Question

实施例：

list = [['1', '13/12/2016', [42, 52]], ['2', '12/12/2016', [36, 46]], ['4', '10/12/2016', [13, 23]], ['4', '10/12/2016', [42, 52]], ['5', '08/12/2016', [42, 52]], ['6', '07/12/2016', [32, 42]], ['7', '12/12/2016', [42, 52]], ['8', '06/12/2016', [42, 52]], ['10', '12/12/2016', [45, 55]], ['11', '08/12/2016', [42, 52]]]

我想检查每个子列表中的所有第二个元素是否重复如果第二个元素在另一个子列表中重复，我想保留整个子列表

预期结果：

endlist = [['2', '12/12/2016', [36, 46]], ['4', '10/12/2016', [13, 23]], ['4', '10/12/2016', [42, 52]], ['5', '08/12/2016', [42, 52]], ['7', '12/12/2016', [42, 52]], ['10', '12/12/2016', [45, 55]], ['11', '08/12/2016', [42, 52]]]

（2016年8月12日| 10/12/2016 | 12/12/2016是双打）

我知道如何使用平面列表([x for x in l if l.count(x) > 1])保留双打，但如何在包含子列表的列表中执行此操作？

Answer 1

您可以使用列表推导：

lst = [['1', '13/12/2016', [42, 52]], ['2', '12/12/2016', [36, 46]], ['4', '10/12/2016', [13, 23]], ['4', '10/12/2016', [42, 52]], ['5', '08/12/2016', [42, 52]], ['6', '07/12/2016', [32, 42]], ['7', '12/12/2016', [42, 52]], ['8', '06/12/2016', [42, 52]], ['10', '12/12/2016', [45, 55]], ['11', '08/12/2016', [42, 52]]]
endlist = [sublist for sublist in lst if sum(x[1] == sublist[1] for x in lst) > 1]
# [['2', '12/12/2016', [36, 46]], ['4', '10/12/2016', [13, 23]], ['4', '10/12/2016', [42, 52]], ['5', '08/12/2016', [42, 52]], ['7', '12/12/2016', [42, 52]], ['10', '12/12/2016', [45, 55]], ['11', '08/12/2016', [42, 52]]]

Answer 2

您可以将计数信息收集到字典中然后使用它。如果列表很大，这将很好地扩展：

myList = [['1', '13/12/2016', [42, 52]], ['2', '12/12/2016', [36, 46]], ['4', '10/12/2016', [13, 23]], ['4', '10/12/2016', [42, 52]], ['5', '08/12/2016', [42, 52]], ['6', '07/12/2016', [32, 42]], ['7', '12/12/2016', [42, 52]], ['8', '06/12/2016', [42, 52]], ['10', '12/12/2016', [45, 55]], ['11', '08/12/2016', [42, 52]]]

d = dict()
for subList in myList:
    if subList[1] in d:
        d[subList[1]] += 1
    else:
        d[subList[1]] = 1

doubles = [subList for subList in myList if d[subList[1]] >= 2]

如果您希望>=2排除三元组等，您当然可以将==2替换为doubles。

编辑：如果要保留每个双重子列表的第一个匹配项，请修改字典，以便存储每个元素的索引。像这样：

d = dict()
for i,subList in enumerate(myList):
    if subList[1] in d:
        d[subList[1]].append(i)
    else:
        d[subList[1]] = [i]

firsts = [subList for i,subList in enumerate(myList) if len(d[subList[1]]) >= 2 and i == d[subList[1]][0]]
print(firsts) #prints [['2', '12/12/2016', [36, 46]], ['4', '10/12/2016', [13, 23]], ['5', '08/12/2016', [42, 52]]]

进一步编辑：以下是删除后续双打的解决方案：

d = dict()

for i,subList in enumerate(myList):
    if not subList[1] in d:
        d[subList[1]] = i #stores first index

noDoubles = [subList for i,subList in enumerate(myList) if i == d[subList[1]]]

如果第二个元素重复，则仅将子列表保留在列表中

2 个答案: