我有一个9列的df。每列的值为0,1。
1-表示离群值。
根据9种不同算法得出的异常值。我想选择那些真正的异常值,以下查询确实起作用。
true_outliers= outliers[
(outliers['isolation_forest_300000']==1) &
(outliers['knn_1000']==1) &
(outliers['knn_10000']==1)&
(outliers['abod_neighbors_5_1000']==1)&
(outliers['abod_neighbors_5_10000']==1)&
(outliers['abod_neighbors_10_1000']==1)&
(outliers['hbos_1000']==1)&
(outliers['hbos_10000']==1)&
(outliers['hbos_100000']==1)]
但是我怎么能这样重构它:
for col in outliers.columns.tolist():
s= outliers[outliers[col] == 1]
我希望它遍历循环并仅选择每列中为'1'的那些行
如果要选择每一列的行都带有1
的行,则最好使用掩码
样本df:
Out[266]:
isolation_forest_300000 knn_1000 knn_10000 abod_neighbors_5_1000 \
0 1 1 1 1
1 0 0 0 1
2 0 0 0 0
3 1 1 1 1
abod_neighbors_5_10000 abod_neighbors_10_1000 hbos_1000 hbos_10000 \
0 1 1 1 1
1 1 0 0 0
2 0 0 0 0
3 1 1 1 1
hbos_100000
0 1
1 0
2 0
3 1
使用eq
和all
创建遮罩和切片
df[df.eq(1).all(1)]
Out[267]:
isolation_forest_300000 knn_1000 knn_10000 abod_neighbors_5_1000 \
0 1 1 1 1
3 1 1 1 1
abod_neighbors_5_10000 abod_neighbors_10_1000 hbos_1000 hbos_10000 \
0 1 1 1 1
3 1 1 1 1
hbos_100000
0 1
3 1
我认为这可以帮助您:
import functools
import operator
import pandas as pd
data = [[0, 0, 0, 0, 0, 0, 0, 0, 0],
[1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 1, 0, 0, 0],
[0, 0, 1, 0, 0, 0, 0, 1, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 1, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0],
[1, 1, 1, 1, 1, 1, 1, 1, 1],
[0, 0, 0, 0, 0, 0, 0, 0, 0]]
df = pd.DataFrame(
data, columns=[str(i) for i in range(9)]
)
condition = functools.reduce(
operator.and_,
(df[col] == 1 for col in df.columns)
)
print(df[condition])