Question

起初，我道歉是因为我是新手...

我有一个3d元组列表：

list=[(0, 1, 6), (5,1,4), (1, 6, 0), (3, 2,1),(4,5,1)]

我想找到重复项并像这样显示，无论数字的位置如何，只具有相同的数字：

{ (0,1,6): 2,
  (4,5,1): 2,
  (3,2,1): 1 }

我想计算相似的元组。

有什么建议吗？

Answer 1

使用Counter中的collections：

>>> from collections import Counter
>>> list=[(0, 1, 6), (5,1,4), (1, 6, 0), (3, 2,1),(4,5,1)]
>>> dict(Counter(map(lambda x: tuple(sorted(x)),list)))
{(0, 1, 6): 2, (1, 4, 5): 2, (1, 2, 3): 1}
>>>

或列表样式的理解：

>>> from collections import Counter
>>> list=[(0, 1, 6), (5,1,4), (1, 6, 0), (3, 2,1),(4,5,1)]
>>> dict(Counter([tuple(sorted(i)) for i in list]))
{(0, 1, 6): 2, (1, 4, 5): 2, (1, 2, 3): 1}
>>>

此外，list是python关键字的副本，在创建了变量之后，您将无法再访问该关键字，因此最好重命名该变量，并在我的计算机中重命名该变量。代码。

编辑：

使用：

dict(Counter([tuple(sorted(x)) for i in list for x in i]))

Answer 2

import pandas as pd
import numpy as np

# make a dataframe out of the list
df = pd.DataFrame(data=ll)

# sort columns and find duplicates
df_dup = df[pd.DataFrame(np.sort(df.values), columns=df.columns, index=df.index)
               .duplicated(keep=False)]

# return duplicates as dictionary of tuples
result = df_dup.T.apply(tuple).to_dict()

# return number of duplicates found
n = len(result)

Answer 3

list=[(0, 1, 6), (5,1,4), (1, 6, 0), (3, 2,1),(4,5,1)]

#first please make a list of sorted tuples to ignore the sequence of numbers
sorted_list = [sorted(x) for x in list]

#then count the tuple in sorted list and ignore count for duplicated items.
list = [ { x: sorted_list.count(sorted(x)) } for i, x in enumerate(list) if sorted(x) not in sorted_list[:i]]

print(list);

Answer 4

您说您正在使用以下方法初始化列表：

mylist = [[[(x,y,z) for x in range(7)] for y in range(7)] for z in range(7)]

此时，您可以更改初始化方式：

from collections import Counter

# take off some square brackets
mylist = [(x,y,z) for x in range(7) for y in range(7) for z in range(7)]

dict(Counter(tuple(sorted(tup)) for tup in mylist))

输出：

{(0, 0, 0): 1, (0, 0, 1): 3, (0, 0, 2): 3, (0, 0, 3): 3, (0, 0, 4): 3, (0, 0, 5): 3, (0, 0, 6): 3, (0, 1, 1): 3, (0, 1, 2): 6, (0, 1, 3): 6, (0, 1, 4): 6, (0, 1, 5): 6, (0, 1, 6): 6, (0, 2, 2): 3, (0, 2, 3): 6, (0, 2, 4): 6, (0, 2, 5): 6, (0, 2, 6): 6, (0, 3, 3): 3, (0, 3, 4): 6, (0, 3, 5): 6, (0, 3, 6): 6, (0, 4, 4): 3, (0, 4, 5): 6, (0, 4, 6): 6, (0, 5, 5): 3, (0, 5, 6): 6, (0, 6, 6): 3, (1, 1, 1): 1, (1, 1, 2): 3, (1, 1, 3): 3, (1, 1, 4): 3, (1, 1, 5): 3, (1, 1, 6): 3, (1, 2, 2): 3, (1, 2, 3): 6, (1, 2, 4): 6, (1, 2, 5): 6, (1, 2, 6): 6, (1, 3, 3): 3, (1, 3, 4): 6, (1, 3, 5): 6, (1, 3, 6): 6, (1, 4, 4): 3, (1, 4, 5): 6, (1, 4, 6): 6, (1, 5, 5): 3, (1, 5, 6): 6, (1, 6, 6): 3, (2, 2, 2): 1, (2, 2, 3): 3, (2, 2, 4): 3, (2, 2, 5): 3, (2, 2, 6): 3, (2, 3, 3): 3, (2, 3, 4): 6, (2, 3, 5): 6, (2, 3, 6): 6, (2, 4, 4): 3, (2, 4, 5): 6, (2, 4, 6): 6, (2, 5, 5): 3, (2, 5, 6): 6, (2, 6, 6): 3, (3, 3, 3): 1, (3, 3, 4): 3, (3, 3, 5): 3, (3, 3, 6): 3, (3, 4, 4): 3, (3, 4, 5): 6, (3, 4, 6): 6, (3, 5, 5): 3, (3, 5, 6): 6, (3, 6, 6): 3, (4, 4, 4): 1, (4, 4, 5): 3, (4, 4, 6): 3, (4, 5, 5): 3, (4, 5, 6): 6, (4, 6, 6): 3, (5, 5, 5): 1, (5, 5, 6): 3, (5, 6, 6): 3, (6, 6, 6): 1}

如果您不想更改初始化方式：

def flatten(alist):
    for item in alist:
        if isinstance(item, list):
            yield from flatten(item)
        else:
            yield item

dict(Counter(tuple(sorted(tup)) for tup in flatten(mylist)))

Answer 5

建议，不要在python中使用关键字作为变量，这是python中的关键字：

import keyword
keyword_list = keyword.kwlist

对于您的问题，这是我的答案：

from collections import Counter

result = dict(Counter([tuple(sorted(item)) for item in your_list ]))

在3维元组列表中查找重复项

5 个答案: