Question

我试图比较跌落高度与反弹高度，并在此处得到一些数据：

drop_heights = [0.2, 0.4, 0.6, 0.8, 1.0, 1.2, 1.4, 1.6, 1.7, 2.0]
rebound_heights = [0.16, 0.30, 0.46, 0.6, 0.74, 0.88, 1.02, 1.15, 1.34, 1.51]

我想从这些变量中选择5个随机数据点，所以我尝试了

smol_drop_heights = []
smol_rebound_heights = []

for each in range(0,5):
     smol_drop_heights.append(drop_heights[randint(0, 9)])
     smol_rebound_heights.append(rebound_heights[randint(0, 9)])
print(smol_drop_heights)
print(smol_rebound_heights)

当他们打印时，他们打印不同的数据集，有时甚至重复数据，我该如何解决？

[0.8, 1.6, 0.6, 0.2, 0.12]
[1.02, 1.15, 0.88, 0.88, 0.6]

这是示例输出，您可以看到重复的.88。

Answer 1

一种方法是：

drop_heights = [0.2, 0.4, 0.6, 0.8, 1.0, 1.2, 1.4, 1.6, 1.7, 2.0]
rebound_heights = [0.16, 0.30, 0.46, 0.6, 0.74, 0.88, 1.02, 1.15, 1.34, 1.51]
indices = [*range(len(drop_heights))]
from random import shuffle
shuffle(indices)
smol_drop_heights = []
smol_rebound_heights = []
for each in indices:
    smol_drop_heights.append(drop_heights[each])
    smol_rebound_heights.append(rebound_heights[each])
print(smol_drop_heights)
print(smol_rebound_heights)

输出：

[1.7, 0.8, 1.6, 1.2, 0.2, 0.4, 1.4, 2.0, 1.0, 0.6]
[1.34, 0.6, 1.15, 0.88, 0.16, 0.3, 1.02, 1.51, 0.74, 0.46]

或更短：

from random import sample

drop_heights = [0.2, 0.4, 0.6, 0.8, 1.0, 1.2, 1.4, 1.6, 1.7, 2.0]
rebound_heights = [0.16, 0.30, 0.46, 0.6, 0.74, 0.88, 1.02, 1.15, 1.34, 1.51]

paired = [*zip(drop_heights, rebound_heights)]
smol_drop_heights, smol_rebound_heights = zip(*sample(paired,5))

print(smol_drop_heights[:5])
print(smol_rebound_heights[:5])

Answer 2

一种避免重复并保持数据点成对并随机对的方式：

from random import random

drop_heights = [0.2, 0.4, 0.6, 0.8, 1.0, 1.2, 1.4, 1.6, 1.7, 2.0]
rebound_heights = [0.16, 0.30, 0.46, 0.6, 0.74, 0.88, 1.02, 1.15, 1.34, 1.51]

pairs = list(sorted(zip(drop_heights, rebound_heights), key=lambda _: random()))[:5]

smol_drop_heights = [d for d, _ in pairs]
smol_rebound_heights = [r for _, r in pairs]

Answer 3

这就是我要做的。

import random
import numpy as np
k=5
drop_heights = np.array([0.2, 0.4, 0.6, 0.8, 1.0, 1.2, 1.4, 1.6, 1.7, 2.0])
rebound_heights = np.array([0.16, 0.30, 0.46, 0.6, 0.74, 0.88, 1.02, 1.15, 1.34, 1.51])
idx = random.sample(range(len(drop_heights )), k)

print(drop_heights[idx])
print(rebound_heights [idx])

Answer 4

您的问题是，当您致电randint时，每次都会给它一个不同的随机数。为了解决这个问题，您需要在每次代码循环时将索引变量保存为一个随机数，以便每次添加相同的随机变量。

for each in range(0, 4):  
    index = randint(0, 9)  
    smol_drop_heights.append(drop_heights[index])  
    smol_rebound_heights.append(rebound_heights[index])  
print(smol_drop_heights)  
print(smol_rebound_heights)

要解决有关重复的问题，只需检查列表中是否已包含要添加的变量，就可以对其中一个变量进行操作，因为它们都没有重复，并且由于可能存在重复，因此循环是不够的，因此您将不得不重复执行，直到列表已满。所以我的最终解决方案是：

while True:  
    index = randint(0, 9)  
    if drop_heights[index] not in smol_drop_heights:  
        smol_drop_heights.append(drop_heights[index])  
        smol_rebound_heights.append(rebound_heights[index])  
    if len(smol_drop_heights) == 4:  
        break  
print(smol_drop_heights)  
print(smol_rebound_heights)

由于您可能想按顺序排列这些值，因此可以这样做：


smol_drop_heights = []  
smol_rebound_heights = []  

while True:  
    index = randint(0, 9)  
    if drop_heights[index] not in smol_drop_heights:  
        smol_drop_heights.append(drop_heights[index])  
        smol_rebound_heights.append(rebound_heights[index])  
    if len(smol_drop_heights) == 4:  
        smol_drop_heights.sort()  
        smol_rebound_heights.sort()  
        break  
print(smol_drop_heights)  
print(smol_rebound_heights)

Answer 5

您可以尝试改组，然后使用原始项目的index，例如

>>> drop_heights = [0.2, 0.4, 0.6, 0.8, 1.0, 1.2, 1.4, 1.6, 1.7, 2.0]
>>> rebound_heights = [0.16, 0.30, 0.46, 0.6, 0.74, 0.88, 1.02, 1.15, 1.34, 1.51]
>>> 
>>> import random 
>>> d = drop_heights[:] # keep a copy to get index for making pairs later
>>> random.shuffle(drop_heights)
>>> # iterate through the new list and get the index of the item 
>>> # from the original lists
>>> nd, nr = zip(*[(x,rebound_heights[d.index(x)]) for x in drop_heights])
>>> nd[:5]
(1.4, 0.6, 1.7, 0.2, 1.0)
>>> nr[:5]
(1.02, 0.46, 1.34, 0.16, 0.74)

或仅使用operator.itemgetter和random.sample，

>>> drop_heights = [0.2, 0.4, 0.6, 0.8, 1.0, 1.2, 1.4, 1.6, 1.7, 2.0]
>>> rebound_heights = [0.16, 0.30, 0.46, 0.6, 0.74, 0.88, 1.02, 1.15, 1.34, 1.51]
>>> 
>>> import random, operator
>>> indexes = random.sample(range(len(drop_heights)), 5)
>>> indexes
[5, 0, 4, 7, 3]
>>> f = operator.itemgetter(*indexes)
>>> f(drop_heights)
(1.2, 0.2, 1.0, 1.6, 0.8)
>>> f(rebound_heights)
(0.88, 0.16, 0.74, 1.15, 0.6)

Answer 6

好吧，您想做两件事，将您的清单配对。惯用的方法是使用zip：

drop_heights = [0.2, 0.4, 0.6, 0.8, 1.0, 1.2, 1.4, 1.6, 1.7, 2.0]
rebound_heights = [0.16, 0.30, 0.46, 0.6, 0.74, 0.88, 1.02, 1.15, 1.34, 1.51]
paired = list(zip(drop_heights, rebound_heights))

然后，您要从中采样五对。因此，请使用random.sample：

sampled = random.sample(paired, 5)

最后，如果您需要将它们放在单独的列表中（可能不需要，但是如果必须），可以像下面这样解压：

smol_drop_heights, smol_rebound_heights = zip(*sampled)

您实际上可以一次全部执行此操作，尽管它可能会变得有点难以理解：

smol_drop_heights, smol_rebound_heights = zip(*random.sample(list(zip(drop_heights, rebound_heights)), 5))

Answer 7

您如何尝试使用集合，然后使用.intersect方法获取常见的集合呢？

如何从配对列表中获取python上的随机元素？

7 个答案: