获取两个列表之间的差异
-
27-09-2019 - |
题
我在 Python 中有两个列表,如下所示:
temp1 = ['One', 'Two', 'Three', 'Four']
temp2 = ['One', 'Two']
我需要创建第三个列表,其中包含第一个列表中的项目,但第二个列表中不存在。从这个例子中我必须得到:
temp3 = ['Three', 'Four']
有没有不需要循环和检查的快速方法?
解决方案
In [5]: list(set(temp1) - set(temp2))
Out[5]: ['Four', 'Three']
要注意的是
In [5]: set([1, 2]) - set([2, 3])
Out[5]: set([1])
在这里你可能期望/想让它等于set([1, 3])
。如果你想set([1, 3])
为你的答案,你需要使用set([1, 2]).symmetric_difference(set([2, 3]))
。
其他提示
现有的解决方案都提供以下之一或另一个:
- 比 O(n*m) 性能更快。
- 保留输入列表的顺序。
但到目前为止还没有解决方案两者兼而有之。如果您两者都想要,请尝试以下操作:
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留顺序也(稍微)比集合减法更快,因为它不需要构造不必要的集合。如果第一个列表比第二个列表长得多并且散列成本很高,则性能差异会更加明显。这是证明这一点的第二个测试:
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
temp3 = [item for item in temp1 if item not in temp2]
两个列表(比如list1的和list2中)之间的差异可以通过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)
或
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,差值可以利用diff(temp2, temp1)
或diff(temp1, temp2)
找到。双方将给出结果['Four', 'Three']
。你不必担心列表的顺序或列表将被首先给出。
如果你想递归的区别,我已经写了一包蟒蛇: https://github.com/seperman/deepdiff
安装
从安装的PyPI:
pip install deepdiff
用法示例
导入
>>> 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))
{}
的项目的类型已经改变了
>>> 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}}}
的项的值已经改变
>>> 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/或除去
>>> 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}}}
的字符串差异
>>> 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'}}}
的字符串差异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
类型变化
>>> 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
>>> 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}}}
列表差忽略顺序或重复:(具有相同的字典如上)
>>> 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)
{}
列表包含词典:
>>> 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}}}
设定:
>>> 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}}}
如果你真的寻找到性能,然后使用numpy的!
下面是完整笔记本作为在github要旨与列表,numpy的,和熊猫之间的比较。
https://gist.github.com/denfromufa/2821ff59b02e9482be15d27f2bbd4451
最简单的方式,
使用的集()。差(设定())强>
list_a = [1,2,3]
list_b = [2,3]
print set(list_a).difference(set(list_b))
答案是set([1])
可以打印作为一个列表,
print list(set(list_a).difference(set(list_b)))
我将在因为没有本发明的溶液的产生元组抛:
temp3 = tuple(set(temp1) - set(temp2))
或者:
#edited using @Mark Byers idea. If you accept this one as answer, just accept his instead.
temp3 = tuple(x for x in temp1 if x not in set(temp2))
如在此方向上的另一非元组产生的答案,它保留顺序
可以使用 python XOR 运算符来完成。
- 这将删除每个列表中的重复项
- 这将显示 temp1 与 temp2 以及 temp2 与 temp1 的差异。
set(temp1) ^ set(temp2)
我想要的东西,将采取两个列表,并可以做什么diff
在bash
一样。由于这个问题弹出第一,当你搜索“蟒蛇DIFF两个列表”,是不是很具体,我会寄我想出了。
使用 SequenceMather
从difflib
你可以比较两个列表像diff
一样。其他的答案都不会告诉你区别在哪里发生的位置,但是这一次呢。一些答案给只在一个方向上的差异。一些重新排序的元素。有些不处理重复。但这种解决方案为您提供了两个列表之间的真正区别:
a = 'A quick fox jumps the lazy dog'.split()
b = 'A quick brown mouse jumps over the dog'.split()
from difflib import SequenceMatcher
for tag, i, j, k, l in SequenceMatcher(None, a, b).get_opcodes():
if tag == 'equal': print('both have', a[i:j])
if tag in ('delete', 'replace'): print(' 1st has', a[i:j])
if tag in ('insert', 'replace'): print(' 2nd has', b[k:l])
这个输出:
both have ['A', 'quick']
1st has ['fox']
2nd has ['brown', 'mouse']
both have ['jumps']
2nd has ['over']
both have ['the']
1st has ['lazy']
both have ['dog']
当然,如果你的应用程序进行同样的假设其他的答案让,你会从中受益最多。但是,如果你正在寻找一个真正的diff
功能,那么这是唯一的出路。
例如,没有其他的答案可以处理:
a = [1,2,3,4,5]
b = [5,4,3,2,1]
但是,这一个作用:
2nd has [5, 4, 3, 2]
both have [1]
1st has [2, 3, 4, 5]
尝试这种情况:
temp3 = set(temp1) - set(temp2)
这可能是速度甚至超过了马克的列表理解:
list(itertools.filterfalse(set(temp2).__contains__, temp1))
下面是最简单的情况下的Counter
答案。
这是短于上面说的人做两路比较,因为它只做这个问题问到底是什么:产生一个什么样的第一列表中,但不是第二列表
from collections import Counter
lst1 = ['One', 'Two', 'Three', 'Four']
lst2 = ['One', 'Two']
c1 = Counter(lst1)
c2 = Counter(lst2)
diff = list((c1 - c2).elements())
或者,根据你的可读性首选项,它使一个体面一行程序:
diff = list((Counter(lst1) - Counter(lst2)).elements())
输出:
['Three', 'Four']
请注意,您可以删除list(...)
通话,如果你只是遍历它。的
由于该解决方案使用计数器,它处理量正确地VS的许多基于集合的答案。例如在此输入:
lst1 = ['One', 'Two', 'Two', 'Two', 'Three', 'Three', 'Four']
lst2 = ['One', 'Two']
的输出是:
['Two', 'Two', 'Three', 'Three', 'Four']
您如果difflist的元素进行排序,并设置可以使用幼稚方法。
list1=[1,2,3,4,5]
list2=[1,2,3]
print list1[len(list2):]
或与天然集方法:
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)
朴素溶液:0.0787101593292
本地组解决方案:0.998837615564
我是小游戏中的太晚了这一点,但你可以做的一些性能比较上面这个代码提到两个速度最快的竞争者是,
list(set(x).symmetric_difference(set(y)))
list(set(x) ^ set(y))
我用于编码的基本水平道歉。
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])
这是另一种解决方案:
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
如果您遇到TypeError: unhashable type: 'list'
您需要打开列表或设置成元组,e.g。
set(map(tuple, list_of_lists1)).symmetric_difference(set(map(tuple, list_of_lists2)))
下面是一些简单的, 保序的版本比较字符串两个列表的方式。
<强>代码强>
使用 pathlib
一个不寻常的方法:
import pathlib
temp1 = ["One", "Two", "Three", "Four"]
temp2 = ["One", "Two"]
p = pathlib.Path(*temp1)
r = p.relative_to(*temp2)
list(r.parts)
# ['Three', 'Four']
此假定这两个列表中包含与相当于开端字符串。见文档了解更多详情。注意,这不是特别快相比的一组操作。
使用直进实施 itertools.zip_longest
:
import itertools as it
[x for x, y in it.zip_longest(temp1, temp2) if x != y]
# ['Three', 'Four']
的单个行版本的 arulmr 强>溶液
def diff(listA, listB):
return set(listA) - set(listB) | set(listA) -set(listB)
如果你想要的东西更像是一个变更......可以使用计数器
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"
我们可以计算出列表的交集减去联合:
temp1 = ['One', 'Two', 'Three', 'Four']
temp2 = ['One', 'Two', 'Five']
set(temp1+temp2)-(set(temp1)&set(temp2))
Out: set(['Four', 'Five', 'Three'])
此可与一种线来解决。 问题是给定的两个列表(temp1中和TEMP2)在第三列表(TEMP3)返回它们的差。
temp3 = list(set(temp1).difference(set(temp2)))
让我们说我们有两个列表
list1 = [1, 3, 5, 7, 9]
list2 = [1, 2, 3, 4, 5]
我们可以从上面的两个列表看到的项目1,3,5存在于list2中和,9项7没有。在另一方面,项目1,3,5存在于list1的和项2,4没有。
什么是返回包含,9,4项目7和2的新列表的最佳解决方案?
上述所有的答案找到解决办法,现在什么最优化?
def difference(list1, list2):
new_list = []
for i in list1:
if i not in list2:
new_list.append(i)
for j in list2:
if j not in list1:
new_list.append(j)
return new_list
与
def sym_diff(list1, list2):
return list(set(list1).symmetric_difference(set(list2)))
使用timeit我们可以看到结果
t1 = timeit.Timer("difference(list1, list2)", "from __main__ import difference,
list1, list2")
t2 = timeit.Timer("sym_diff(list1, list2)", "from __main__ import sym_diff,
list1, list2")
print('Using two for loops', t1.timeit(number=100000), 'Milliseconds')
print('Using two for loops', t2.timeit(number=100000), 'Milliseconds')
返回
[7, 9, 2, 4]
Using two for loops 0.11572412995155901 Milliseconds
Using symmetric_difference 0.11285737506113946 Milliseconds
Process finished with exit code 0
下面是如下所示来区分两个列表(无论内容是),则可以得到的结果的简单的方法:
>>> from sets import Set
>>>
>>> l1 = ['xvda', False, 'xvdbb', 12, 'xvdbc']
>>> l2 = ['xvda', 'xvdbb', 'xvdbc', 'xvdbd', None]
>>>
>>> Set(l1).symmetric_difference(Set(l2))
Set([False, 'xvdbd', None, 12])
希望这将有帮助。
(list(set(a)-set(b))+list(set(b)-set(a)))