如何高效生成特定范围内的唯一随机非零整数？

huangapple
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英文:

How to efficiently generate unique random non-zero integers from specific spaces in a range?

问题

以下是您要翻译的代码部分：

  1. import random
  3. my_list = []
  4. c = 0
  5. while c < 100:
  6.     if c < 5:
  7.         # Both values are unique.
  8.         first_num, second_num = random.sample(range(-5, 6), 2)
  9.         # Exclusion of 0.
  10.         while first_num == 0 or second_num == 0:
  11.             first_num, second_num = random.sample(range(-5, 6), 2)
  12.         c += 1
  13.     elif 5 <= c < 80:
  14.         first_num, second_num = random.sample(range(-100, 101), 2)
  15.         while first_num == 0 or second_num == 0:
  16.             first_num, second_num = random.sample(range(-100, 101), 2)
  17.         c += 1
  18.     else:
  19.         first_num, second_num = random.sample(range(-150, 151), 2)
  20.         while first_num == 0 or second_num == 0:
  21.             first_num, second_num = random.sample(range(-150, 151), 2)
  22.         c += 1
  24.     random_nums = (first_num, second_num)
  25.     if random_nums not in my_list:
  26.         my_list.append(random_nums)
  27.     else:
  28.         c -= 1

如果您有任何其他需要翻译的内容，请随时提出。

英文:

I want to generate 100 pairs of unique random non-zero integers from the range (-150, 151). But I want my code to generate from specific areas in (-150, 151). I have coded it as follows:

  1. import random
  3. my_list = []
  4. c = 0
  5. while c &lt; 100:
  6.     if c &lt; 5:
  7.         # Both values are unique.
  8.         first_num, second_num = random.sample(range(-5, 6), 2)
  9.         # Exclusion of 0.
  10.         while first_num == 0 or second_num == 0:
  11.             first_num, second_num = random.sample(range(-5, 6), 2)
  12.         c += 1
  13.     elif 5 &lt;= c &lt; 80:
  14.         first_num, second_num = random.sample(range(-100, 101), 2)
  15.         while first_num == 0 or second_num == 0:
  16.             first_num, second_num = random.sample(range(-100, 101), 2)
  17.         c += 1
  18.     else:
  19.         first_num, second_num = random.sample(range(-150, 151), 2)
  20.         while first_num == 0 or second_num == 0:
  21.             first_num, second_num = random.sample(range(-150, 151), 2)
  22.         c += 1
  24.     random_nums = (first_num, second_num)
  25.     if random_nums not in my_list:
  26.         my_list.append(random_nums)
  27.     else:
  28.         c -= 1

Is there a more elegant or efficient way alternative to what I have coded above?

UPDATE

After discussing with MatBailie in the comments, I am updating the requirements:

If the whole selected range is (-150 to 151), then in the final my_list:

  • (5,6), (6,5) are allowed but (5,6), (5,6) are not.
  • If (-1, 1) is generated once from range (-5 to 6) then it should not be generated again from the bigger ranges such as (-100 to 101) or (-150 to 151)
  • It is also good to have pairs such as (-1, 140) or (140, -1).

UPDATE

Benchmarking MatBailie, Alain T. and Samwise's answers on perfpy with 100 pairs, I got the following result:

如何高效生成特定范围内的唯一随机非零整数？

The same benchmark with generating 1000 pairs and the same ratio of areas, I got this:

如何高效生成特定范围内的唯一随机非零整数？

答案1

得分: 2

以下是代码的中文翻译：

  1. # 可以构建一个包含100个范围的列表，以满足所需大小的分组。这将使剩下的逻辑更加通用和简单：
  3. import random
  5. ranges = [range(-5,6)]*5 + [range(-100,101)]*75 + [range(-150,151)]*20
  6. pairs  = []
  7. for R in ranges:
  8.     a = 0
  9.     while not a or not b or (a,b) in pairs:
  10.         a,b = random.sample(R,2)  # random.choices(R,k=2)
  11.     pairs.append((a,b))
  13. # 输出:
  15. print(pairs)
  17. # [(-3, 2), (-3, -5), (3, 1), (-4, -3), (2, 4), (-73, -9), (16, -56),
  18. #  (12, -81), (5, -50), (99, -6), (35, 71), (30, -75), (98, 25), (55, -58), 
  19. #  (73, -24), (-65, 4), (75, -96), (-6, -90), (-80, 22), (-93, -39), 
  20. #  (-69, -48), (-30, 25), (85, -11), (37, 60), (91, 96), (98, 100), 
  21. #  (-100, -54), (58, 20), (-14, -95), (-76, -12), (-5, -84), (70, -53), 
  22. #  (91, -66), (-61, 2), (4, -42), (-15, -70), (-52, -6), (-5, 93), (26, 76),
  23. #  (-8, -79), (63, -7), (-23, -27), (56, 13), (46, 27), (80, 94), (-94, -66), 
  24. #  (-46, -19), (-4, -87), (-92, -48), (24, 32), (10, -89), (-50, -96),   
  25. #  (-5, -85), (-32, 69), (-60, 69), (87, -81), (-100, -91), (87, 37), 
  26. #  (-60, 45), (-70, -84), (-98, 80), (-88, 57), (-67, -44), (-72, 4), 
  27. #  (-76, 5), (-79, -16), (-9, 80), (76, -41), (15, 77), (-47, -1), (58, 12), 
  28. #  (66, -2), (21, 16), (-24, -12), (54, -69), (58, -73), (61, 68), (-89, -37), 
  29. #  (-85, 68), (-7, -21), (-86, -31), (48, 136), (137, 86), (-61, -32), 
  30. #  (104, -144), (-24, -103), (80, 135), (-2, 39), (49, -21), (15, -103), 
  31. #  (-14, -145), (-48, -61), (142, -90), (93, -132), (-68, -111), (-80, 2), 
  32. #  (6, 45), (-89, 75), (-146, -76), (-94, -24)]
  34. # 你也可以只使用一个列表来实现，通过初始化包含范围的结果列表，然后用生成的对替换范围：
  36. pairs = [range(-5,6)]*5 + [range(-100,101)]*75 + [range(-150,151)]*20
  37. for i,R in enumerate(pairs):
  38.     pairs[i] = (0,0)
  39.     while not all(pairs[i]) or pairs.index(pairs[i])&lt;i:
  40.         pairs[i] = random.sample(R,2) # random.choices(R,k=2)

注意：请确保在代码中不要使用HTML实体"lt"，而应使用小于号"<"。

英文:

You could build a list of the 100 ranges repeating the same ones in groups of the required size. This will make the rest of the logic more generalized and simpler:

  1. import random
  2.    
  3. ranges = [range(-5,6)]*5 + [range(-100,101)]*75 + [range(-150,151)]*20
  4. pairs  = []
  5. for R in ranges:
  6.     a = 0
  7.     while not a or not b or (a,b) in pairs:
  8.         a,b = random.sample(R,2)  # random.choices(R,k=2)
  9.     pairs.append((a,b))

output:

  1. print(pairs)
  3. [(-3, 2), (-3, -5), (3, 1), (-4, -3), (2, 4), (-73, -9), (16, -56),
  4.  (12, -81), (5, -50), (99, -6), (35, 71), (30, -75), (98, 25), (55, -58), 
  5.  (73, -24), (-65, 4), (75, -96), (-6, -90), (-80, 22), (-93, -39), 
  6.  (-69, -48), (-30, 25), (85, -11), (37, 60), (91, 96), (98, 100), 
  7.  (-100, -54), (58, 20), (-14, -95), (-76, -12), (-5, -84), (70, -53), 
  8.  (91, -66), (-61, 2), (4, -42), (-15, -70), (-52, -6), (-5, 93), (26, 76),
  9.  (-8, -79), (63, -7), (-23, -27), (56, 13), (46, 27), (80, 94), (-94, -66), 
  10.  (-46, -19), (-4, -87), (-92, -48), (24, 32), (10, -89), (-50, -96),   
  11.  (-5, -85), (-32, 69), (-60, 69), (87, -81), (-100, -91), (87, 37), 
  12.  (-60, 45), (-70, -84), (-98, 80), (-88, 57), (-67, -44), (-72, 4), 
  13.  (-76, 5), (-79, -16), (-9, 80), (76, -41), (15, 77), (-47, -1), (58, 12), 
  14.  (66, -2), (21, 16), (-24, -12), (54, -69), (58, -73), (61, 68), (-89, -37), 
  15.  (-85, 68), (-7, -21), (-86, -31), (48, 136), (137, 86), (-61, -32), 
  16.  (104, -144), (-24, -103), (80, 135), (-2, 39), (49, -21), (15, -103), 
  17.  (-14, -145), (-48, -61), (142, -90), (93, -132), (-68, -111), (-80, 2), 
  18.  (6, 45), (-89, 75), (-146, -76), (-94, -24)]

You could also do this using only one list by initializing the resulting list with ranges that you replace with the generated pairs:

  1. pairs = [range(-5,6)]*5 + [range(-100,101)]*75 + [range(-150,151)]*20
  2. for i,R in enumerate(pairs):
  3.     pairs[i] = (0,0)
  4.     while not all(pairs[i]) or pairs.index(pairs[i])&lt;i:
  5.         pairs[i] = random.sample(R,2) # random.choices(R,k=2)

答案2

得分: 2

这是你要翻译的内容：

"已经过了午夜，但我将尽力解释...

基本原则是我要计算在给定范围内（不包括零）有多少不同的配对，然后在该范围上运行一个样本，以确保没有重复项。

然后，我编写一个函数来“解码”这些整数为配对。

所以，如果我想要一个元素在范围-4到+4之间的配对，不包括零，不包括范围-2到+2的配对，我可以将其映射到可能配对的网格...

  1.     -4  -3  -2  -1  +0  +1  +2  +3  +4
  3. +4   O   O   O   O   X   O   O   O   O
  4. +3   O   O   O   O   X   O   O   O   O
  5. +2   O   O   X   X   X   X   X   O   O 
  6. +1   O   O   X   X   X   X   X   O   O 
  7. +0   X   X   X   X   X   X   X   X   X
  8. -1   O   O   X   X   X   X   X   O   O 
  9. -2   O   O   X   X   X   X   X   O   O 
  10. -3   O   O   O   O   X   O   O   O   O
  11. -4   O   O   O   O   X   O   O   O   O

O = 可能的配对
X = 排除的配对

有48种允许的可能配对。

  • 48 = (4² - 2²) * 4

然后，我编写一个函数，可以将值0..47中的每个值转换为这些允许的可能配对之一。

  1. import random
  2. from pprint import pprint
  4. def decode(val, lower, upper):
  5.   q = val % 4
  6.   v = val >> 2
  7.   
  8.   x = v % (upper + lower) - lower
  9.   if x >= 0:
  10.     x += 1
  11.   y = v // (upper + lower) + lower + 1
  12.   
  13.   if q in (1,3):
  14.     x = -x
  15.     y = -y
  16.   
  17.   if q in (2,3):
  18.     x,y = -y,x
  20.   return (x,y)
  22. def pairs(n, lower, upper):
  23.   return [
  24.     decode(id, lower, upper)
  25.       for id in random.sample(
  26.         range((upper**2 - lower**2) * 4),
  27.         n
  28.       )
  29.   ]
  31. # 较小的范围用于使其可打印和可测试
  32. result = [
  33.   *pairs( 5,  0,  5),  # 在范围-5到5的情况下有5对，不包括范围-0到0的情况
  34.   *pairs( 5,  5, 10),  # 在范围-10到10的情况下有5对，不包括范围-5到5的情况 
  35.   *pairs( 5, 10, 20)   # 在范围-20到20的情况下有5对，不包括范围-10到10的情况 
  36. ]
  38. pprint(result)

演示：https://trinket.io/python3/f9e65fe9ef"

英文:

It's gone past midnight, but I'll do my best to explain...

The basic principle is that I work out how many different pairs there are in a given range (excluding any zeros), and run a sample on that range, so I know there are no duplicates.

Then I code a function to "decode" those integers in to a pairs.

So, if I want a pair with elements in the range -4 to +4, excluding zeros, excluding pairs from the range -2 to +2, I can map that to a grid of possible pairs...

  1.     -4  -3  -2  -1  +0  +1  +2  +3  +4
  3. +4   O   O   O   O   X   O   O   O   O
  4. +3   O   O   O   O   X   O   O   O   O
  5. +2   O   O   X   X   X   X   X   O   O 
  6. +1   O   O   X   X   X   X   X   O   O 
  7. +0   X   X   X   X   X   X   X   X   X
  8. -1   O   O   X   X   X   X   X   O   O 
  9. -2   O   O   X   X   X   X   X   O   O 
  10. -3   O   O   O   O   X   O   O   O   O
  11. -4   O   O   O   O   X   O   O   O   O

O = Possible pair
X = Excluded pair

There are 48 allowed possible pairs.

  • 48 = (4² - 2²) * 4

I then write a function that can turn each of the values 0..47 in to a different one of those allowed possible pairs.

  1. import random
  2. from pprint import pprint
  4. def decode(val, lower, upper):
  5.   q = val % 4
  6.   v = val &gt;&gt; 2
  7.   
  8.   x = v % (upper + lower) - lower
  9.   if x &gt;= 0:
  10.     x += 1
  11.   y = v // (upper + lower) + lower + 1
  12.   
  13.   if q in (1,3):
  14.     x = -x
  15.     y = -y
  16.   
  17.   if q in (2,3):
  18.     x,y = -y,x
  20.   return (x,y)
  22. def pairs(n, lower, upper):
  23.   return [
  24.     decode(id, lower, upper)
  25.       for id in random.sample(
  26.         range((upper**2 - lower**2) * 4),
  27.         n
  28.       )
  29.   ]
  31. # smaller ranges used to make it printable and testable
  32. result = [
  33.   *pairs( 5,  0,  5),  # 5 pairs with values in the range - 5 to  5, except pairs using the range - 0 to 0
  34.   *pairs( 5,  5, 10),  # 5 pairs with values in the range -10 to 10, except pairs using the range - 5 to 5 
  35.   *pairs( 5, 10, 20)   # 5 pairs with values in the range -20 to 20, except pairs using the range -10 to 10 
  36. ]
  38. pprint(result)

Demo : https://trinket.io/python3/f9e65fe9ef

答案3

得分: 1

以下是您要翻译的内容：

  1. 我可以这样做，使用一个数字列表来获取“sample”，而不必多余地删除零，使用一个字典来强制唯一性（使用集合会更简单，但字典会保留插入顺序，如果您希望最终列表中的不同样本的排列方式相同，并且较小的数字位于前面，这很重要）：
  3. ```python
  4. import random
  6. max_num = 150
  7. nums = list(range(-max_num, max_num))
  8. nums.remove(0)
  10. my_list = list({
  11.     tuple(
  12.         random.sample(nums[max_num-r:r-max_num or None], 2)
  13.     ): 0
  14.     for c, r in ((5, 5), (75, 100), (20, 150))
  15.     for _ in range(c)
  16. })

生成一个单独的nums列表比为每个样本群体生成一个新列表稍微复杂一些，但效率更高，因为我们不需要反复重新分配它，只需根据需要取片段。

请注意，max_num需要至少与for c, r 迭代中所有r值一样大，否则我们将无法通过使用max_num-r等来切片nums来获得所需的范围。

  2. <details>
  3. <summary>英文:</summary>
  5. I might do it like this, using a list of numbers to be able to take the `sample` without having to take an extra step of dropping the zeroes, and using a dict to enforce uniqueness (a set would be simpler, but a dict preserves insertion order, which is important if you want the same arrangement of the different samples in the final list, with smaller numbers toward the beginning):

import random

max_num = 150
nums = list(range(-max_num, max_num))
nums.remove(0)

my_list = list({
tuple(
random.sample(nums[max_num-r:r-max_num or None], 2)
): 0
for c, r in ((5, 5), (75, 100), (20, 150))
for _ in range(c)
})

  2. Generating a single `nums` list is a little more complicated than generating a new list for each sample population, but is more efficient since then we don&#39;t need to repeatedly re-allocate it, and can simply take slices as needed.  
  4. Note that `max_num` needs to be at least as big as all of the `r` values in the `for c, r` iteration, or we won&#39;t be able to get the desired range by slicing `nums` with `max_num-r` etc.
  6. </details>

huangapple
  • 本文由 发表于 2023年3月7日 05:36:46
  • 转载请务必保留本文链接：https://go.coder-hub.com/75656074.html
  • python
  • python-3.x
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