2024-10-29 07:00:04

计算列表差值

在Python中,计算两个列表之间的差值的最佳方法是什么?

例子

A = [1,2,3,4]
B = [2,5]

A - B = [1,3,4]
B - A = [5]

当前回答

有三种选择,其中两种是可以接受的,另一种不应该这样做。

在较高的级别上,这3个选项是:

减去两组(有时最好) 检查每个列表项是否存在于一个集合中(大多数情况下最好) 检查每个列表项是否存在于列表中(不做)

选项3)永远不应该超过选项2)。根据应用程序的需要,您可能更喜欢选项1)或2),而在大多数用例中,2)可能是首选方法。2)与1)的性能非常相似,因为两者都具有O(m + n)个时间复杂度。相比之下,2)在空间复杂度上比1)有边际优势,并且既保持了原始列表的顺序,又保持了原始列表中的任何重复。

如果你想删除重复,不关心顺序,那么1)可能是最适合你的。

import time

def fun1(l1, l2):
    # Order and duplications in l1 are both lost, O(m) + O(n)
    return set(l1) - set(l2)

def fun2(l1, l2):
    # Order and duplications in l1 are both preserved, O(m) + O(n)
    l2_set = set(l2)
    return [item for item in l1 if item not in l2_set]

def fun3(l1, l2):
    # Order and duplications in l1 are both preserved, O(m * n)
    # Don't do
    return [item for item in l1 if item not in l2]

A = list(range(7500))
B = list(range(5000, 10000))

loops = 100

start = time.time()
for _ in range(loops):
    fun1(A, B)
print(f"fun1 time: {time.time() - start}")

start = time.time()
for _ in range(loops):
    fun2(A, B)
print(f"fun2 time: {time.time() - start}")

start = time.time()
for _ in range(loops):
    fun3(A, B)
print(f"fun3 time: {time.time() - start}")
fun1 time: 0.03749704360961914
fun2 time: 0.04109621047973633
fun3 time: 32.55076885223389

其他回答

A = [1,2,3,4]
B = [2,5]

#A - B
x = list(set(A) - set(B))
#B - A 
y = list(set(B) - set(A))

print x
print y 

添加一个答案来处理我们想要严格区别重复的情况,也就是说,在第一个列表中有我们想要保留在结果中的重复。例如,得到,

[1, 1, 1, 2] - [1, 1] --> [1, 2]

我们可以用一个额外的计数器来得到一个优雅的差分函数。

from collections import Counter

def diff(first, second):
    secondCntr = Counter(second)
    second = set(second)
    res = []
    for i in first:
        if i not in second:
            res.append(i)
        elif i in secondCntr:
            if secondCntr[i] > 0:
                secondCntr[i] -= 1
            else:
                res.append(i)        
    return res

您可能希望使用集合而不是列表。

有三种选择,其中两种是可以接受的,另一种不应该这样做。

在较高的级别上,这3个选项是:

减去两组(有时最好) 检查每个列表项是否存在于一个集合中(大多数情况下最好) 检查每个列表项是否存在于列表中(不做)

选项3)永远不应该超过选项2)。根据应用程序的需要,您可能更喜欢选项1)或2),而在大多数用例中,2)可能是首选方法。2)与1)的性能非常相似,因为两者都具有O(m + n)个时间复杂度。相比之下,2)在空间复杂度上比1)有边际优势,并且既保持了原始列表的顺序,又保持了原始列表中的任何重复。

如果你想删除重复,不关心顺序,那么1)可能是最适合你的。

import time

def fun1(l1, l2):
    # Order and duplications in l1 are both lost, O(m) + O(n)
    return set(l1) - set(l2)

def fun2(l1, l2):
    # Order and duplications in l1 are both preserved, O(m) + O(n)
    l2_set = set(l2)
    return [item for item in l1 if item not in l2_set]

def fun3(l1, l2):
    # Order and duplications in l1 are both preserved, O(m * n)
    # Don't do
    return [item for item in l1 if item not in l2]

A = list(range(7500))
B = list(range(5000, 10000))

loops = 100

start = time.time()
for _ in range(loops):
    fun1(A, B)
print(f"fun1 time: {time.time() - start}")

start = time.time()
for _ in range(loops):
    fun2(A, B)
print(f"fun2 time: {time.time() - start}")

start = time.time()
for _ in range(loops):
    fun3(A, B)
print(f"fun3 time: {time.time() - start}")
fun1 time: 0.03749704360961914
fun2 time: 0.04109621047973633
fun3 time: 32.55076885223389

如果你不关心项目的顺序或重复,请使用set。使用列表推导式:

>>> def diff(first, second):
        second = set(second)
        return [item for item in first if item not in second]

>>> diff(A, B)
[1, 3, 4]
>>> diff(B, A)
[5]
>>>