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Description

Your goal in this kata is to implement a difference function, which subtracts one list from another and returns the result.

• It should remove all values from list a, which are present in list b,such as array_diff([1,2],[1]) == [2]
• If a value is present in b, all of its occurrences must be removed from the other: such as array_diff([1,2,2,2,3],[2]) == [1,3]

Solution

my code

def array_diff(a, b):
    for num in b:
        while num in a:
            a.remove(num)
    return a

others‘ code

def array_diff(a, b):
    return [x for x in a if x not in b]

def array_diff(a, b):
    return [x for x in a if x not in set(b)]

def array_diff(a, b):
    #your code here
    return filter(lambda i: i not in b, a)

Conclusion

• set()具有去重功能

• 列表推导式

• list & dict & set的时间复杂度
  
  
  

• filter(function,iterable)用于过滤序列，过滤掉不符合条件的元素，返回一个迭代器对象，如果要转换为列表，可以使用list()来转换

fliter使用示例

#过滤出列表中的奇数
def is_odd(n):
    return n % 2 == 1 
tmplist = filter(is_odd, [1, 2, 3, 4, 5, 6, 7, 8, 9, 10])
newlist = list(tmplist)
print(newlist)

#过滤出1~100内的平方数
import math
def is_sqr(x):
    return math.sqrt(x) % 1 == 0 
tmplist = filter(is_sqr, range(1, 101))
newlist = list(tmplist)
print(newlist)

备注：部分内容参考这个博客，仅供学习交流使用，侵删。

[1].Array.diff

标签：元素   lang   present   blog   log   python   转换   示例   loading   

原文地址：https://www.cnblogs.com/zjx-pku/p/13174177.html