如何在Python中生成带有重复数字的随机列表 [英] How do I generate a random list in Python with duplicates numbers
问题描述
所以几天前我才开始用Python编程.现在,我试图制作一个生成随机列表的程序,然后选择重复元素.问题是,我的列表中没有重复的数字.
So I just started programming in Python a few days ago. And now, im trying to make a program that generates a random list, and then, choose the duplicates elements. The problem is, I dont have duplicate numbers in my list.
这是我的代码:
import random
def generar_listas (numeros, rango):
lista = [random.sample(range(numeros), rango)]
print("\n", lista, sep="")
return
def texto_1 ():
texto = "Debes de establecer unos parámetros para generar dos listas aleatorias"
print(texto)
return
texto_1()
generar_listas(int(input("\nNumero maximo: ")), int(input("Longitud: ")))
例如,我为random.sample选择20和20,它将为我生成一个从0到20的列表,但位置随机.我想要一个带有随机数且重复的列表.
And for example, I choose 20 and 20 for random.sample, it generates me a list from 0 to 20 but in random position. I want a list with random numbers and duplicated.
推荐答案
您想要的非常简单.您要生成一个随机数字列表,其中包含一些重复项.如果您使用诸如numpy之类的方法,则这样做很容易.
What you want is fairly simple. You want to generate a random list of numbers that contain some duplicates. The way to do that is easy if you use something like numpy.
- 生成一个0到10的列表(范围).
- 从该列表中随机抽样(替换).
赞:
import numpy as np
print np.random.choice(10, 10, replace=True)
结果:
[5 4 8 7 0 8 7 3 0 0]
如果要排序列表,只需使用内置函数"sorted(list)"
If you want the list to be ordered just use the builtin function "sorted(list)"
sorted([5 4 8 7 0 8 7 3 0 0])
[0 0 0 3 4 5 7 7 8 8]
如果您不想使用numpy,则可以使用以下命令:
If you don't want to use numpy you can use the following:
print [random.choice(range(10)) for i in range(10)]
[7, 3, 7, 4, 8, 0, 4, 0, 3, 7]
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