我在Python中有两个列表:
temp1 = ['One', 'Two', 'Three', 'Four']
temp2 = ['One', 'Two']
假设每个列表中的元素都是唯一的，我想用第一个列表中的项创建第三个列表，这些项不在第二个列表中:
temp3 = ['Three', 'Four']
有没有没有周期和检查的快速方法?
当前回答
def diffList(list1, list2): # returns the difference between two lists.
if len(list1) > len(list2):
return (list(set(list1) - set(list2)))
else:
return (list(set(list2) - set(list1)))
例如，如果list1 =[10, 15, 20, 25, 30, 35, 40]和list2 =[25, 40, 35]，那么返回的列表将输出= [10,20,30,15]
其他回答
以下是@SuperNova的回答的修改版本
def get_diff(a: list, b: list) -> list:
return list(set(a) ^ set(b))
最简单的方法，
使用设置().difference(设置())
list_a = [1,2,3]
list_b = [2,3]
print set(list_a).difference(set(list_b))
答案设置([1])
可以打印为列表，
print list(set(list_a).difference(set(list_b)))
如果你遇到TypeError: unhashable type: 'list'，你需要将列表或集转换为元组。
set(map(tuple, list_of_lists1)).symmetric_difference(set(map(tuple, list_of_lists2)))
参见如何比较python中的列表/集列表?
我在这个游戏中有点晚了，但你可以做一个性能的比较，上面提到的一些代码和这个，两个最快的竞争者是，
list(set(x).symmetric_difference(set(y)))
list(set(x) ^ set(y))
我为我的初级编码水平道歉。
import time
import random
from itertools import filterfalse
# 1 - performance (time taken)
# 2 - correctness (answer - 1,4,5,6)
# set performance
performance = 1
numberoftests = 7
def answer(x,y,z):
if z == 0:
start = time.clock()
lists = (str(list(set(x)-set(y))+list(set(y)-set(y))))
times = ("1 = " + str(time.clock() - start))
return (lists,times)
elif z == 1:
start = time.clock()
lists = (str(list(set(x).symmetric_difference(set(y)))))
times = ("2 = " + str(time.clock() - start))
return (lists,times)
elif z == 2:
start = time.clock()
lists = (str(list(set(x) ^ set(y))))
times = ("3 = " + str(time.clock() - start))
return (lists,times)
elif z == 3:
start = time.clock()
lists = (filterfalse(set(y).__contains__, x))
times = ("4 = " + str(time.clock() - start))
return (lists,times)
elif z == 4:
start = time.clock()
lists = (tuple(set(x) - set(y)))
times = ("5 = " + str(time.clock() - start))
return (lists,times)
elif z == 5:
start = time.clock()
lists = ([tt for tt in x if tt not in y])
times = ("6 = " + str(time.clock() - start))
return (lists,times)
else:
start = time.clock()
Xarray = [iDa for iDa in x if iDa not in y]
Yarray = [iDb for iDb in y if iDb not in x]
lists = (str(Xarray + Yarray))
times = ("7 = " + str(time.clock() - start))
return (lists,times)
n = numberoftests
if performance == 2:
a = [1,2,3,4,5]
b = [3,2,6]
for c in range(0,n):
d = answer(a,b,c)
print(d[0])
elif performance == 1:
for tests in range(0,10):
print("Test Number" + str(tests + 1))
a = random.sample(range(1, 900000), 9999)
b = random.sample(range(1, 900000), 9999)
for c in range(0,n):
#if c not in (1,4,5,6):
d = answer(a,b,c)
print(d[1])
我知道这个问题已经得到了很好的答案，但我希望使用numpy添加以下方法。
temp1 = ['One', 'Two', 'Three', 'Four']
temp2 = ['One', 'Two']
list(np.setdiff1d(temp1,temp2))
['Four', 'Three'] #Output
