如何在python中比较数据帧的行以获得相等性

Question

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

我有一个数据帧（见上文）。我需要比较它的行以得到匹配的行。所以对于上面的数据，我应该在比较后得到row1=row4=row6和row2=row5。在python中有没有有效的方法来进行行比较呢。在

Answer 1

下面是我的想法。在

import pandas as pd


df = pd.DataFrame({0: {0: 1, 1: 0, 2: 0, 3: 1, 4: 0, 5: 1}, 
                   1: {0: 0, 1: 0, 2: 1, 3: 0, 4: 0, 5: 0}, 
                   2: {0: 0, 1: 0, 2: 0, 3: 0, 4: 0, 5: 0}, 
                   3: {0: 0, 1: 0, 2: 0, 3: 0, 4: 0, 5: 0}, 
                   4: {0: 1, 1: 0, 2: 1, 3: 1, 4: 0, 5: 1}})

groups = df.groupby(df.columns.tolist())
df.loc[:, 'group_num'] = None


for num, group in enumerate(groups):
    df.loc[group[1].index, 'group_num'] = num

收益率。。。在

^{pr2}$

为什么第[1]组在最后一行？在

因为您正在迭代一个形式为（group_name，group_table）的元组。组[1]访问实际分组的数据帧。在

Answer 2

使用：

import pandas as pd


df = pd.DataFrame({0: {0: 1, 1: 0, 2: 0, 3: 1, 4: 0, 5: 1}, 
                   1: {0: 0, 1: 0, 2: 1, 3: 0, 4: 0, 5: 0}, 
                   2: {0: 0, 1: 0, 2: 0, 3: 0, 4: 0, 5: 0}, 
                   3: {0: 0, 1: 0, 2: 0, 3: 0, 4: 0, 5: 0}, 
                   4: {0: 1, 1: 0, 2: 1, 3: 1, 4: 0, 5: 1}})
print df
   0  1  2  3  4
0  1  0  0  0  1
1  0  0  0  0  0
2  0  1  0  0  1
3  1  0  0  0  1
4  0  0  0  0  0
5  1  0  0  0  1

^{pr2}$

#print groups    
for i, g in df.groupby((~((df2 == df2.shift(1)).all(1))).cumsum()):
    print g

   0  1  2  3  4
1  0  0  0  0  0
4  0  0  0  0  0
   0  1  2  3  4
0  1  0  0  0  1
3  1  0  0  0  1
5  1  0  0  0  1

#dict comprehension for storing groups
dfs = {i-1: g for i,g in df.groupby((~((df2 == df2.shift(1)).all(1))).cumsum())}
print dfs
{0.0:    0  1  2  3  4
1  0  0  0  0  0
4  0  0  0  0  0, 1.0:    0  1  2  3  4
0  1  0  0  0  1
3  1  0  0  0  1
5  1  0  0  0  1}

print dfs[0]
   0  1  2  3  4
1  0  0  0  0  0
4  0  0  0  0  0

print dfs[1]
   0  1  2  3  4
0  1  0  0  0  1
3  1  0  0  0  1
5  1  0  0  0  1