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.php.cn/ -"
[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)
])
# 输出结果
for k,v in result.items():
print(k)
print(v)
print('='*80)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：
Splitting the data into groups based on some criteria
Applying a function to each group independently
Combining the results into a data structure
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 -0.136473
6 foo one 0.007207 -0.561757
7 foo three 1.928123 -1.623033group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 3.146492 -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 -0.136473 foo one -1.195665 -0.616981 three 1.928123 -1.623033 two 2.414034 1.600434
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