使用python绘制人人网好友关系图示例

代码依赖：networkx matplotlib

代码如下:
 #! /bin/env python
# -*- coding: utf-8 -*-

import urllib
import urllib2
import cookielib
import re
import cPickle as p
import networkx as nx
import matplotlib.pyplot as plt

__author__ = """Reverland (lhtlyy@gmail.com)"""

# Control parameters,EDIT it here
## Login
username = 'None'
password = 'None'
## Control Graphs, Edit for better graphs as you need
label_flag = True # Whether shows labels.NOTE: configure your matplotlibrc for Chinese characters.
remove_isolated = True # Whether remove isolated nodes(less than iso_level connects)
different_size = True # Nodes for different size, bigger means more shared friends
iso_level = 10
node_size = 40 # Default node size

 
def login(username, password):
    """log in and return uid"""
    logpage = "http://www.renren.com/ajaxLogin/login"
    data = {'email': username, 'password': password}
    login_data = urllib.urlencode(data)
    cj = cookielib.CookieJar()
    opener = urllib2.build_opener(urllib2.HTTPCookieProcessor(cj))
    urllib2.install_opener(opener)
    res = opener.open(logpage, login_data)
    print "Login now ..."
    html = res.read()
    #print html

    # Get uid
    print "Getting user id of you now"
    res = urllib2.urlopen("http://www.renren.com/home")
    html = res.read()
    # print html
    uid = re.search("'ruid':'(//d+)'", html).group(1)
    # print uid
    print "Login and got uid successfully"
    return uid

 
def getfriends(uid):
    """Get the uid's friends and return the dict with uid as key,name as value."""
    print "Get %s 's friend list" % str(uid)
    pagenum = 0
    dict1 = {}
    while True:
        targetpage = "http://friend.renren.com/GetFriendList.do?curpage=" + str(pagenum) + "&id=" + str(uid)
        res = urllib2.urlopen(targetpage)
        html = res.read()

        pattern = '<a href="http://www//.renren//.com/profile//.do//?id=(//d+)"><img src="[//S]*" alt="[//S]*[//s]//((.*)//)" />'

        m = re.findall(pattern, html)
        #print len(m)
        if len(m) == 0:
            break
        for i in range(0, len(m)):
            no = m[i][0]