我有一个有9列的df。每列都有值0,1。
1个-means异常值。
这是根据9种不同算法得出的异常值。我想选择那些真正的异常值,下面的查询是可行的。
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,我如何像这样重构它:
for col in outliers.columns.tolist():
s= outliers[outliers[col] == 1]我希望它遍历循环,并且只选择每列中为'1‘的那些行
发布于 2019-11-18 13:29:23
如果希望在每列上都使用1选择行,则使用掩码更好
示例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创建蒙版和切片
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发布于 2019-11-18 14:56:53
我认为这可以帮助你:
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])https://stackoverflow.com/questions/58908304
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