在python中：两个列表之间的区别

I have two lists like so

我有两个这样的清单

found = ['CG', 'E6', 'E1', 'E2', 'E4', 'L2', 'E7', 'E5', 'L1', 'E2BS', 'E2BS', 'E2BS', 'E2', 'E1^E4', 'E5']
expected = ['E1', 'E2', 'E4', 'E1^E4', 'E6', 'E7', 'L1', 'L2', 'CG', 'E2BS', 'E3']

I want to find the differences between both lists.
I have done

我想找到两个列表之间的差异。我已经做好了

list(set(expected)-set(found))

and

list(set(found)-set(expected))

Which returns ['E3'] and ['E5'] respectively.

它分别返回['E3']和['E5']。

However, the answers I need are:

但是,我需要的答案是:

'E3' is missing from found.
'E5' is missing from expected.
There are 2 copies of 'E5' in found.
There are 3 copies of 'E2BS' in found.
There are 2 copies of 'E2' in found.

Any help/suggestions are welcome!

欢迎任何帮助/建议!

3 个解决方案

#1

The collections.Counter class will excel at enumerating the differences between multisets:

collections.Counter类将擅长枚举多重集之间的差异:

>>> from collections import Counter
>>> found = Counter(['CG', 'E6', 'E1', 'E2', 'E4', 'L2', 'E7', 'E5', 'L1', 'E2BS', 'E2BS', 'E2BS', 'E2', 'E1^E4', 'E5'])
>>> expected = Counter(['E1', 'E2', 'E4', 'E1^E4', 'E6', 'E7', 'L1', 'L2', 'CG', 'E2BS', 'E3'])
>>> list((found - expected).elements())
['E2', 'E2BS', 'E2BS', 'E5', 'E5']
>>> list((expected - found).elements())

You might also be interested in difflib.Differ:

您可能也对difflib.Differ感兴趣:

>>> from difflib import Differ
>>> found = ['CG', 'E6', 'E1', 'E2', 'E4', 'L2', 'E7', 'E5', 'L1', 'E2BS', 'E2BS', 'E2BS', 'E2', 'E1^E4', 'E5']
>>> expected = ['E1', 'E2', 'E4', 'E1^E4', 'E6', 'E7', 'L1', 'L2', 'CG', 'E2BS', 'E3']
>>> for d in Differ().compare(expected, found):
...     print(d)

+ CG
+ E6
  E1
  E2
  E4
+ L2
+ E7
+ E5
+ L1
+ E2BS
+ E2BS
+ E2BS
+ E2
  E1^E4
+ E5
- E6
- E7
- L1
- L2
- CG
- E2BS
- E3

#2

Leverage the Python set class and Counter class instead of rolling your own solution:

利用Python集合类和Counter类,而不是滚动自己的解决方案:

1. symmetric_difference: finds elements that are either in one set or the other, but not both.

symmetric_difference:查找一组或另一组中的元素,但不是两者。

2. intersection: finds elements in common with the two sets.

intersection:找到两个集合的共同元素。

3. difference: which is essentially what you did by subtracting one set from another

差异:这基本上是你从另一个中减去一个集合所做的

Code examples

• found.difference(expected) # set(['E5'])
  

found.difference(expected)#set(['E5'])

• expected.difference(found) # set(['E3'])
  

expected.difference(found)#set(['E3'])

• found.symmetric_difference(expected) # set(['E5', 'E3'])
  

found.symmetric_difference(expected)#set(['E5','E3'])

• Finding copies of objects: this question was already referenced. Using that technique gets you all duplicates, and using the resultant Counter object, you can find how many duplicates. For example:

  查找对象的副本:此问题已被引用。使用该技术可以获得所有重复项,并使用生成的Counter对象,您可以找到多少重复项。例如:

  collections.Counter(found)['E5'] # 2
  

#3

You've already answered the first two:

你已经回答了前两个问题:

print('{0} missing from found'.format(list(set(expected) - set(found)))
print('{0} missing from expected'.format(list(set(found) - set(expected)))

The second two require you to look at counting duplicates in lists, for which there are many solutions to be found online (including this one: Find and list duplicates in a list?).

后两个要求你看一下计算列表中的重复项,有很多解决方案可以在网上找到(包括这一个:在列表中查找和列出重复项?)。