【每日一题20220331】列表运算

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:mage:‍ 给定两个数字列表，请编写一个函数，从第一个列表nums_a中移除所有在第二个列表nums_b中出现的元素，返回剩下元素组成的列表。

【示例】
输入：[1, 1, 2 ,3 ,1 ,2 ,3 ,4], [1, 3]
输出：[2, 2, 4]
解释：第一个列表剔除所有的数字1和3之后，最终剩下[2, 2, 4]

题目难度：简单
题目来源：CodeWars

def solution(nums_a: list, nums_b: list)-> list:
  # your code here

assert solution([1, 1, 2 ,3 ,1 ,2 ,3 ,4], [1, 3]) == [2, 2, 4]
assert solution([8, 2, 7, 2, 3, 4, 6, 5, 4, 4, 1, 2 , 3], [2, 4, 3]) == [8, 7, 6, 5, 1]
assert solution([1, 1, 2 ,3 ,1 ,2 ,3 ,4, 4, 3 ,5, 6, 7, 2, 8], [1, 3, 4, 2]) == [5, 6 ,7 ,8]
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def solution(nums_a, nums_b):
    return [item for item in nums_a if item not in nums_b]
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def solution(nums_a: list, nums_b: list) -> list:
    for x in nums_b:
        while x in nums_a:
            nums_a.remove(x)
    return nums_a
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def solution(nums_a, nums_b):
    map=Counter(nums_a)
    for i in nums_b:
        if i in map.keys():
            map.pop(i)
    return list(map.elements())
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def solution(nums_a: list, nums_b: list)-> list:

from collections import Counter
for x in nums_b:
    for i in range(0,Counter(nums_a).get(x)):
        nums_a.remove(x)
return nums_a

assert solution([1, 1, 2 ,3 ,1 ,2 ,3 ,4], [1, 3]) == [2, 2, 4]
assert solution([8, 2, 7, 2, 3, 4, 6, 5, 4, 4, 1, 2 , 3], [2, 4, 3]) == [8, 7, 6, 5, 1]
assert solution([1, 1, 2 ,3 ,1 ,2 ,3 ,4, 4, 3 ,5, 6, 7, 2, 8], [1, 3, 4, 2]) == [5, 6 ,7 ,8] ```
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    def solution(nums_a: list, nums_b: list) -> list:
        return [i for i in nums_a if i not in nums_b]
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def solution(nums_a:list, nums_b:list)->list:
  lst = [i for i in nums_a if i not in nums_b]
  return lst

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截屏2022-04-01 下午3.36.31

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def solution(nums_a: list, nums_b: list)-> list:
    for _ in range(len(nums_a)):
        for i in nums_b:
            for j in nums_a:
                if i==j:
                    nums_a.remove(i)
    return nums_a
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def solution(nums_a: list, nums_b: list)-> list:
    return [i for i in nums_a if i not in nums_b]
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def solution(nums_a: list, nums_b: list)-> list:
  # your code here
     return [var for var in nums_a if var not in nums_b]

assert solution([1, 1, 2 ,3 ,1 ,2 ,3 ,4], [1, 3]) == [2, 2, 4]
assert solution([8, 2, 7, 2, 3, 4, 6, 5, 4, 4, 1, 2 , 3], [2, 4, 3]) == [8, 7, 6, 5, 1]
assert solution([1, 1, 2 ,3 ,1 ,2 ,3 ,4, 4, 3 ,5, 6, 7, 2, 8], [1, 3, 4, 2]) == [5, 6 ,7 ,8]
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def solution(nums_a: list, nums_b: list)-> list:

your code here

for y in range(0,len(nums_b)):
    for i in nums_a[:]:
        if i == nums_b[y]:
            nums_a.remove(i)
return nums_a
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def solution(nums_a: list, nums_b: list) -> list:
    return [x for x in nums_a if x not in nums_b]


assert solution([1, 1, 2, 3, 1, 2, 3, 4], [1, 3]) == [2, 2, 4]
assert solution([8, 2, 7, 2, 3, 4, 6, 5, 4, 4, 1, 2, 3], [2, 4, 3]) == [8, 7, 6, 5, 1]
assert solution([1, 1, 2, 3, 1, 2, 3, 4, 4, 3, 5, 6, 7, 2, 8], [1, 3, 4, 2]) == [5, 6, 7, 8]
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def solution(nums_a: list, nums_b: list) -> list:
   new_list = []
   for i in nums_a:
      if i not in nums_b:
         new_list.append(i)
   return new_list
assert solution([1, 1, 2, 3, 1, 2, 3, 4], [1, 3]) == [2, 2, 4]
assert solution([8, 2, 7, 2, 3, 4, 6, 5, 4, 4, 1, 2, 3], [2, 4, 3]) == [8, 7, 6, 5, 1]
assert solution([1, 1, 2, 3, 1, 2, 3, 4, 4, 3, 5, 6, 7, 2, 8], [1, 3, 4, 2]) == [5, 6, 7, 8]
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def solution(nums_a: list, nums_b: list) -> list:
    # your code here
    for i in nums_b:
        while i in nums_a:
            nums_a.remove(i)
    return nums_a


def solution1(nums_a: list, nums_b: list) -> list:
    nums = []
    for i in nums_a:
        if i not in nums_b:
            nums.append(i)
    return nums