Python – 删除列表中索引处的元素

给定列表，删除索引列表中存在的所有元素。

Input : test_list = [5, 6, 3, 7, 8, 1, 2, 10], idx_list = [2, 4, 5] 
Output : [5, 6, 7, 2, 10] 
Explanation : 3, 6, and 1 has been removed.
Input : test_list = [5, 6, 3, 7, 8, 1, 2, 10], idx_list = [2] 
Output : [5, 6, 7, 8, 1, 2, 10] 
Explanation : 3 has been removed. 
 

方法 #1：使用 enumerate() + 循环

在此，我们迭代所有元素，如果索引存在于列表中，则该索引元素从结果列表中省略。

Python3

# Python3 code to demonstrate working of
# Remove elements at Indices in List
# Using loop
 
# initializing list
test_list = [5, 6, 3, 7, 8, 1, 2, 10]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing idx list
idx_list = [2, 4, 5, 7]
 
res = []
for idx, ele in enumerate(test_list):
     
    # checking if element not present in index list
    if idx not in idx_list:
        res.append(ele)
 
# printing results
print("Filtered List after removal : " + str(res))

输出

The original list is : [5, 6, 3, 7, 8, 1, 2, 10]
Filtered List after removal : [5, 6, 7, 2]

方法 #2：使用 enumerate() + 列表推导

在此，我们以紧凑的方式使用列表推导执行迭代任务，其余方法与上述类似。

Python3

# Python3 code to demonstrate working of
# Remove elements at Indices in List
# Using enumerate() + list comprehension
 
# initializing list
test_list = [5, 6, 3, 7, 8, 1, 2, 10]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing idx list
idx_list = [2, 4, 5, 7]
 
# one-liner to test for element in index list
res = [ele for idx, ele in enumerate(test_list) if idx not in idx_list]
 
# printing results
print("Filtered List after removal : " + str(res))

输出

The original list is : [5, 6, 3, 7, 8, 1, 2, 10]
Filtered List after removal : [5, 6, 7, 2]