Python从满足特定条件的列表中查找数字[关闭]答案

【问题标题】：Python finding numbers from a list that satisfy a specific condition [closed]Python从满足特定条件的列表中查找数字[关闭]
【发布时间】：2020-09-14 16:09:43
【问题描述】：

晚上好，我目前在实现二进制搜索算法时遇到了一些问题，该算法提取列表中满足此条件的数字数量

假设 Sorted 列表 A = [1,2,4,5,6,7,8] 我需要找出满足这个条件的数字的数量。

Absolute(A[i] - i)

|A[i] -i|

例子：

    A = [1, 2, 4, 8, 16, 32, 64]
    c = 4

    A[0] = | 1 - 0 |  = 1.
    A[1] = | 2 - 1 |  = 1.
    A[2] = | 4 - 2 |  = 2.
    A[3] = | 8 - 3 |  = 5.
    A[4] = | 16 - 4 | = 12.
    A[5] = | 32 - 5 | = 27.
    A[6] = | 64 - 6 | = 58.

现在我意识到我需要使用二进制搜索来确保我的运行时间在 O(log N) 时间内，但我不确定将条件/if 语句放在哪里。

谁能告诉我这在 python 代码中的样子。非常感谢您的帮助。

【问题讨论】：

到目前为止你有什么尝试？
我看不出二分搜索在这里是如何工作的，因为您的条件取决于项目在列表中的位置。
我尝试过使用 pythons bisect 模块进行二分搜索，但没有运气:(

标签： python algorithm binary

【解决方案1】：

使用 Python Bisect 模块

使用key with binary_search module 允许函数评估

from bisect import bisect_left

class KeyifyList(object):
    " Allows specifying a key with binary search module"
    def __init__(self, inner, key):
        self.inner = inner
        self.key = key

    def __len__(self):
        return len(self.inner)

    def __getitem__(self, k):
        return self.key((k, self.inner[k]))

def bin_search(a, c):
  # Binary search for placement
  # Using key function to allow binary search using a function
  # Computes abs(a[i] - i) at places where binary search is evaluated
  # key computes abs(a[k]-k)
  # Binary search so O(log(n)) time complexity
  i = bisect_left(KeyifyList(a, lambda kv: abs(kv[1]-kv[0])), c)

  if i == len(a):
    last_index = len(a) -1
    if abs(a[last_index] - last_index) <= c:
      return len(a)  # all indices satisfy
    else:
      i = last_index

  while i >= 0 and abs(a[i]-i) > c:
    # this is normally a one point move over
    # so O(1) rather than O(n) in time complexity
    i -= 1
  
  # number of points is one more than index to satisfy
  return i + 1

测试

A = [1, 2, 4, 8, 16, 32, 64]
c = 4

测试 c 从 0 到 63

for c in range(65):
  print(f'c = {c}, number of points = {bin_search(A, c)}')

输出

c = 0, number of points = 0
c = 1, number of points = 1
c = 2, number of points = 3
c = 3, number of points = 3
c = 4, number of points = 3
c = 5, number of points = 4
c = 6, number of points = 4
c = 7, number of points = 4
c = 8, number of points = 4
c = 9, number of points = 4
c = 10, number of points = 4
c = 11, number of points = 4
c = 12, number of points = 5
c = 13, number of points = 5
c = 14, number of points = 5
c = 15, number of points = 5
c = 16, number of points = 5
c = 17, number of points = 5
c = 18, number of points = 5
c = 19, number of points = 5
c = 20, number of points = 5
c = 21, number of points = 5
c = 22, number of points = 5
c = 23, number of points = 5
c = 24, number of points = 5
c = 25, number of points = 5
c = 26, number of points = 5
c = 27, number of points = 6
c = 28, number of points = 6
c = 29, number of points = 6
c = 30, number of points = 6
c = 31, number of points = 6
c = 32, number of points = 6
c = 33, number of points = 6
c = 34, number of points = 6
c = 35, number of points = 6
c = 36, number of points = 6
c = 37, number of points = 6
c = 38, number of points = 6
c = 39, number of points = 6
c = 40, number of points = 6
c = 41, number of points = 6
c = 42, number of points = 6
c = 43, number of points = 6
c = 44, number of points = 6
c = 45, number of points = 6
c = 46, number of points = 6
c = 47, number of points = 6
c = 48, number of points = 6
c = 49, number of points = 6
c = 50, number of points = 6
c = 51, number of points = 6
c = 52, number of points = 6
c = 53, number of points = 6
c = 54, number of points = 6
c = 55, number of points = 6
c = 56, number of points = 6
c = 57, number of points = 6
c = 58, number of points = 7
c = 59, number of points = 7
c = 60, number of points = 7
c = 61, number of points = 7
c = 62, number of points = 7
c = 63, number of points = 7
c = 64, number of points = 7

性能测试

与列表理解比较（O(n) 算法）

def list_comprehension_method(a, c):
  " Use list comprehension to find number of points "
  return len([1 for i, v in enumerate(A) if abs(v - i) <= c])

计时测试

创建一个大的随机数组

n = 10000   # number of points in array
c = n // 4  # c value
A = sorted([randint(1, n) for _ in range(n)])

print(timeit(lambda: bin_search(A, c), number=100))
# Time: 0.00173 seconds
print(timeit(lambda: list_comprehension_method(A, c), number=100))
# Time: 0.49982 seconds

对于 n = 10, 000，二分搜索 ~289X 快

【讨论】：

很遗憾，您无意中帮助了某人作弊，请参阅OP trying to hide evidence。这就是为什么最好只认真回答问题。
@AndrasDeak——你可能是对的。主要回答帮助其他人，例如来自 cmets 尝试过的 DeepAlgo212345 和 JohnGordon，但无法弄清楚如何使用 Python 的 bisect 方法使用函数（在我遇到如何使用 Python bisect 键入函数之前，我之前也想知道这一点方法所以想帮助传播这个词）。令人惊讶的是家庭作业、编码挑战等问题的数量，而海报显然没有付出任何努力。为了帮助清理网站——我应该在将来将这些标记为需要管理员干预吗？
不，这些并不是版主真正关心的问题。如果您看到一个问题显然是有人试图让其他人为他们完成他们的工作/家庭作业/考试，只需投票关闭“需要更多关注”（以前是“太宽泛”），并且在提问者表现出真诚之前不要回答尝试解决问题:)

【解决方案2】：

lambda 和过滤器应该可以带你到那里

A = [1, 2, 4, 8, 16, 32, 64]
c = 4
res = list(filter(lambda x: A[x] - x <= c, range(len(A))))
print(res)

[0, 1, 2]

【讨论】：

我认为这不会在日志 N 时间内运行。
很遗憾，您无意中帮助了某人作弊，请参阅OP trying to hide evidence。这就是为什么最好只用努力回答问题。