寻找自然分钟内访问量大于100的ip

example.py

# https://www.v2ex.com/t/565534#reply8

import random
import datetime


def genlog():
    ips = []
    for _ in range(100):
        ips.append(".".join([
            str(random.randint(0, 255)),
            str(random.randint(0, 255)),
            str(random.randint(0, 255)),
            str(random.randint(0, 255)),
        ]))
    f = open("sample.log", "w")
    t = datetime.datetime(2019, 5, 19)
    for i in range(1440):
        for i in ips:
            if random.randint(0, 100) < 95:
                continue
            count = random.randint(0, 110)
            while count > 0:
                f.write("{}\t{}\n".format(t.strftime("%Y-%m-%d %H:%M:%S"), i))
                count -= 1
        t = t+datetime.timedelta(minutes=1)
    f.close()


def main():
    current = None
    container = {}
    exceed = set()
    with open("sample.log", "r") as f:
        for i in f:
            ds, ip = i.strip().split("\t")
            minute = ds[:16]
            if minute != current:
                container = {}
                current = minute
            try:
                container[ip] += 1
            except:
                container[ip] = 1
            if container[ip] > 100:
                exceed.add(ip)
    return exceed


if __name__ == "__main__":
    genlog()
    print("日志生成完成")
    print(main())