前言

最近工作工作中遇到一个需求，是要根据CDN日志过滤一些数据，例如流量、状态码统计，TOP IP、URL、UA、Referer等。以前都是用 bash shell 实现的，但是当日志量较大，日志文件数G、行数达数千万亿级时，通过 shell 处理有些力不从心，处理时间过长。于是研究了下Python pandas这个数据处理库的使用。一千万行日志，处理完成在40s左右。

代码

#!/usr/bin/python
# -*- coding: utf-8 -*-
# sudo pip install pandas
author = 'Loya Chen'
import sys
import pandas as pd
from collections import OrderedDict
"""
Description: This script is used to analyse qiniu cdn log.
================================================================================
日志格式
IP - ResponseTime [time +0800] "Method URL HTTP/1.1" code size "referer" "UA"
================================================================================
日志示例
 [0] [1][2] [3] [4] [5]
101.226.66.179 - 68 [16/Nov/2016:04:36:40 +0800] "GET http://www.qn.com/1.jpg -" 
[6] [7] [8] [9]
200 502 "-" "Mozilla/5.0 (compatible; MSIE 9.0; Windows NT 6.1; Trident/5.0)"
================================================================================
"""
if len(sys.argv) != 2:
 print('Usage:', sys.argv[0], 'file_of_log')
 exit() 
else:
 log_file = sys.argv[1] 
# 需统计字段对应的日志位置 
ip = 0
url = 5
status_code = 6
size = 7
referer = 8
ua = 9
# 将日志读入DataFrame
reader = pd.read_table(log_file, sep=' ', names=[i for i in range(10)], iterator=True)
loop = True
chunkSize = 10000000
chunks = []
while loop:
 try:
 chunk = reader.get_chunk(chunkSize)
 chunks.append(chunk)
 except StopIteration:
 #Iteration is stopped.
 loop = False
df = pd.concat(chunks, ignore_index=True)
byte_sum = df[size].sum() #流量统计
top_status_code = pd.DataFrame(df[6].value_counts()) #状态码统计
top_ip = df[ip].value_counts().head(10) #TOP IP
top_referer = df[referer].value_counts().head(10) #TOP Referer
top_ua = df[ua].value_counts().head(10) #TOP User-Agent
top_status_code['persent'] = pd.DataFrame(top_status_code/top_status_code.sum()*100)
top_url = df[url].value_counts().head(10) #TOP URL
top_url_byte = df[[url,size]].groupby(url).sum().apply(lambda x:x.astype(float)/1024/1024) 
 .round(decimals = 3).sort_values(by=[size], ascending=False)[size].head(10) #请求流量最大的URL
top_ip_byte = df[[ip,size]].groupby(ip).sum().apply(lambda x:x.astype(float)/1024/1024) 
 .round(decimals = 3).sort_values(by=[size], ascending=False)[size].head(10) #请求流量最多的IP
# 将结果有序存入字典
result = OrderedDict([("流量总计[单位:GB]:" , byte_sum/1024/1024/1024),
 ("状态码统计[次数|百分比]:" , top_status_code),
 ("IP TOP 10:" , top_ip),
 ("Referer TOP 10:" , top_referer),
 ("UA TOP 10:" , top_ua),
 ("URL TOP 10:" , top_url),
 ("请求流量最大的URL TOP 10[单位:MB]:" , top_url_byte), 
 ("请求流量最大的IP TOP 10[单位:MB]:" , top_ip_byte)
])
# 

pandas 学习笔记

Pandas 中有两种基本的数据结构，Series 和 Dataframe。 Series 是一种类似于一维数组的对象，由一组数据和索引组成。 Dataframe 是一个表格型的数据结构，既有行索引也有列索引。

from pandas import Series, DataFrame
import pandas as pd

Series

In [1]: obj = Series([4, 7, -5, 3])
In [2]: obj
Out[2]: 
0 4
1 7
2 -5
3 3

Series的字符串表现形式为：索引在左边，值在右边。没有指定索引时，会自动创建一个0到N-1（N为数据的长度）的整数型索引。可以通过Series的values和index属性获取其数组表示形式和索引对象:

In [3]: obj.values
Out[3]: array([ 4, 7, -5, 3])
In [4]: obj.index
Out[4]: RangeIndex(start=0, stop=4, step=1)

通常创建Series时会指定索引:

In [5]: obj2 = Series([4, 7, -5, 3], index=['d', 'b', 'a', 'c'])
In [6]: obj2
Out[6]: 
d 4
b 7
a -5
c 3

通过索引获取Series中的单个或一组值：

In [7]: obj2['a']
Out[7]: -5
In [8]: obj2[['c','d']]
Out[8]: 
c 3
d 4

排序

In [9]: obj2.sort_index()
Out[9]: 
a -5
b 7
c 3
d 4
In [10]: obj2.sort_values()
Out[10]: 
a -5
c 3
d 4
b 7

筛选运算

In [11]: obj2[obj2 > 0]
Out[11]: 
d 4
b 7
c 3
In [12]: obj2 * 2
Out[12]: 
d 8
b 14
a -10
c 6

成员

In [13]: 'b' in obj2
Out[13]: True
In [14]: 'e' in obj2
Out[14]: False

通过字典创建Series

In [15]: sdata = {'Shanghai':35000, 'Beijing':40000, 'Nanjing':26000, 'Hangzhou':30000}
In [16]: obj3 = Series(sdata)
In [17]: obj3
Out[17]: 
Beijing 40000
Hangzhou 30000
Nanjing 26000
Shanghai 35000

如果只传入一个字典，则结果Series中的索引就是原字典的键（有序排列）

In [18]: states = ['Beijing', 'Hangzhou', 'Shanghai', 'Suzhou']
In [19]: obj4 = Series(sdata, index=states)
In [20]: obj4
Out[20]: 
Beijing 40000.0
Hangzhou 30000.0
Shanghai 35000.0
Suzhou NaN

当指定index时，sdata中跟states索引相匹配的3个值会被找出并放到响应的位置上，但由于‘Suzhou'所对应的sdata值找不到，所以其结果为NaN(not a number),pandas中用于表示缺失或NA值

pandas的isnull和notnull函数可以用于检测缺失数据:

In [21]: pd.isnull(obj4)
Out[21]: 
Beijing False
Hangzhou False
Shanghai False
Suzhou True
In [22]: pd.notnull(obj4)
Out[22]: 
Beijing True
Hangzhou True
Shanghai True
Suzhou False

Series也有类似的实例方法

In [23]: obj4.isnull()
Out[23]: 
Beijing False
Hangzhou False
Shanghai False
Suzhou True

Series的一个重要功能是，在数据运算中，自动对齐不同索引的数据

In [24]: obj3
Out[24]: 
Beijing 40000
Hangzhou 30000
Nanjing 26000
Shanghai 35000
In [25]: obj4
Out[25]: 
Beijing 40000.0
Hangzhou 30000.0
Shanghai 35000.0
Suzhou NaN
In [26]: obj3 + obj4
Out[26]: 
Beijing 80000.0
Hangzhou 60000.0
Nanjing NaN
Shanghai 70000.0
Suzhou NaN

Series的索引可以通过复制的方式就地修改

In [27]: obj.index = ['Bob', 'Steve', 'Jeff', 'Ryan']
In [28]: obj
Out[28]: 
Bob 4
Steve 7
Jeff -5
Ryan 3

DataFrame

pandas读取文件

In [29]: df = pd.read_table('pandas_test.txt',sep=' ', names=['name', 'age'])
In [30]: df
Out[30]: 
 name age
0 Bob 26
1 Loya 22
2 Denny 20
3 Mars 25

DataFrame列选取

df[name]

In [31]: df['name']
Out[31]: 
0 Bob
1 Loya
2 Denny
3 Mars
Name: name, dtype: object

DataFrame行选取

df.iloc[0,:] #第一个参数是第几行，第二个参数是列。这里指第0行全部列
df.iloc[:,0] #全部行，第0列

In [32]: df.iloc[0,:]
Out[32]: 
name Bob
age 26
Name: 0, dtype: object
In [33]: df.iloc[:,0]
Out[33]: 
0 Bob
1 Loya
2 Denny
3 Mars
Name: name, dtype: object

获取一个元素，可以通过iloc，更快的方式是iat

In [34]: df.iloc[1,1]
Out[34]: 22
In [35]: df.iat[1,1]
Out[35]: 22

DataFrame块选取

In [36]: df.loc[1:2,['name','age']]
Out[36]: 
 name age
1 Loya 22
2 Denny 20

根据条件过滤行

在方括号中加入判断条件来过滤行，条件必需返回 True 或者 False

In [37]: df[(df.index >= 1) & (df.index <= 3)]
Out[37]: 
 name age city
1 Loya 22 Shanghai
2 Denny 20 Hangzhou
3 Mars 25 Nanjing
In [38]: df[df['age'] > 22]
Out[38]: 
 name age city
0 Bob 26 Beijing
3 Mars 25 Nanjing

增加列

In [39]: df['city'] = ['Beijing', 'Shanghai', 'Hangzhou', 'Nanjing']
In [40]: df
Out[40]: 
 name age city
0 Bob 26 Beijing
1 Loya 22 Shanghai
2 Denny 20 Hangzhou
3 Mars 25 Nanjing

排序

按指定列排序

In [41]: df.sort_values(by='age')
Out[41]: 
 name age city
2 Denny 20 Hangzhou
1 Loya 22 Shanghai
3 Mars 25 Nanjing
0 Bob 26 Beijing

# 引入numpy 构建 DataFrame
import numpy as np

In [42]: df = pd.DataFrame(np.arange(8).reshape((2, 4)), index=['three', 'one'], columns=['d', 'a', 'b', 'c'])
In [43]: df
Out[43]: 
 d a b c
three 0 1 2 3
one 4 5 6 7

# 以索引排序
In [44]: df.sort_index()
Out[44]: 
 d a b c
one 4 5 6 7
three 0 1 2 3
In [45]: df.sort_index(axis=1)
Out[45]: 
 a b c d
three 1 2 3 0
one 5 6 7 4
# 降序
In [46]: df.sort_index(axis=1, ascending=False)
Out[46]: 
 d c b a
three 0 3 2 1
one 4 7 6 5

查看

# 查看表头5行 
df.head(5)
# 查看表末5行
df.tail(5) 
# 查看列的名字
In [47]: df.columns
Out[47]: Index(['name', 'age', 'city'], dtype='object')
# 查看表格当前的值
In [48]: df.values
Out[48]: 
array([['Bob', 26, 'Beijing'],
 ['Loya', 22, 'Shanghai'],
 ['Denny', 20, 'Hangzhou'],
 ['Mars', 25, 'Nanjing']], dtype=object)

转置

df.T
Out[49]: 
 0 1 2 3
name Bob Loya Denny Mars
age 26 22 20 25
city Beijing Shanghai Hangzhou Nanjing

使用isin

In [50]: df2 = df.copy()
In [51]: df2[df2['city'].isin(['Shanghai','Nanjing'])]
Out[52]: 
 name age city
1 Loya 22 Shanghai
3 Mars 25 Nanjing

运算操作：

In [53]: df = pd.DataFrame([[1.4, np.nan], [7.1, -4.5], [np.nan, np.nan], [0.75, -1.3]], 
 ...: index=['a', 'b', 'c', 'd'], columns=['one', 'two'])
In [54]: df
Out[54]: 
 one two
a 1.40 NaN
b 7.10 -4.5
c NaN NaN
d 0.75 -1.3

#按列求和
In [55]: df.sum()
Out[55]: 
one 9.25
two -5.80
# 按行求和
In [56]: df.sum(axis=1)
Out[56]: 
a 1.40
b 2.60
c NaN
d -0.55

group

group 指的如下几步：

See the Grouping section

In [57]: df = pd.DataFrame({'A' : ['foo', 'bar', 'foo', 'bar',
 ....: 'foo', 'bar', 'foo', 'foo'],
 ....: 'B' : ['one', 'one', 'two', 'three',
 ....: 'two', 'two', 'one', 'three'],
 ....: 'C' : np.random.randn(8),
 ....: 'D' : np.random.randn(8)})
 ....: 
In [58]: df
Out[58]: 
 A B C D
0 foo one -1.202872 -0.055224
1 bar one -1.814470 2.395985
2 foo two 1.018601 1.552825
3 bar three -0.595447 0.166599
4 foo two 1.395433 0.047609
5 bar two -0.392670 -0.1373
6 foo one 0.007207 -0.561757
7 foo three 1.928123 -1.623033

group一下，然后应用sum函数

In [59]: df.groupby('A').sum()
Out[59]: 
 C D
A 
bar -2.802588 2.42611
foo 3.1492 -0.63958
In [60]: df.groupby(['A','B']).sum()
Out[60]: 
 C D
A B 
bar one -1.814470 2.395985
 three -0.595447 0.166599
 two -0.392670 -0.1373
foo one -1.195665 -0.616981
 three 1.928123 -1.623033
 two 2.414034 1.600434

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