如何检查列表列表中是否存在重复列表？

Question

我有一个列表RESULT包含列表。我想在RESULT中添加一个列表，如果它不存在的话。

所以

input = [1,2,3]
RESULT = [[5,6], [4,5,8]]

现在，RESULT.append(input)给出了

RESULT = [[5,6], [4,5,8], [1,2,3]]

现在，如果我尝试append [5,6]它不应该添加，因为它已经存在。

我不能在这里使用set那么替代方案是什么？

Answer 1

def add(data_, value):
    if value not in data_:
        data_.append(value)

data = [[5, 6], [4, 5, 8]]
print(data)  # [[5, 6], [4, 5, 8]]
add(data, [1, 2, 3])
print(data)  # {(5, 6), (4, 5, 8), (1, 2, 3)}
add(data, [5, 6])
print(data)  # {(5, 6), (4, 5, 8), (1, 2, 3)}

Answer 2

你可以使用itertools.groupby()：

no_dupes = list(ls for ls, _ in itertools.groupby(ls))

然后检查它：

if ls == no_dupes:
     # Do x

Answer 3

最简单的解决方案可能是使用if语句首先检查[5,6]是否已经在RESULT中，如果没有，append它，否则继续，可能向用户报告它是重复的而不是附加：

myinput = [1,2,3]
RESULT = [[5,6], [4,5,8]]

RESULT.append(myinput)

l = [5,6]

if l not in RESULT:
    RESULT.append(l)
else:
    # Do something with RESULT 
    pass # or
    # print('Duplicate not appended')
    print(f'RESULT: {RESULT}')
    raise(Exception(f'{l} is a duplicate and thus was not appended'))

输出：

RESULT: [[5, 6], [4, 5, 8], [1, 2, 3]]
Traceback (most recent call last):
  File "main.py", line 15, in <module>
    raise(Exception(f'{l} is a duplicate and thus was not appended'))
Exception: [5, 6] is a duplicate and thus was not appended

Answer 4

input = [1,2,3]
RESULT = [[5,6], [4,5,8]]

假设在追加RESULT.append(input)之后

RESULT=[[5,6], [4,5,8], [1,2,3]]

此特定代码的基本思路：

用于检查：

i=0    
count=0
while i<3:
  if input == RESULT[i]:
     count=count+1
  i = i + 1
if count==0:
    RESULT.append(input)
print(RESULT)