区分两个列表-Java 学习之路

569

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不存在于第二个列表中的项目 . 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？

25 回答

如果您想以递归方式获得差异，我已经为python编写了一个包：https://github.com/seperman/deepdiff

安装

从PyPi安装：

pip install deepdiff

示例用法

输入

>>> from deepdiff import DeepDiff
>>> from pprint import pprint
>>> from __future__ import print_function # In case running on Python 2

同一对象返回空

>>> t1 = {1:1, 2:2, 3:3}
>>> t2 = t1
>>> print(DeepDiff(t1, t2))
{}

项目类型已更改

>>> 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}}}

商品的 Value 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}}}

添加和/或删除项目

>>> 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

输入更改

>>> 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}}

清单差异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}}}

列出差异忽略顺序或重复:(使用与上面相同的词典）

>>> 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)
{}

包含字典的列表：

>>> 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}}}

添加了对象属性：

>>> t2.c = "new attribute"
>>> pprint(DeepDiff(t1, t2))
{'attribute_added': ['root.c'],
 'values_changed': {'root.b': {'newvalue': 2, 'oldvalue': 1}}}

回复于 2024-05-16T11:51:09+08:00

16
我想要一些需要两个列表的东西，并且可以在 bash 中执行 diff . 由于当您搜索"python diff two lists"并且不是非常具体时会首先弹出此问题，我将发布我想出的内容 .

使用 difflib 中的SequenceMather，您可以比较两个列表，如 diff . 其他答案都没有告诉你差异发生的位置，但这个答案会发生 . 有些答案只能在一个方向上产生差异 . 有些人重新排序元素 . 有些人不处理重复 . 但是这个解决方案为您提供了两个列表的真正区别：
```
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，如果您的应用程序做出与其他答案相同的假设，您将从中受益最多 . 但如果您正在寻找真正的 diff 功能，那么这是唯一的出路 .

例如，其他答案都无法处理：
```
a = [1,2,3,4,5]
b = [5,4,3,2,1]
```
但是这个做了：
```
2nd has [5, 4, 3, 2]
both have [1]
  1st has [2, 3, 4, 5]
```
回复于 2024-05-16T11:51:09+08:00
3
可以使用以下简单函数找到两个列表（例如list1和list2）之间的差异 .
```
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) 找到差异 . 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 .

Python doc reference
回复于 2024-05-16T11:51:09+08:00
5
对于最简单的情况，这是一个 Counter 答案 .

这比执行双向差异的那个要短，因为它只能完成问题所要求的：生成第一个列表中的内容而不是第二个列表中的内容 .
```
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，根据您的可读性偏好，它可以提供一个像样的单行：
```
diff = list((Counter(lst1) - Counter(lst2)).elements())
```
输出：
```
['Three', 'Four']
```
请注意，如果您只是迭代它，则可以删除 list(...) 调用 .

因为此解决方案使用计数器，所以它与许多基于集合的答案相比正确处理数量 . 例如，在此输入上：
```
lst1 = ['One', 'Two', 'Two', 'Two', 'Three', 'Three', 'Four']
lst2 = ['One', 'Two']
```
输出是：
```
['Two', 'Two', 'Three', 'Three', 'Four']
```
回复于 2024-05-16T11:51:09+08:00

(list(set(a)-set(b))+list(set(b)-set(a)))

回复于 2024-05-16T11:51:09+08:00

3

如果你真的在研究性能，那么使用numpy！

这是完整的笔记本作为github上的要点，列表，numpy和pandas之间的比较 .

https://gist.github.com/denfromufa/2821ff59b02e9482be15d27f2bbd4451

回复于 2024-05-16T11:51:09+08:00
884
这可以用一行解决 . 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)))
```
回复于 2024-05-16T11:51:09+08:00
8
因为现有的解决方案都没有产生元组，所以我会抛弃：
```
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))
```
就像其他非元组在这个方向上产生答案一样，它保留了秩序
回复于 2024-05-16T11:51:09+08:00

现有的解决方案都提供以下一种或另一种：

比O（n * m）性能快 .
保留输入列表的顺序 .

但到目前为止，没有任何解决方案 . 如果你想要两者，试试这个：

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

Performance test

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（稍微）比set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是昂贵的，则性能差异会更明显 . 这是第二次测试，证明了这一点：

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

回复于 2024-05-16T11:51:09+08:00

这是区分两个列表（无论内容是什么）的简单方法，您可以得到如下所示的结果：

>>> 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])

希望这会有所帮助 .

回复于 2024-05-16T11:51:09+08:00

我在游戏中为此已经太晚了，但你可以对上面提到的一些代码的性能进行比较，两个最快的竞争者是，

list(set(x).symmetric_difference(set(y)))
list(set(x) ^ set(y))

我为编码的基本级别道歉 .

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])

回复于 2024-05-16T11:51:09+08:00

0
可以使用python XOR运算符完成 .
- 这将删除每个列表中的重复项
- 这将显示temp1与temp1和temp2与temp1的区别 .
```
set(temp1) ^ set(temp2)
```
回复于 2024-05-16T11:51:09+08:00

单行版 arulmr 解决方案

def diff(listA, listB):
    return set(listA) - set(listB) | set(listA) -set(listB)

回复于 2024-05-16T11:51:09+08:00

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

回复于 2024-05-16T11:51:09+08:00

假设我们有两个列表

list1 = [1, 3, 5, 7, 9]
list2 = [1, 2, 3, 4, 5]

我们可以从上面的两个列表中看到，项目1,3,5存在于list2中，而项目7,9则不存在 . 上另一方面，项目1,3,5存在于list1中，而项目2,4则不存在 .

返回包含项目7,9和2,4的新列表的最佳解决方案是什么？

上面的所有答案都找到了解决方案，现在最优化的是什么？

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')

回报

[7, 9, 2, 4]
Using two for loops 0.11572412995155901 Milliseconds
Using symmetric_difference 0.11285737506113946 Milliseconds

Process finished with exit code 0

回复于 2024-05-16T11:51:09+08:00

temp3 = [item for item in temp1 if item not in temp2]

回复于 2024-05-16T11:51:09+08:00

8
试试这个：
```
temp3 = set(temp1) - set(temp2)
```
回复于 2024-05-16T11:51:09+08:00

如果difflist的元素已排序并设置，则可以使用naive方法 .

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

print list1[len(list2):]

或使用本机设置方法：

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)

天真的解决方案：0.0787101593292

原生集解决方案：0.998837615564

回复于 2024-05-16T11:51:09+08:00

10
如果您遇到 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)))
```
另见How to compare a list of lists/sets in python?
回复于 2024-05-16T11:51:09+08:00

我们可以计算交集减去联合列表：

temp1 = ['One', 'Two', 'Three', 'Four']
temp2 = ['One', 'Two', 'Five']

set(temp1+temp2)-(set(temp1)&set(temp2))

Out: set(['Four', 'Five', 'Three'])

回复于 2024-05-16T11:51:09+08:00

2
这可能比马克的列表理解速度更快：
```
list(itertools.filterfalse(set(temp2).__contains__, temp1))
```
回复于 2024-05-16T11:51:09+08:00
4
以下是一些简单的方法来区分两个字符串列表 .

Code

使用pathlib的一种不寻常的方法：
```
import pathlib


temp1 = ["One", "Two", "Three", "Four"]
temp2 = ["One", "Two"]

p = pathlib.Path(*temp1)
r = p.relative_to(*temp2)
list(r.parts)
# ['Three', 'Four']
```
这假设两个列表都包含具有等效开头的字符串 . 有关详细信息，请参阅docs . 注意，与设置操作相比，它不是特别快 .

使用itertools.zip_longest的直接实现：
```
import itertools as it


[x for x, y in it.zip_longest(temp1, temp2) if x != y]
# ['Three', 'Four']
```
回复于 2024-05-16T11:51:09+08:00
2
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))
```
答案是 set([1])

可以打印为列表，
```
print list(set(list_a).difference(set(list_b)))
```
回复于 2024-05-16T11:51:09+08:00

400

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"

回复于 2024-05-16T11:51:09+08:00

-1
```
In [5]: list(set(temp1) - set(temp2))
Out[5]: ['Four', 'Three']
```
要小心
```
In [5]: set([1, 2]) - set([2, 3])
Out[5]: set([1])
```
您可能期望/希望它等于 set([1, 3]) . 如果您确实想要 set([1, 3]) 作为答案，则需要使用 set([1, 2]).symmetric_difference(set([2, 3])) .
回复于 2024-05-16T11:51:09+08:00

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