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from itertools import groupby
NAME_IP_MAPPING = {
    '202.108.11.103':'百度蜘蛛',
    '220.181.32.137': '百度蜘蛛',
}
spiders = [
    {'ip':'202.108.11.103','count':123}, 
    {'ip':'220.181.32.137','count':345}
]
# 先用ip通過映射得到名字，再根據(jù)名字將spiders里的item分組，之后各自求和存入新的dict中。
{k: sum(s['count'] for s in g)
    for k, g in groupby(spiders, lambda s:NAME_IP_MAPPING.get(s['ip']))}
# output: {'百度蜘蛛': 468}

You can try to build a large dictionary with the dictionary as the key and the crawler name as the value;

ip_map = {
    '202.108.11.103': 'baidu-spider',
    '220'.181.32.137: 'baidu-spider',
    '192.168.1.1': 'other'
    ....
}
sum = {}
for ip in source_ip:
    print ip
    sum[ip_mapping.get(ip, 'other')] = sum.get(ip, 0) + source_ip[ip]
print sum

Pivot table using pandas

How tiring it is!
Why not create a separate table for this IP group, named IPGroup (id, ip, groupname)

After that, it can be done with just one SQL, how easy it is (let the poster use IPStastics)

SELECT b.groupName, SUM(a.count)
FROM IPStastics a 
  INNER JOIN IPGroup b
  ON a.ip = b.ip
GROUP BY b.groupName