In a recent experiment, the data needed to be randomly split into two parts according to a certain proportion. The core of this problem is actually the problem of generating non-repeating random numbers. The first thing I thought of was the recursive method, and then I discovered that Python actually already provides a function for this method, which can be used directly. The specific code is as follows:
#生成某区间内不重复的N个随机数的方法 import random; #1、利用递归生成 resultList=[];#用于存放结果的List A=1; #最小随机数 B=10 #最大随机数 COUNT=10 #生成随机数的递归数学,参数counter表示当前准备要生成的第几个有效随机数 def generateRand(counter): tempInt=random.randint(A,B); # 生成一个范围内的临时随机数, if(counter<=COUNT): # 先看随机数的总个数是不是够了,如果不够 if(tempInt not in resultList): # 再检查当前已经生成的临时随机数是不是已经存在,如果不存在 resultList.append(tempInt); #则将其追加到结果List中 counter+=1;# 然后将表示有效结果的个数加1. 请注意这里,如果临时随机数已经存在,则此if不成立,那么将直接执行16行,counter不用再加1 generateRand(counter); # 不管上面的if是否成立,都要递归。如果上面的临时随机数有效,则这里的conter会加1,如果上面的临时随机数已经存在了,则需要重新再生成一次随机数,counter不能变化 generateRand(1);#调用递归函数,并给当前要生成的有效随机数的个序号置为1,因为要从第一个开始嘛 print(resultList)# 打印结果 #2、利用Python中的randomw.sample()函数实现 resultList=random.sample(range(A,B+1),COUNT); # sample(x,y)函数的作用是从序列x中,随机选择y个不重复的元素。上面的方法写了那么多,其实Python一句话就完成了。 print(resultList)# 打印结果
Result:
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