学习Pandas，第 7 课

英文原文: 07 - Lesson

离群值 (Outlier)

import pandas as pd
import sys

print('Python version ' + sys.version)
print('Pandas version ' + pd.__version__)

Python version 3.6.1 | packaged by conda-forge | (default, Mar 23 2017, 21:57:00) 
[GCC 4.2.1 Compatible Apple LLVM 6.1.0 (clang-602.0.53)]
Pandas version 0.19.2

# 创建一个 dataframe，用日期作为索引
States = ['NY', 'NY', 'NY', 'NY', 'FL', 'FL', 'GA', 'GA', 'FL', 'FL'] 
data = [1.0, 2, 3, 4, 5, 6, 7, 8, 9, 10]
idx = pd.date_range('1/1/2012', periods=10, freq='MS')
df1 = pd.DataFrame(data, index=idx, columns=['Revenue'])
df1['State'] = States
# 创建第二个 dataframe
data2 = [10.0, 10.0, 9, 9, 8, 8, 7, 7, 6, 6]
idx2 = pd.date_range('1/1/2013', periods=10, freq='MS')
df2 = pd.DataFrame(data2, index=idx2, columns=['Revenue'])
df2['State'] = States

# 把两个 dataframe 合并起来
df = pd.concat([df1,df2])
df

计算离群值的方法

注意: 均值(average)和标准差(Standard Deviation)只对高斯分布(gaussian distribution)有意义。

# 方法 1
# 原始的 df 拷贝一份
newdf = df.copy()
newdf['x-Mean'] = abs(newdf['Revenue'] - newdf['Revenue'].mean())
newdf['1.96*std'] = 1.96*newdf['Revenue'].std()  
newdf['Outlier'] = abs(newdf['Revenue'] - newdf['Revenue'].mean()) > 1.96*newdf['Revenue'].std()
newdf

# 方法 2
# 分组的方法
# 原始的 df 拷贝一份
newdf = df.copy()
State = newdf.groupby('State')
newdf['Outlier'] = State.transform( lambda x: abs(x-x.mean()) > 1.96*x.std() )
newdf['x-Mean'] = State.transform( lambda x: abs(x-x.mean()) )
newdf['1.96*std'] = State.transform( lambda x: 1.96*x.std() )
newdf

# 方法 2
# 多个条件分组
# 原始 df 拷贝一份
newdf = df.copy()
StateMonth = newdf.groupby(['State', lambda x: x.month])
newdf['Outlier'] = StateMonth.transform( lambda x: abs(x-x.mean()) > 1.96*x.std() )
newdf['x-Mean'] = StateMonth.transform( lambda x: abs(x-x.mean()) )
newdf['1.96*std'] = StateMonth.transform( lambda x: 1.96*x.std() )
newdf

# 方法 3
# 分组的方法
# 原始 df 拷贝一份
newdf = df.copy()
State = newdf.groupby('State')
def s(group):
    group['x-Mean'] = abs(group['Revenue'] - group['Revenue'].mean())
    group['1.96*std'] = 1.96*group['Revenue'].std()  
    group['Outlier'] = abs(group['Revenue'] - group['Revenue'].mean()) > 1.96*group['Revenue'].std()
    return group
Newdf2 = State.apply(s)
Newdf2

# 方法 3
# 多个条件分组
# 原始 df 拷贝一份
newdf = df.copy()
StateMonth = newdf.groupby(['State', lambda x: x.month])
def s(group):
    group['x-Mean'] = abs(group['Revenue'] - group['Revenue'].mean())
    group['1.96*std'] = 1.96*group['Revenue'].std()  
    group['Outlier'] = abs(group['Revenue'] - group['Revenue'].mean()) > 1.96*group['Revenue'].std()
    return group
Newdf2 = StateMonth.apply(s)
Newdf2

假设是一个非高斯分布 (如果你绘制出图形，看上去不像是一个正态分布)

# 原始的 df 拷贝一份
newdf = df.copy()
State = newdf.groupby('State')
newdf['Lower'] = State['Revenue'].transform( lambda x: x.quantile(q=.25) - (1.5*(x.quantile(q=.75)-x.quantile(q=.25))) )
newdf['Upper'] = State['Revenue'].transform( lambda x: x.quantile(q=.75) + (1.5*(x.quantile(q=.75)-x.quantile(q=.25))) )
newdf['Outlier'] = (newdf['Revenue'] < newdf['Lower']) | (newdf['Revenue'] > newdf['Upper']) 
newdf

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