Comparison of two python lottery codes

This article has made two very simple codes to show that using different ideas can achieve the same effect, but the execution efficiency and code readability are quite different.

Require:

A company has 300 employees, and the annual meeting requires a lottery. The awards are as follows:
First prize: 3
second prizes: 6
third prizes: 30
winners. Last prize draw. Each employee can only win the prize once and cannot repeat it.

Code one:

import random
person = 300
alllist = []
for i in range(1,person + 1) :
    alllist.append('per'+str(i))
jiang = ['三等奖','二等奖','一等奖']
no = 0
for i in [30,6,3] :
    print(f'正在抽取{jiang[no]}。')
    for b in range(i) :
        while True :
            num = 'per' + str(random.randrange(person))  # 随机生成中奖员工编号
            
            if num in alllist :  #如果随机生成的员工未被删除
                break
            else :
                continue
        print(f'第{b+1}名{jiang[no]}是{num}')
        alllist.remove(num)  # 删除已经中奖的员工
    no += 1

code two

import random
person = 300
alllist = []
for i in range(1,person + 1) :
    alllist.append('per'+str(i))
jiang = ['三等奖','二等奖','一等奖']
getpernum = [30,6,3]
no = 0

for i in getpernum :
    getper = random.sample(alllist, i)  #直接生成i个中奖员工的列表
    print(f'获得{jiang[no]}的有：',getper)
    for num in getper :
        alllist.remove(num)  #循环删除已经中奖的员工
    print(f'还剩下{len(alllist)}人')  #此句为调试用。
    person -= getpernum[no]
    no += 1

The advantage of code 2 is: use the sample() method of the random module to directly generate a list of winning employees in batches without generating the winning employees one by one, which improves the readability and efficiency of the code.