我有一个 Pandas 数据框,如下所示:
import pandas as pd
import numpy as np
df = pd.DataFrame({"Dummy_Var": [1]*12,
"B": [6, 143.3, 143.3, 143.3, 3, 4, 93.9, 93.9, 93.9, 2, 2, 7],
"C": [4.1, 23.2, 23.2, 23.2, 4.3, 2.5, 7.8, 7.8, 2, 7, 7, 7]})
B C Dummy_Var
0 6.0 4.1 1
1 143.3 23.2 1
2 143.3 23.2 1
3 143.3 23.2 1
4 3.0 4.3 1
5 4.0 2.5 1
6 93.9 7.8 1
7 93.9 7.8 1
8 93.9 2.0 1
9 2.0 7.0 1
10 2.0 7.0 1
11 7.0 7.0 1
每当相同的数字连续出现三次或更多次时,该数据应替换为 NAN
。所以结果应该是:
B C Dummy_Var
0 6.0 4.1 1
1 NaN NaN 1
2 NaN NaN 1
3 NaN NaN 1
4 3.0 4.3 1
5 4.0 2.5 1
6 NaN 7.8 1
7 NaN 7.8 1
8 NaN 2.0 1
9 2.0 NaN 1
10 2.0 NaN 1
11 7.0 NaN 1
我已经编写了一个函数来执行此操作:
def non_sense_remover(df, examined_columns, allowed_repeating):
def count_each_group(grp, column):
grp['Count'] = grp[column].count()
return grp
for col in examined_columns:
sel = df.groupby((df[col] != df[col].shift(1)).cumsum()).apply(count_each_group, column=col)["Count"] > allowed_repeating
df.loc[sel, col] = np.nan
return df
df = non_sense_remover(df, ["B", "C"], 2)
然而,我的真实数据框有 200 万行和 18 列!在 2M 行上运行此函数非常非常慢。有没有更有效的方法来做到这一点?我错过了什么吗?提前致谢。
最佳答案
在这种情况下构建 bool 掩码将比基于 apply()
的解决方案更有效,特别是对于大型数据集。这是一种方法:
cols = df[['B', 'C']]
mask = (cols.shift(-1) == cols) & (cols.shift(1) == cols)
df[mask | mask.shift(1).fillna(False) | mask.shift(-1).fillna(False)] = np.nan
编辑:
对于更通用的方法,将长度为 N
的序列替换为 NaN
,您可以这样做:
from functools import reduce
from operator import or_, and_
def replace_sequential_duplicates_with_nan(df, N):
mask = reduce(and_, [cols.shift(i) == cols.shift(i + 1)
for i in range(N - 1)])
full_mask = reduce(or_, [mask.shift(-i).fillna(False)
for i in range(N)])
df[full_mask] = np.nan
关于python - 如何删除 Pandas 中连续的坏数据点,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/47780904/