查找第二个(第三个…)的*索引*的最快方法矢量或列中的最高/最低值

def find_nth_index(arr, n, value):
    count = 0
    for i in range(len(arr)):
        if arr[i] == value:
            count += 1
            if count == n:
                return i
    return -1  # 如果没有找到，返回-1

# 示例
arr = [1, 2, 3, 2, 4, 2, 5]
value = 2
n = 3
index = find_nth_index(arr, n, value)
print(f"第{n}个{value}的索引是: {index}")

def find_nth_index(arr, n, value):
    indices = [i for i, x in enumerate(arr) if x == value]
    if len(indices) >= n:
        return indices[n-1]
    return -1

# 示例
arr = [1, 2, 3, 2, 4, 2, 5]
value = 2
n = 3
index = find_nth_index(arr, n, value)
print(f"第{n}个{value}的索引是: {index}")

# 最高值
max_value = max(arr)

# 最低值
min_value = min(arr)

print(f"最高值是: {max_value}")
print(f"最低值是: {min_value}")

import numpy as np

arr = np.array([1, 2, 3, 4, 5])
max_value = np.max(arr)
min_value = np.min(arr)

print(f"最高值是: {max_value}")
print(f"最低值是: {min_value}")

try:
    index = find_nth_index(arr, n, value)
    if index == -1:
        print("未找到指定元素")
    else:
        print(f"第{n}个{value}的索引是: {index}")
except IndexError:
    print("索引越界")

def binary_search(arr, n, value):
    left, right = 0, len(arr) - 1
    count = 0
    while left <= right:
        mid = (left + right) // 2
        if arr[mid] == value:
            count += 1
            if count == n:
                return mid
            elif count < n:
                left = mid + 1
            else:
                right = mid - 1
        elif arr[mid] < value:
            left = mid + 1
        else:
            right = mid - 1
    return -1

# 示例
arr = [1, 2, 2, 2, 3, 4, 5]
value = 2
n = 3
index = binary_search(arr, n, value)
print(f"第{n}个{value}的索引是: {index}")