在python中优化标签编码的时间迭代

Question

我在python的数据框架中有以下列，名为admission：

Patient ID, Regular ward, Semi-intensive, Intensive
1                0               0            0
2                1               0            0
3                0               1            0
4                0               1            0
5                0               0            1

我已经完成了以下转换：

 Patient ID, Admission type
 1                 0
 2                 1
 3                 2
 4                 2
 5                 3

基本上，为了实现这一转换，我创建了一个名为“No admission”的新列，其中包含1，如果三个列（常规病房、半强化病房、强化病房）为0，则返回0

之后，我将每列中的值映射到一列中，以返回列的名称，而不是0或1

Patient ID, Admission type
 1          No admission
 2          Regular ward
 3          Semi-intensive
 4          Semi-intensive
 5          Intensive

对于不入院=0，普通病房=1，半强化病房=2，强化病房=3，再次进行映射

下面是我的完整代码转换

df['No admission'] = None
admission= df.iloc[:,3:6] # Selecting only the three columns mentioned above.

for row in df.index:
    if admission.apply(any,axis=1).values[row]:
        df.loc[row,'No admission'] = 0
    else:
        df.loc[row,'No admission'] = 1

cols = [4,5,6,106]
df.insert(loc = 1,
          column = 'Admission type',
          value = df.iloc[:,cols].apply(np.argmax, axis=1))

df.drop(df.columns[cols],axis=1,inplace=True)

le = LabelEncoder()
le.fit(df['Admission type'])
df['Admission type'] = le.transform(df['Admission type'])

关键是：我的原始数据集有5644行和111列。解释上述内容的执行时间大约为10分钟

我知道这是一个更好的方法来改善它，我感谢任何帮助

Answer 1

使用np.where()如何提高性能？我想会快得多

输入：

Patient ID  Regular ward   Semi-intensive  Intensive
1                0               0            0
2                1               0            0
3                0               1            0
4                0               1            0
5                0               0            1

代码：

df=pd.read_clipboard(sep='\\s\\s+'))
df['Admission type'] = ''
df['Admission type'] = np.where(df['Regular ward'] == 1, 'Regular ward', df['Admission type'])
df['Admission type'] = np.where(df['Semi-intensive'] == 1, 'Semi-intensive', df['Admission type'])
df['Admission type'] = np.where(df['Intensive'] == 1, 'Intensive', df['Admission type'])
df['Admission type'] = np.where((df['Regular ward'] + df['Semi-intensive'] + df['Intensive'] == 0),
                              'No admission', df['Admission type'])
df=df[['Patient ID', 'Admission type']]
df 

输出：

    Patient ID   Admission type
0   1          No admission
1   2          Regular ward
2   3          Semi-intensive
3   4          Semi-intensive
4   5          Intensive