Python|检查列表列表中是否存在元素

给定一个列表列表，任务是确定给定元素是否存在于任何子列表中。下面给出了一些解决给定任务的方法。方法 #1：使用 any()

只要给定迭代器中存在特定元素， any()方法就会返回 true。

# Python code to demonstrate
# finding whether element
# exists in listof list
  
# initialising nested lists
ini_list = [[1, 2, 5, 10, 7],
            [4, 3, 4, 3, 21],
            [45, 65, 8, 8, 9, 9]]
  
elem_to_find = 8
elem_to_find1 = 0
  
# element exists in listof listor not?
res1 = any(elem_to_find in sublist for sublist in ini_list)
res2 = any(elem_to_find1 in sublist for sublist in ini_list)
  
# printing result
print(str(res1), "\n", str(res2))

输出：

True 
 False

方法#2：使用运算符符

'in'运算符用于检查一个值是否存在于序列中。如果找到指定序列中的变量，则评估为true ，否则评估为false 。

# Python code to demonstrate
# finding whether element
# exists in listof list
  
# initialising nested lists
ini_list = [[1, 2, 5, 10, 7],
            [4, 3, 4, 3, 21],
            [45, 65, 8, 8, 9, 9]]
  
elem = 8
elem1 = 0
  
# element exists in listof listor not?
res1 = elem in (item for sublist in ini_list for item in sublist)
res2 = elem1 in (item for sublist in ini_list for item in sublist)
  
# printing result
print(str(res1), "\n", str(res2))

输出：

True 
 False

方法 #3：使用itertools.chain()

创建一个迭代器，从第一个迭代器返回元素，直到它用尽，然后继续到下一个迭代器，直到所有的迭代器都用尽。

# Python code to demonstrate
# finding whether element
# exists in listof list
from  itertools import chain
  
# initialising nested lists
ini_list = [[1, 2, 5, 10, 7], 
            [4, 3, 4, 3, 21],
            [45, 65, 8, 8, 9, 9]]
  
elem_to_find = 8
elem_to_find1 = 0
  
# element exists in listof listor not?
res1 = elem_to_find in chain(*ini_list)
res2 = elem_to_find1 in chain(*ini_list)
  
# printing result
print(str(res1), "\n", str(res2))

输出：

True 
 False