将 Pandas 函数同时应用于行和列以进行置信区间计算

Question

我是 python 编程新手。我正在尝试确定数据集中的异常值。我已将数据集转换为 pandas 数据框，然后应用 IQR 原理。之后我想用零替换我的 OUTLIERS，然后计算平均值和标准偏差作为异常值，因为偏离平均值和 SD。

数据集代码如下：

import pandas as pd
data = [[123,100,1200,800,800,1200,900,1400],[246,15,16,45,15,45,11,55],[234,90,105,180,90,180,100,220],[236,100,90,9000,90,9000,70,140]]
df = pd.DataFrame(data,columns=['ID','Store1','Store2','Store3','Min','Max','Lower_Limit','Upper_limit'])
print (df)

数据集片段：

    ID  Store1  Store2  Store3  Min   Max  Lower_Limit Upper_limit
  123     100    1200     800  800  1200          900        1400
  246      15      16      45   15    45           11        55
  234      90     105     180   90   180          100          220
  236     100      90    9000   90  9000           70          140

如果 Store1、Store2、Store3 的值小于 Lower_limit(['Store1'] < ['Lower_limit']) 或更大，我想将它们更新为零 (0)比 Upper_limit(['Store1'] > ['Upper_limit']).

下面是我的函数：

def calculate_Outliers(row):
    if row['Store1'] < row['Lower_limit'] or row['Store1'] > row['Upper_limit']:
        return 0
    else:
        return row['Store1']
    if row['Store2'] < row['Lower_limit'] or row['Store2'] > row['Upper_limit']:
        return 0
    else:
        return row['Store2']
    if row['Store3'] < row['Lower_limit'] or row['Store3'] > row['Upper_limit']:
        return 0
    else:
        return row['Store3']

我是这样应用的：

df['Store1','Store3','Store3'] = df.apply(calculate_Outliers, axis=1)

下面是错误的结果...

    ID  Store1 Store2 Store3(Store1 Store2 Store3)
ID                  
123 NaN NaN NaN NaN 1000
246 NaN NaN NaN NaN 15
234 NaN NaN NaN NaN 0
236 NaN NaN NaN NaN 0

Desired Result:
ID  Store1  Store2  Store3  Min Max Lower_Limit Upper_limit
123 100    1200     800     800 1200    900      1400
246 15     16       45      15  45      11       55
234 0      105      180     90  180    100       220
236 100    90       0       90  9000    70       140

有什么方法可以修改我的原始代码来实现这个目标吗？

Answer 1

下面的函数应该可以完成工作：

def calculate_outliers(df):
    df['Store1'][(df['Store1']<df['Lower_Limit']) | (df['Store1'] > df['Upper_limit'])] = 0
    df['Store2'][(df['Store2']<df['Lower_Limit']) | (df['Store2'] > df['Upper_limit'])] = 0
    df['Store3'][(df['Store3']<df['Lower_Limit']) | (df['Store3'] > df['Upper_limit'])] = 0

Answer 2

df.loc[(df['Store1']<df['Lower_Limit']) | (df['Store1']>df['Upper_limit']),['Store1'] ] = 0

对其他商店重复上述步骤。

Answer 3

试试这个：

m=df.filter(like='Store').lt(df.Lower_Limit,axis=0)|df.filter(like='Store').\
                                                     gt(df.Upper_limit,axis=0)

df.update(df.where(~m,0).filter(like='Store'))
print(df)

    ID  Store1  Store2  Store3  Min   Max  Lower_Limit  Upper_limit
0  123       0    1200       0  800  1200          900         1400
1  246      15      16      45   15    45           11           55
2  234       0     105     180   90   180          100          220
3  236     100      90       0   90  9000           70          140

编辑如果列名没有通用字符串，您可以使用 iloc[]：

m=df.iloc[:,1:4].lt(df.Lower_Limit,axis=0)|df.iloc[:,1:4].gt(df.Upper_limit,axis=0)
df.update(df.where(~m,0).iloc[:,1:4])
print(df)

    ID  Store1  Store2  Store3  Min   Max  Lower_Limit  Upper_limit
0  123       0    1200       0  800  1200          900         1400
1  246      15      16      45   15    45           11           55
2  234       0     105     180   90   180          100          220
3  236     100      90       0   90  9000           70          140

包装在函数中：

def calculate_Outliers(df):
    m1= df['Store1'].lt(df['Lower_limit'])|df['Store1'].gt(df['Upper_limit'])
    m2 = df['Store2'].lt(df['Lower_limit'])|df['Store2'].gt(df['Upper_limit'])
    m3= df['Store3'].lt(df['Lower_limit'])|df['Store3'].gt(df['Upper_limit'])
    df.loc[m1,'Store1']=0
    df.loc[m1,'Store2']=0
    df.loc[m1,'Store3']=0
    print(df)
calculate_Outliers(df)

    ID  Store1  Store2  Store3  Min   Max  Lower_limit  Upper_limit
0  123       0       0       0  800  1200          900         1400
1  246      15      16      45   15    45           11           55
2  234       0       0       0   90   180          100          220
3  236     100      90    9000   90  9000           70          140

将 Pandas 函数同时应用于行和列以进行置信区间计算

Applying Pandas Function to rows and columns simultaneously for Confidence Interval Calculations

python

outliers

dataframe

pandas

iqr