Python random.sample与带有IF语句的random.choice相同

Question

I'm working on a random suggestion script. I came across random.sample in my search for how to make sure I don't have repeats in my generated list. Is it the same thing as if I were to use an if statement to test the resultant against the existing list? The number of outputs is set by the user, and the output should be random but not duplicate. Here is my code:

import random
def myRandom():
    myOutput = input("How many suggestions would you like?")
    outList = list()
    counter = 1
    myDict = {
    "The First":1988,
    "The Second:": 1992,
    "The Third": 1974,
    "The Fourth": 1935,
    "The Fifth":2012,
    "The Six":2001,
    "The Seventh": 1994,
    "The Eighth":2004,
    "The Ninth": 2010,
    "The Tenth": 2003}
    while counter <= myOutput:
        thePick = random.choice(myDict.keys())
        if thePick in outList:
            pass
        else:
            outList.append(thePick)
            counter = counter + 1
    print "These are your suggestions: {0}".format(outList)
myRandom()

I'm not receiving any duplicates in my output list, so is this what random.sample is supposed to do as well?

Answer 1

Yes, a single call to random.sample() will do the trick:

outList = random.sample(myDict.keys(), myOutput)

You can remove the following lines from your code:

outList = list()
counter = 1

as well as the entire while loop.

Answer 2

Yeah, you should use random.sample() . It will make your code cleaner and as a bonus you get a performance increase.

Performance issues with your loop solution:
a) It has to check in the output list before choosing any number
b) does rejection sampling, so the expected time will be higher as the input number approaches the length of the list.

def myRandom():
    myOutput = input("How many suggestions would you like?")
    myDict = {
    "The First":1988,
    "The Second:": 1992,
    "The Third": 1974,
    "The Fourth": 1935}
    outList = random.sample(myDict, myOutput)
    print "These are your suggestions: {0}".format(outList)