Question

我需要减去两个这样的2D列表：

list1= [['some',2],['other',1],['thing',5]]

list2= [['some',1],['thing',5]]

结果应该是这样的：

result= [['some',1],['other',1],['thing',0]]

或

result= [['some',1],['other',1]]

它应该是一个列表，而不是字典;订单并不重要。

Answer 1

from collections import Counter
list1= [['some',2],['other',1],['thing',5]]

list2= [['some',1],['thing',5]]
c1 = Counter({item[0]: item[1] for item in list1})
c2 = Counter({item[0]: item[1] for item in list2})
result = [[key, value] for key, value in (c1-c2).items()]

result
[['other', 1], ['some', 1]]

Answer 2

我得到了这样的人：

tmp=[[k,v] for k,v in (Counter(dict(list1)) - Counter(dict(list2))).items()]

Ty伙伴们寻求答案！

Answer 3

>>> d1, d2 = dict(list1), dict(list2)
>>> d = {key: d1[key] - d2.get(key, 0) for key in d1}
>>> d.items()
[('thing', 0), ('other', 1), ('some', 1)]
>>> map(list, d.items())
[['thing', 0], ['other', 1], ['some', 1]]

Answer 4

代码：

list1 = [['some',2], ['other',1], ['thing',5]]
list2 = [['some',1], ['thing',5]]

def list_substractor(l1, l2):
    d1 = dict(l1)
    d2 = dict(l2)

    for key, value in d1.items():
        if key in d2.keys():
            d1[key] = d1[key] - d2[key]

    for item in set( set(d2.keys()) - set(d1.keys())):
        d1[item] = -d2[item]

    return [[key, value] for key, value in d1.items()]

print list_substractor(list1, list2)
print list_substractor(list2, list1)

输出：

[['thing', 0], ['other', 1], ['some', 1]]
[['thing', 0], ['other', -1], ['some', -1]]

如何在python中减去两个2D列表？

4 个答案: