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,我希望创建第三个列表,其中包含第一个列表中不在第二个列表中的项目:
假设每个列表中的元素都是唯一的,我想要创建一个新的列表,其中包含来自第一个列表的但不在第二个列表中的项目:
temp3 = ['Three', 'Four']
有没有不需要循环和检查的快速方式?
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,我希望创建第三个列表,其中包含第一个列表中不在第二个列表中的项目:
假设每个列表中的元素都是唯一的,我想要创建一个新的列表,其中包含来自第一个列表的但不在第二个列表中的项目:
temp3 = ['Three', 'Four']
有没有不需要循环和检查的快速方式?
s = set(temp2);
[x for x in temp1 if x not in s]
表现结果
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))
print(timeit.timeit('list(set(temp1) ^ set(temp2))', init, number = 100000))
结果
0.3485999000258744
0.3314229999668896
2.067719299986493
0.3791518000070937
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)))
list1 = [10, 15, 20, 25, 30, 35, 40]
,而list2 = [25, 40, 35]
,则返回的列表将是output = [10, 20, 30, 15]
。继@arkolec的回答之后,这里提供了一个用于比较列表、元组和集合的实用类:
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__())
使用方法:
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']
我无法对性能进行评估,但您可以使用ld.only_left()
来获得您想要的结果。
temp1 = ['One', 'One', 'One']
和temp2 = ['One']
,你希望得到['One', 'One']
还是[]
? - Michael Mrozek