问题

我有一个包含超过20行的二维列表。我想要对每个子列表进行排序并返回其中重复次数最多的元素。例如，以下是我的列表：

list1 = [[a,a,b, b, b, v,v,v,v,x,x,p], [b, c, d, c], [a, j,j,j,c,c,f,f,h,h,h,h]]

我希望我的结果如下：

List2 = [[b,v,a,x],[c],[h,j,c,f]]

 list1 = [[a,a,b, b, b, v,v,v,v,x,x,p], [b, c, d, c], [a, j,j,j,c,c,f,f,h,h,h,h]]

 List2 = [[b,v,a,x],[c],[h,j,c,f]]

答案1

得分: 3

你可以使用 `Counter.most_common` 来按频率顺序获取子列表中的项目。 在这里，`v` 比 `b` 更常见：

```py
from collections import Counter

a="a"; b="b"; c="c"; d="d"; f="f"; h="h"; j="j"; p="p"; v="v"; x="x"
ll = [[a,a,b, b, b, v,v,v,v,x,x,p], [b, c, d, c], [a, j,j,j,c,c,f,f,h,h,h,h]]

ls = [[e for e, n in Counter(items).most_common() if n >= 2] for items in ll]
print(ls)  # [['v', 'b', 'a', 'x'], ['c'], ['h', 'j', 'c', 'f']]

<details>
<summary>英文:</summary>

You can use `Counter.most_common` to get the items in the sublist in order of frequency. `v` is more common than `b` here:

```py
from collections import Counter

a=&quot;a&quot;; b=&quot;b&quot;; c=&quot;c&quot;; d=&quot;d&quot;; f=&quot;f&quot;; h=&quot;h&quot;; j=&quot;j&quot;; p=&quot;p&quot;; v=&quot;v&quot;; x=&quot;x&quot;
ll = [[a,a,b, b, b, v,v,v,v,x,x,p], [b, c, d, c], [a, j,j,j,c,c,f,f,h,h,h,h]]

ls = [[e for e, n in Counter(items).most_common() if n &gt;= 2] for items in ll]
print(ls)  # [[&#39;v&#39;, &#39;b&#39;, &#39;a&#39;, &#39;x&#39;], [&#39;c&#39;], [&#39;h&#39;, &#39;j&#39;, &#39;c&#39;, &#39;f&#39;]]

答案2

得分: 2

这是使用Counter的解决方案的代码：

import collections

list1 = [['a', 'a', 'b', 'b', 'b', 'v', 'v', 'v', 'v', 'x', 'x', 'p'], ['b', 'c', 'd', 'c'], ['a', 'j', 'j', 'j', 'c', 'c', 'f', 'f', 'h', 'h', 'h', 'h']]
list2 = []
for l in list1:
    c = collections.Counter(l)
    list2.append([k for k, v in c.items() if v > 1])
print(list2)

输出结果为：

[['a', 'b', 'v', 'x'], ['c'], ['j', 'c', 'f', 'h']]

import collections

list1 = [["a","a","b","b","b","v","v","v","v","x","x","p"],["b","c","d","c"],["a","j","j","j","c","c","f","f","h","h","h","h"]]
list2 = []
for l in list1:
    c = collections.Counter(l)
    list2.append( [k for k,v in c.items() if v > 1] )
print(list2)

[['a', 'b', 'v', 'x'], ['c'], ['j', 'c', 'f', 'h']]