如何在不使用.count的情况下计算列表列表的特定数量？

def count_elements(lst, target):
    count = 0
    for element in lst:
        if element == target:
            count += 1
    return count

# 示例
my_list = [1, 2, 3, 2, 4, 2, 5]
target_element = 2
print(count_elements(my_list, target_element))  # 输出: 3

def count_elements(lst, target):
    return len([element for element in lst if element == target])

# 示例
my_list = [1, 2, 3, 2, 4, 2, 5]
target_element = 2
print(count_elements(my_list, target_element))  # 输出: 3

from collections import Counter

def count_elements(lst, target):
    counter = Counter(lst)
    return counter[target]

# 示例
my_list = [1, 2, 3, 2, 4, 2, 5]
target_element = 2
print(count_elements(my_list, target_element))  # 输出: 3

import numpy as np

def count_elements(lst, target):
    arr = np.array(lst)
    return np.sum(arr == target)

# 示例
my_list = [1, 2, 3, 2, 4, 2, 5]
target_element = 2
print(count_elements(my_list, target_element))  # 输出: 3

def flatten(lst):
    for item in lst:
        if isinstance(item, list):
            yield from flatten(item)
        else:
            yield item

def count_elements(lst, target):
    flat_list = list(flatten(lst))
    return flat_list.count(target)

# 示例
my_list = [1, [2, 3], [2, [4, 2]], 5]
target_element = 2
print(count_elements(my_list, target_element))  # 输出: 3

def count_elements(lst, target):
    target = tuple(target) if isinstance(target, list) else target
    return sum(1 for element in lst if tuple(element) == target)

# 示例
my_list = [[1, 2], [2, 3], [2, [4, 2]], 5]
target_element = [2, 3]
print(count_elements(my_list, target_element))  # 输出: 1