如果散差表的元素是排序和集合的，你可以使用朴素方法。

list1=[1,2,3,4,5]
list2=[1,2,3]

print list1[len(list2):]

或者使用本机set方法:

subset=set(list1).difference(list2)

print subset

import timeit
init = 'temp1 = list(range(100)); temp2 = [i * 2 for i in range(50)]'
print "Naive solution: ", timeit.timeit('temp1[len(temp2):]', init, number = 100000)
print "Native set solution: ", timeit.timeit('set(temp1).difference(temp2)', init, number = 100000)

朴素解:0.0787101593292

本机集解决方案:0.998837615564

这可能比马克的清单理解还要快:

list(itertools.filterfalse(set(temp2).__contains__, temp1))

现有的解决方案都提供了其中的一个或另一个:

比O(n*m)性能快。 保持输入列表的顺序。

但到目前为止，还没有一种解决方案兼而有之。如果你两者都想要，试试这个:

s = set(temp2)
temp3 = [x for x in temp1 if x not in s]

性能测试

import timeit
init = 'temp1 = list(range(100)); temp2 = [i * 2 for i in range(50)]'
print timeit.timeit('list(set(temp1) - set(temp2))', init, number = 100000)
print timeit.timeit('s = set(temp2);[x for x in temp1 if x not in s]', init, number = 100000)
print timeit.timeit('[item for item in temp1 if item not in temp2]', init, number = 100000)

结果:

4.34620224079 # ars' answer
4.2770634955  # This answer
30.7715615392 # matt b's answer

我提出的方法以及保持顺序也比集合减法(稍微)快，因为它不需要构造一个不必要的集合。如果第一个列表比第二个列表长得多，并且散列代价昂贵，那么性能差异将更加明显。下面是第二个测试:

init = '''
temp1 = [str(i) for i in range(100000)]
temp2 = [str(i * 2) for i in range(50)]
'''

结果:

11.3836875916 # ars' answer
3.63890368748 # this answer (3 times faster!)
37.7445402279 # matt b's answer

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

两个列表(例如list1和list2)之间的差异可以使用下面的简单函数找到。

def diff(list1, list2):
    c = set(list1).union(set(list2))  # or c = set(list1) | set(list2)
    d = set(list1).intersection(set(list2))  # or d = set(list1) & set(list2)
    return list(c - d)

or

def diff(list1, list2):
    return list(set(list1).symmetric_difference(set(list2)))  # or return list(set(list1) ^ set(list2))

通过使用上述函数，可以使用diff(temp2, temp1)或diff(temp1, temp2)找到差异。两者都会给出结果['Four'， 'Three']。你不需要担心列表的顺序或者先给出哪个列表。

Python文档参考

下面是@arkolec的回答，下面是一个用于比较列表、元组和集的实用程序类:

from difflib import SequenceMatcher

class ListDiffer:

    def __init__(self, left, right, strict:bool=False):
        assert isinstance(left, (list, tuple, set)), "left must be list, tuple or set"
        assert isinstance(right, (list, tuple, set)), "right must be list, tuple or set"
        self.l = list(left) if isinstance(left, (tuple, set)) else left
        self.r = list(right) if isinstance(left, (tuple, set)) else right

        if strict:
            assert isinstance(left, right.__class__), \
                f'left type ({left.__class__.__name__}) must equal right type ({right.__class__.__name__})'

        self.diffs = []
        self.equal = []

        for tag, i, j, k, l in SequenceMatcher(None, self.l, self.r).get_opcodes():
            if tag in ['delete', 'replace', 'insert']:
                self.diffs.append((tag, i, j, k, l))
            elif tag == 'equal':
                [self.equal.append(v) for v in left[i:j]]
                


    def has_diffs(self):
        return len(self.diffs) > 0


    def only_left(self):
        a = self.l[:]
        [a.remove(v) for v in self.equal]
        return a

    def only_right(self):
        a = self.r[:]
        [a.remove(v) for v in self.equal]
        return a


    def __str__(self, verbose:bool=False):
        iD = 0
        sb = []
        if verbose:
            sb.append(f"left: {self.l}\n")
            sb.append(f"right: {self.r}\n")
            sb.append(f"diffs: ")
        for tag, i, j, k, l in self.diffs:
            s = f"({iD})"
            if iD > 0: sb.append(' | ')
            if tag in ('delete', 'replace'): s = f'{s} l:{self.l[i:j]}'
            if tag in ('insert', 'replace'): s = f'{s} r:{self.r[k:l]}'
            sb.append(s)
            iD = iD + 1

        if verbose:
            sb.append(f"\nequal: {self.equal}")
        return ''.join(sb)

    def __repr__(self) -> str:
        return "<ListDiffer> {}".format(self.__str__())

用法:

left = ['a','b','c']
right = ['aa','b','c','d']
# right = ('aa','b','c','d')
ld = ListDiffer(left, right, strict=True)
print(f'ld.has_diffs(): {ld.has_diffs()}')
print(f'ld: {ld}')
print(f'ld.only_left(): {ld.only_left()}')
print(f'ld.only_right(): {ld.only_right()}')

输出:

ld.has_diffs(): True
ld: (0) l:['a'] r:['aa'] | (1) r:['d']
ld.only_left(): ['a']
ld.only_right(): ['aa', 'd']

我不能说性能，但你可以使用ld.only_left()来获得你正在寻找的结果。