[英]in python: difference between two lists
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']
我想找到两个列表之间的差异。
我已经做好了
list(set(expected)-set(found))
和
list(set(found)-set(expected))
它分别返回['E3']
和['E5']
。
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.
欢迎任何帮助/建议!
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())
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
利用Python set
类和Counter
类,而不是滚动自己的解决方案:
symmetric_difference
:查找一组或另一组中的元素,但不是两者。 intersection
:找到与两个集合相同的元素。 difference
:这基本上是你从另一个中减去一个集合所做的 found.difference(expected) # set(['E5'])
expected.difference(found) # set(['E3'])
found.symmetric_difference(expected) # set(['E5', 'E3'])
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
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)))
后两个要求你看看列表中的重复计数,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? )。
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