基于布尔条件的范畴分配

Question

我试图找出是否有更好的方法根据条件对数据进行分类。

示例数据：查看所识别的地方是否具有物理、社会和/或经济角色。如果有任何/多个角色存在，该位置将标记为"1“。

import pandas as pd
df = pd.DataFrame([[0, 1, 0], [0, 1, 1], [0, 1, 0], [0, 0, 1], [1,1,1], [1,1,0], columns=["PHYSICAL", "SOCIAL", "ECONOMIC"])

数据

|   | PHYSICAL | SOCIAL | ECONOMIC | 
| - | -------- | ------ | -------- | 
| 0 | 0        | 1      | 0        | 
| 1 | 0        | 1      | 1        | 
| 2 | 0        | 1      | 0        | 
| 3 | 0        | 0      | 1        | 
| 4 | 1        | 1      | 1        |       
| 5 | 1        | 1      | 0        |      

我想知道的：如何创建一个新列，根据True/False值为每一行分配一个类别。

所有可能的类别：

• 物质(仅)
• 社会(仅)
• 经济(仅)
• 物质与社会
• 物理与经济
• 社会经济
• 物质、社会和经济(全部)

预期结果

|   | PHYSICAL | SOCIAL | ECONOMIC | CATEGORY        |
| - | -------- | ------ | -------- | --------------- |
| 0 | 0        | 1      | 0        | social          |
| 1 | 0        | 1      | 1        | social_economic |
| 2 | 0        | 1      | 0        | social          |
| 3 | 0        | 0      | 1        | economic        |
| 4 | 1        | 1      | 1        | all_cat         |
| 5 | 1        | 1      | 0        | physical_social |

我尝试过的：

df['CATEGORY'] = np.where(df['PHYSICAL'], np.where(df['SOCIAL'], 
np.where(df['ECONOMIC'], 'All', 'FALSE'), 'FALSE'), 'FALSE')

谢谢!

Answer 1

你可以使用df.dot

df['CATEGORY'] = df.dot(df.columns + '_').str.rstrip('_').str.lower()

输出：

   PHYSICAL  SOCIAL  ECONOMIC                  CATEGORY
0         0       1         0                    social
1         0       1         1           social_economic
2         0       1         0                    social
3         0       0         1                  economic
4         1       1         1  physical_social_economic
5         1       1         0           physical_social

Answer 2

也许你也可以试试这个解决方案

import pandas as pd
df = pd.DataFrame([[0, 1, 0], [0, 1, 1], [0, 1, 0], [0, 0, 1], [1,1,1], [1,1,0]], columns=["PHYSICAL", "SOCIAL", "ECONOMIC"])

def get_category(physical,social,economic):
    if physical ==1 and social ==1 and economic ==1:
        return 'all_cat'
    elif physical ==0 and social ==0 and economic ==0:
        return 'None'
    elif physical == 1 and social == 1 and economic == 0:
        return 'physical_social'
    elif physical == 1 and social == 0 and economic == 1:
        return 'physical_economic'
    elif physical == 0 and social == 1 and economic == 1:
        return 'social_economic'
    elif physical ==1:
        return 'physical'
    elif social == 1:
        return 'social'
    elif economic == 1:
        return 'economic'

df['category'] = df.apply(lambda x:get_category(x['PHYSICAL'],x['SOCIAL'],x['ECONOMIC']),axis=1)

​

​

Answer 3

这是否理想的结果？如果你不介意的话，这是一种很残忍的方法。

df['CATEGORY'] = ''
for col_name in df.columns:
    df.ix[df[col_name]==1, 'CATEGORY'] = df['CATEGORY'] + '_' + col_name
df['CATEGORY'] = df['CATEGORY'].apply(lambda x: x.lower().strip('_'))
df

输出

    PHYSICAL    SOCIAL  ECONOMIC    CATEGORY
0   0           1       0           social
1   0           1       1           social_economic
2   0           1       0           social
3   0           0       1           economic
4   1           1       1           physical_social_economic
5   1           1       0           physical_social