将表转换为存在/不存在矩阵python

Question

来自两个表格文件：

FILE1.TXT

name1  house1
name2  house1
name3  house1
name4  house2
name5  house2
name6  house2

和file2.txt

name1  car
name2  bike
name3  skate
name4  car
name5  motorcycle
name6  boat

我想使用两个字典中的信息来创建这样的存在/不存在矩阵。

       car  motorcycle  bike  boat  skate
house1  1       0         1     0     1
house2  1       1         0     1     0

这是我的代码：

import pandas as pd

with open('file1.txt', 'r') as file1:
    col_names = ['name', 'house']
    df1 = pd.read_csv(file1, sep='\t', header=None, names=col_names)

with open('file2.txt', 'r') as file2:
    col_names = ['name', 'transport']
    df2 = pd.read_csv(file2, sep='\t', header=None, names=col_names)

    # include the values from df1 into the df2 creating a new column        
    df2['house'] = df2['name'].map(df1.set_index('name')['house'])

    g = df2.groupby('house')['transport'].apply(list).reset_index()


    g.join(pd.get_dummies(g['transport'].apply(pd.Series).stack()).sum(level=0)).drop('transport', 1)

    print g

这样做，我获得了这个输出：

    house                transport
0  house1       [car, bike, skate]
1  house2  [car, motorcycle, boat]

Answer 1

这是一种方式。

<强>设置

# df1
  individual   house
0      name1  house1
1      name2  house1
2      name3  house1
3      name4  house2
4      name5  house2
5      name6  house2

# df2
  individual   transport
0      name1         car
1      name2        bike
2      name3       skate
3      name4         car
4      name5  motorcycle
5      name6        boat

<强>解决方案

df2['house'] = df2['individual'].map(df1.set_index('individual')['house'])
g = df2.groupby('house')['transport'].apply(list).reset_index()

g.join(pd.get_dummies(g['transport']\
 .apply(pd.Series).stack()).sum(level=0)).drop('transport', 1)

<强>结果

    house  bike  boat  car  motorcycle  skate
0  house1     1     0    1           0      1
1  house2     0     1    1           1      0

<强>解释

有3个步骤：

1. 将“住宅”列添加到df2，从df1。
映射
2. 集体运输。
3. 使用pd.get_dummies将transports列扩展为虚拟列。