如何用基础Python找出一个列表中不在另一个列表里的元素

方法1：列表推导式（推荐，保留元素顺序与重复项） #

直接遍历目标列表，筛选出不在对比列表中的元素，逻辑清晰且能保留原顺序和重复元素：

arr1 = [1,2,3,4,5,6,7]
arr2 = [1,2,3,4,5,6,7,8,9,10]
diff = [x for x in arr2 if x not in arr1]
print(diff)  # 输出 [8,9,10]

方法2：集合差集（适合无重复元素的高效场景） #

如果两个列表均无重复元素，可利用集合的差集操作提升效率，若需保持原顺序则结合列表推导式：

arr1 = [1,2,3,4,5,6,7]
arr2 = [1,2,3,4,5,6,7,8,9,10]
# 基础差集（无序）
diff = list(set(arr2) - set(arr1))
# 保持原顺序的高效写法（集合查找为O(1)，比列表查找更快）
diff = [x for x in arr2 if x not in set(arr1)]

修复嵌套循环写法 #

你之前得到连续数字串，是因为错误地拼接字符串而非收集元素到列表。正确的嵌套循环实现如下：

arr1 = [1,2,3,4,5,6,7]
arr2 = [1,2,3,4,5,6,7,8,9,10]
diff = []
for num in arr2:
    exists = False
    for target in arr1:
        if num == target:
            exists = True
            break
    if not exists:
        diff.append(num)
print(diff)  # 输出 [8,9,10]

内容的提问来源于stack exchange，提问作者Pepwave Dave