python 获取两个列表之间的区别
现有的解决方案都提供以下两种方法之一:
- 比O(n * m)的性能更快。
- 保留输入列表的顺序。
但迄今为止没有解决方案。 如果你想要两个,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存顺序也比设置的减法(略快)快,因为它不需要构建不必要的集合。 如果第一个列表比第二个列表长得多,并且如果哈希值很高,则性能差异将更加明显。 这是第二个测试,演示了这一点:
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
我有两个Python列表,如下所示:
temp1 = ['One', 'Two', 'Three', 'Four']
temp2 = ['One', 'Two']
我需要创建第三个列表,其中第一个列表中的项目不在第二个列表中。 从我必须得到的例子:
temp3 = ['Three', 'Four']
有没有循环和检查的快速方法?
如果你想递归的区别,我写了一个python包: https://github.com/seperman/deepdiff : https://github.com/seperman/deepdiff
安装
从PyPi安装:
pip install deepdiff
用法示例
输入
>>> from deepdiff import DeepDiff
>>> from pprint import pprint
>>> from __future__ import print_function # In case running on Python 2
相同的对象返回空
>>> t1 = {1:1, 2:2, 3:3}
>>> t2 = t1
>>> print(DeepDiff(t1, t2))
{}
项目的类型已更改
>>> t1 = {1:1, 2:2, 3:3}
>>> t2 = {1:1, 2:"2", 3:3}
>>> pprint(DeepDiff(t1, t2), indent=2)
{ 'type_changes': { 'root[2]': { 'newtype': <class 'str'>,
'newvalue': '2',
'oldtype': <class 'int'>,
'oldvalue': 2}}}
一个项目的价值已经改变
>>> t1 = {1:1, 2:2, 3:3}
>>> t2 = {1:1, 2:4, 3:3}
>>> pprint(DeepDiff(t1, t2), indent=2)
{'values_changed': {'root[2]': {'newvalue': 4, 'oldvalue': 2}}}
项目添加和/或删除
>>> t1 = {1:1, 2:2, 3:3, 4:4}
>>> t2 = {1:1, 2:4, 3:3, 5:5, 6:6}
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff)
{'dic_item_added': ['root[5]', 'root[6]'],
'dic_item_removed': ['root[4]'],
'values_changed': {'root[2]': {'newvalue': 4, 'oldvalue': 2}}}
字符串差异
>>> t1 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":"world"}}
>>> t2 = {1:1, 2:4, 3:3, 4:{"a":"hello", "b":"world!"}}
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff, indent = 2)
{ 'values_changed': { 'root[2]': {'newvalue': 4, 'oldvalue': 2},
"root[4]['b']": { 'newvalue': 'world!',
'oldvalue': 'world'}}}
字符串差异2
>>> t1 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":"world!\nGoodbye!\n1\n2\nEnd"}}
>>> t2 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":"world\n1\n2\nEnd"}}
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff, indent = 2)
{ 'values_changed': { "root[4]['b']": { 'diff': '--- \n'
'+++ \n'
'@@ -1,5 +1,4 @@\n'
'-world!\n'
'-Goodbye!\n'
'+world\n'
' 1\n'
' 2\n'
' End',
'newvalue': 'world\n1\n2\nEnd',
'oldvalue': 'world!\n'
'Goodbye!\n'
'1\n'
'2\n'
'End'}}}
>>>
>>> print (ddiff['values_changed']["root[4]['b']"]["diff"])
---
+++
@@ -1,5 +1,4 @@
-world!
-Goodbye!
+world
1
2
End
类型更改
>>> t1 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":[1, 2, 3]}}
>>> t2 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":"world\n\n\nEnd"}}
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff, indent = 2)
{ 'type_changes': { "root[4]['b']": { 'newtype': <class 'str'>,
'newvalue': 'world\n\n\nEnd',
'oldtype': <class 'list'>,
'oldvalue': [1, 2, 3]}}}
列表差异
>>> t1 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":[1, 2, 3, 4]}}
>>> t2 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":[1, 2]}}
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff, indent = 2)
{'iterable_item_removed': {"root[4]['b'][2]": 3, "root[4]['b'][3]": 4}}
列表差异2:
>>> t1 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":[1, 2, 3]}}
>>> t2 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":[1, 3, 2, 3]}}
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff, indent = 2)
{ 'iterable_item_added': {"root[4]['b'][3]": 3},
'values_changed': { "root[4]['b'][1]": {'newvalue': 3, 'oldvalue': 2},
"root[4]['b'][2]": {'newvalue': 2, 'oldvalue': 3}}}
列表差异忽略顺序或重复:(使用与上面相同的字典)
>>> t1 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":[1, 2, 3]}}
>>> t2 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":[1, 3, 2, 3]}}
>>> ddiff = DeepDiff(t1, t2, ignore_order=True)
>>> print (ddiff)
{}
包含词典的列表:
>>> t1 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":[1, 2, {1:1, 2:2}]}}
>>> t2 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":[1, 2, {1:3}]}}
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (ddiff, indent = 2)
{ 'dic_item_removed': ["root[4]['b'][2][2]"],
'values_changed': {"root[4]['b'][2][1]": {'newvalue': 3, 'oldvalue': 1}}}
集:
>>> t1 = {1, 2, 8}
>>> t2 = {1, 2, 3, 5}
>>> ddiff = DeepDiff(t1, t2)
>>> pprint (DeepDiff(t1, t2))
{'set_item_added': ['root[3]', 'root[5]'], 'set_item_removed': ['root[8]']}
命名的元组:
>>> from collections import namedtuple
>>> Point = namedtuple('Point', ['x', 'y'])
>>> t1 = Point(x=11, y=22)
>>> t2 = Point(x=11, y=23)
>>> pprint (DeepDiff(t1, t2))
{'values_changed': {'root.y': {'newvalue': 23, 'oldvalue': 22}}}
自定义对象:
>>> class ClassA(object):
... a = 1
... def __init__(self, b):
... self.b = b
...
>>> t1 = ClassA(1)
>>> t2 = ClassA(2)
>>>
>>> pprint(DeepDiff(t1, t2))
{'values_changed': {'root.b': {'newvalue': 2, 'oldvalue': 1}}}
添加对象属性:
>>> t2.c = "new attribute"
>>> pprint(DeepDiff(t1, t2))
{'attribute_added': ['root.c'],
'values_changed': {'root.b': {'newvalue': 2, 'oldvalue': 1}}}
两个列表(比如list1和list2)之间的区别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)
要么
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)
来找到diff(temp2, temp1)
。 两者都会给出结果['Four', 'Three']
。 您不必担心列表的顺序或首先给出哪个列表。
可以使用python XOR操作符完成。
- 这将删除每个列表中的重复项
- 这将显示temp1与temp1的temp1和temp2的差异。
set(temp1) ^ set(temp2)
这可能比Mark的列表理解速度更快:
filterfalse(set(temp2).__contains__, temp1)
最简单的方法,
使用set()。difference(set())
list_a = [1,2,3]
list_b = [2,3]
print set(list_a).difference(set(list_b))
答案已set([1])
可以打印列表,
print list(set(list_a).difference(set(list_b)))
如果遇到TypeError: unhashable type: 'list'
,则需要将列表或集合转换为元组,例如
set(map(tuple, list_of_lists1)).symmetric_difference(set(map(tuple, list_of_lists2)))
这是另一个解决方案:
def diff(a, b):
xa = [i for i in set(a) if i not in b]
xb = [i for i in set(b) if i not in a]
return xa + xb
如果你想要更像变更集的东西......可以使用Counter
from collections import Counter
def diff(a, b):
""" more verbose than needs to be, for clarity """
ca, cb = Counter(a), Counter(b)
to_add = cb - ca
to_remove = ca - cb
changes = Counter(to_add)
changes.subtract(to_remove)
return changes
lista = ['one', 'three', 'four', 'four', 'one']
listb = ['one', 'two', 'three']
In [127]: diff(lista, listb)
Out[127]: Counter({'two': 1, 'one': -1, 'four': -2})
# in order to go from lista to list b, you need to add a "two", remove a "one", and remove two "four"s
In [128]: diff(listb, lista)
Out[128]: Counter({'four': 2, 'one': 1, 'two': -1})
# in order to go from listb to lista, you must add two "four"s, add a "one", and remove a "two"
这可以用一条线解决。 问题是两个列表(temp1和temp2)在第三个列表(temp3)中返回它们的差异。
temp3 = list(set(temp1).difference(set(temp2)))
(list(set(a)-set(b))+list(set(b)-set(a)))