Question

我有一个包含列表列表的字典，如下所示：

{'S26': [['2016-03-18', '2016-03-28'], ['2016-03-18', '2016-03-28']], 'S24': [['2016-03-19', '2016-03-25'], ['2016-03-25', '2016-04-03']], 'S25': [['2016-03-18', '2016-03-25'], ['2016-03-18', '2016-03-25'], ['2016-03-20', '2016-03-25'], ['2016-03-20', '2016-03-25'], ['2016-03-25', '2016-04-03']]}

我想浏览每个列表并删除任何重复的子列表。因此，例如，键S26在列表中有两个重复：

'S26': [['2016-03-18', '2016-03-28'], ['2016-03-18', '2016-03-28']]

此密钥现在应为：

'S26': [['2016-03-18', '2016-03-28']]

这可能吗？该系统上的python版本是2.6.6

Answer 1

for val in dict:
    tmp=[]
    for l in dict[val]:
        if l not in tmp: tmp.append(l)
    dict[val]=tmp

Answer 2

这个也适用：

>>> from collections import defaultdict
>>>
>>> d = {'S26': [['2016-03-18', '2016-03-28'], ['2016-03-18', '2016-03-28']], 'S24': [['2016-03-19', '2016-03-25'], ['2016-03-25', '2016-04-03']], 'S25': [['2016-03-18', '2016-03-25'], ['2016-03-18', '2016-03-25'], ['2016-03-20', '2016-03-25'], ['2016-03-20', '2016-03-25'], ['2016-03-25', '2016-04-03']]}
>>> 
>>> out_d = defaultdict(list)
>>> for k,v in d.items():
        for vv in v:
            if vv not in out_d[k]:
                out_d[k].append(vv)

>>> out_d
defaultdict(<class 'list'>, {'S24': [['2016-03-19', '2016-03-25'], ['2016-03-25', '2016-04-03']], 'S26': [['2016-03-18', '2016-03-28']], 'S25': [['2016-03-18', '2016-03-25'], ['2016-03-20', '2016-03-25'], ['2016-03-25', '2016-04-03']]})

Answer 3

您可以使用以下内容：

tnsnames.ora

<强>输出

.exe

这将迭代字典中的每个d = {'S26': [['2016-03-18', '2016-03-28'], ['2016-03-18', '2016-03-28']], 'S24': [['2016-03-19', '2016-03-25'], ['2016-03-25', '2016-04-03']], 'S25': [['2016-03-18', '2016-03-25'], ['2016-03-18', '2016-03-25'], ['2016-03-20', '2016-03-25'], ['2016-03-20', '2016-03-25'], ['2016-03-25', '2016-04-03']]} answer = {k:[list(el) for el in set([tuple(sublist) for sublist in v])] for k, v in d.items()} print(answer)元素对，并从列表值中删除重复的子列表。为此，我们使用列表推导将每个{'S24': [['2016-03-19', '2016-03-25'], ['2016-03-25', '2016-04-03']], 'S25': [['2016-03-20', '2016-03-25'], ['2016-03-18', '2016-03-25'], ['2016-03-25', '2016-04-03']], 'S26': [['2016-03-18', '2016-03-28']]}转换为(k, v)，以便我们可以创建sublist（因为列表不可用）。

Answer 4

如果您需要保留列表的顺序，可以使用以下方法：

from collections import OrderedDict

dlol = {
    'S26': [['2016-03-18', '2016-03-28'], ['2016-03-18', '2016-03-28']], 
    'S24': [['2016-03-19', '2016-03-25'], ['2016-03-25', '2016-04-03']], 
    'S25': [['2016-03-18', '2016-03-25'], ['2016-03-18', '2016-03-25'], ['2016-03-20', '2016-03-25'], ['2016-03-20', '2016-03-25'], ['2016-03-25', '2016-04-03']]}

output = {}

for k, lol in dlol.items():
    output[k] = OrderedDict([(tuple(e), None) for e in lol])

output = {k: list(list(e) for e in v) for k, v in output.items()}    
print(output)

给你：

{'S25': [['2016-03-18', '2016-03-25'], ['2016-03-20', '2016-03-25'], ['2016-03-25', '2016-04-03']], 
'S24': [['2016-03-19', '2016-03-25'], ['2016-03-25', '2016-04-03']], 
'S26': [['2016-03-18', '2016-03-28']]}

从字典中的每个值中删除重复项

4 个答案: