从 python 嵌套列表中以任意顺序删除重复字符串

Question

[["hello", "bye", "start"], ["bye", "start", "hello"], ["john", "riya", "tom"], ["riya","john", "tom"].....]

我有一个这样的列表。我想从 Python 中的嵌套列表中删除重复元素，其中元素应按任何顺序排列。

输出应该是：-

[["hello", "bye", "start"], ["john", "riya", "tom"]]

3 个字符串在任何列表中只能出现一次。如何实现？

Answer 1

一个简单而天真的解决方案

arr = [["hello", "bye", "start"], ["bye", "start", "hello"], ["john", "riya", "tom"], ["riya","john", "tom"]]

seen, tmp, result = list()

for i in arr:
  for j in i:
    if j not in seen:
      seen.append(j)
      tmp.append(j)

      if len(tmp) == 3:
        result.append(tmp)
        tmp = list()

print result

请记住，这只会附加包含 3 个元素的列表，因此某些情况下可能会导致输出列表中缺少元素（根据您的示例和给出的说明）

Answer 2

我认为这是一个非常简单的解决方案：

a = [['hello', 'bye', 'start'], ['bye', 'start', 'hello'], ['john', 'riya', 'tom'], ['riya', 'john', 'tom']]
for i in range(len(a)):
    a[i].sort()
for i in a:
    if i not in b:
        b.append(i)

它输出：

 [['bye', 'hello', 'start'], ['john', 'riya', 'tom']]

Answer 3

试试这个：-

a = [["hello", "bye", "start"], ["bye", "start", "hello"], ["john", "riya", "tom"], ["riya","john", "tom"]]

ls = []
for i in a:
    i = sorted(i)
    if i not in ls:
        ls.append(i)
print(ls)

正如你所说，想要像列表一样的输出然后尝试这个棘手的方法但是它不会是 pythonic 方式:-

ls = []
ind = []
for i,j in enumerate(a):
    j = sorted(j)
    if j not in ls:
        ind.append(i)
        ls.append(j)
ls1 = [a[x] for x in ind]
print(ls1)

输出：-

[['hello', 'bye', 'start'], ['john', 'riya', 'tom']]

Answer 4

只需转换成 set 它会自动删除项目。

a = [list(i) for i in {frozenset(k) for k in a}]

Answer 5

您可以使用 enumerate:

s = [["hello", "bye", "start"], ["bye", "start", "hello"], ["john", "riya", "tom"], ["riya","john", "tom"]]
new_s = [a for i, a in enumerate(s) if not any(all(c in h for c in a) for h in s[:i])]

输出：

[['hello', 'bye', 'start'], ['john', 'riya', 'tom']]

从 python 嵌套列表中以任意顺序删除重复字符串

Removing duplicate strings from python nested list present in any order

python

nested

list

duplicates

nested-lists