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
我有一个数据框(见上文)。我需要比较它的行以获得匹配的行。因此,对于上面的df,我应该在比较后得到row1 = row4 = row6和row2 = row5。有什么有效的方法可以在python中进行行比较。
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
#first select only all duplicated rows
df1 = df[df.duplicated(keep=False)]
print df1
0 1 2 3 4
0 1 0 0 0 1
1 0 0 0 0 0
3 1 0 0 0 1
4 0 0 0 0 0
5 1 0 0 0 1
#sort values by all columns
df2 = df1.sort_values(by=df.columns.tolist())
print df2
0 1 2 3 4
1 0 0 0 0 0
4 0 0 0 0 0
0 1 0 0 0 1
3 1 0 0 0 1
5 1 0 0 0 1
#find groups
print (~((df2 == df2.shift(1)).all(1))).cumsum()
1 1
4 1
0 2
3 2
5 2
dtype: int32
#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
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