什么。

Answer 1

df2=df.apply(lambda x:x>0)# come up with a boolean dataframe

l=df.columns.to_numpy() put column names into a numpy array

#Calculate column `labels` using list comprehension in a `pd.DataFrame()` method.
df['labels']=pd.DataFrame({'a':[l[i] for i in df2.to_numpy()]})

Answer 2

一种 Numpy 方法

i, j = np.where(df)
a = df.columns.to_numpy()[j]
b = np.flatnonzero(np.diff(i)) + 1
df.assign(labels=np.split(a, b))

   a  b  c  d  e     labels
0  1  1  0  0  1  [a, b, e]
1  0  0  1  0  1     [c, e]
2  1  0  0  0  0        [a]
3  0  1  1  0  1  [b, c, e]

Answer 3

让我们试试dot 和str.split

df['labels'] = df.dot(df.columns+',').str[:-1].str.split(',')
0    ["a", "b", "e"]
1         ["c", "e"]
2              ["a"]
3    ["b", "c", "e"]
dtype: object

Answer 4

你可以stack数据，过滤值，分组：

df['labels'] = (df.stack()
   .loc[lambda x: x>0]
   .reset_index()
   .groupby('level_0')
   .agg({'level_1':list})
)

输出：

   "a"  "b"  "c"  "d"  "e"           labels
0    1    1    0    0    1  ["a", "b", "e"]
1    0    0    1    0    1       ["c", "e"]
2    1    0    0    0    0            ["a"]
3    0    1    1    0    1  ["b", "c", "e"]