Question

Python的新手。我在列表中有20个数字，每次都想要2个唯一值。因此，理想情况下，我最终有10行，并且每行都有两个以前从未使用过的唯一编号。我有这样的东西，但是它显示的数字不止一次。

numberList=["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20"]
myset = set(numberList)
rows = len(numberList)//2
i = 0
while i < rows:
    random_nums = random.sample(myset,2)

    print(random_nums)
    i += 1

OUTPUT: 
['13', '8']
['19', '8']
['11', '9']
['16', '7']
['1', '10']
['11', '20']
['16', '18']
['18', '2']
['20', '10']
['7', '4']

Answer 1

假设：

您以后不需要numberList
您需要的对数除以列表的长度而无需提醒

您可以使用这种非常简单有效的方法。从列表中抽样2个随机元素，并在列表不为空时将其删除。

import random

numberList = ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15", "16", "17", "18",
              "19", "20"]

while numberList:
    a, b = random.sample(numberList, 2)
    numberList.remove(a)
    numberList.remove(b)
    print(a, b)

示例输出：

如果稍后在代码中确实需要使用numberList，则下一个有效的方法是处理一组拾取的元素。您将需要更复杂的逻辑：

import random

numberList = ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15", "16", "17", "18",
              "19", "20"]

taken = set()

pairs = len(numberList) // 2

for _ in range(pairs):
    a, b = None, None
    while a is None or a in taken:
        a = random.choice(numberList)
    taken.add(a)
    while b is None or b in taken:
        b = random.choice(numberList)
    taken.add(b)
    print(a, b)

Answer 2

您可以从“ numberList”或更好的“ myset”中删除项目。

import random

numberList=["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20"]
myset = set(numberList)
rows = len(numberList)//2
i = 0
while i < rows:
    random_nums = random.sample(myset,2)

    myset.remove(random_nums[0])
    myset.remove(random_nums[1])

    print(random_nums)
    i += 1

Answer 3

您可能还有一个对象：

seen = set()

在while循环的主体上：

while i < rows:
    random_nums = set(random.sample(myset,2))
    if not random_nums.issubset(seen):
        print(random_nums)
        seen.update(random_nums)
        i += 1

Answer 4

如果我说对了，那么您就不想重复这些数字。就像以前出现过一个数字一样，它不应再出现。然后，您可以相应地删除myset的元素。像这样：

import random
numberList=["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20"]
myset = set(numberList)
rows = len(numberList)//2
i = 0
while i < rows:

    random_nums = random.sample(myset,2)
    myset.remove(random_nums[0])
    myset.remove(random_nums[1])
    print(random_nums)
    i += 1

Answer 5

这应该可以解决问题

from random import choice

numberList = 
["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17",
"18","19","20"]

for i in range(10):
    a = choice(numberList)
    numberList.pop(numberList.index(a))
    b = choice(numberList)
    numberList.pop(numberList.index(b))
    print([a, b])

Answer 6

随着列表的增加，采样和删除将比您需要做的工作更多，并且效率很低。而是使用np.random.shuffle

numberList= ["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20"]
np.random.shuffle(numberList)
for r in np.reshape(numberList, (10,2)):
   print(r)

输出：

['18' '20']
['17' '16']
['11' '1']
['5' '13']
['7' '14']
['19' '15']
['6' '8']
['9' '12']

Answer 7

您可以按照以下方式使用random.shuffle来完成这项工作：

import random
numberList=["1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20"]
random.shuffle(numberList)
rows = len(numberList)//2
for i in range(rows):
    random_nums = numberList[2*i:2*i+2]
    print(random_nums)

请注意，random.shuffle会改变列表的位置，即，如果您在代码末尾添加print(numberList)，则会打印20个数字，但顺序是随机的。

Answer 8

首先，如果您知道将始终使用整数，那么我认为将它们全部转换为整数会更容易，就像在下面的示例中使用列表推导那样。由于您恰好指定了20个数字和10对，因此我在脚本中对这些数字进行了硬编码，但是您可以轻松地设置变量。

import random


numberList = ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10",
              "11", "12", "13", "14", "15", "16", "17", "18", "19", "20"]

# If you know you always have numbers, convert them all to integers → easier
numberList = [int(x) for x in numberList]

myset = set()  # Will be used to add any number NOT seen before
numberList_pairs = []  # Will be used to build a list of unique pairs

while len(numberList_pairs) < 10:  # You need 10 pairs in that new list
    pair = []

    # While you don't have a unique pair, keep generating random numbers
    while len(pair) < 2:
        random_num = random.randint(1, 20)  # 1 to 20. Both ends are included

        # If the number hasn't been seen before
        if random_num not in myset:
            myset.add(random_num)
            pair.append(random_num)
    numberList_pairs.extend([pair])  # Add the pair as a list

for pair in numberList_pairs:
    print(pair)

这将是一个示例输出：

[20, 14]
[16, 2]
[13, 10]
[6, 12]
[9, 18]
[3, 19]
[17, 11]
[1, 5]
[8, 4]
[15, 7]

注意：如果您确实需要输出为字符串列表，则可以在最终循环中将其转换回字符串，如下所示：

for a, b in numberList_pairs:
    a, b = str(a), str(b)
    print([a, b])

您将获得与最初发布的输出相同的输出：

['11', '9']
['2', '15']
['7', '12']
['13', '3']
['19', '20']
['1', '14']
['16', '5']
['18', '8']
['10', '6']
['4', '17']

循环遍历列表并输出两个唯一数字

8 个答案: