如何计算列表列表中相同值的最长序列，然后在元组中输出最大序列 [英] how to count the longest sequence of the same value in a list of lists, and then output the largest sequence in a tuple

问题描述

我有一个文本文件中的列表9的列表)，其值类似于以下内容:

I have a list of lists of lists 9in a text file) with values similar to what is below:

L = [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]

这是我正在使用的功能:

This is the function I'm using:

def longest_sequence(l):
        counter = 0
        sl = []
        sublists = []
        for i in l:
            if (l[counter+1]==l[counter]):
                sl.append(l[counter])
                counter = counter + 1
            else:
                counter = 0
                sublists.append([sl[i], len(sl)])
        return sublists

现在，在这种情况下，右侧仅计数一个值，该值首先出现(1)，然后转到下一个类似于序列的行，而我得到的输出就是这个值:

Right now counts only one value in this case the one that appears first (1) then it goes to the next line which is similar a sequence and the output i'm getting is this one:

returns the sub lists [[1, 111], [1, 222], [1, 333], [1, 444], [1, 555], [1, 666], [1, 777], [1, 888]] 

基本上，我想做的是检查列表，并从该列表中验证哪个子列表的长度最长，所以我应该得到如下所示的东西:

basically what I'm trying to do is checking the list and from that list verifying which sub-list has the longest length, so I should be getting something like this instead:

sl = [(1, 111), (0, 395), (1, 65), (2, 358), (1, 71)]

作为第二个元组，返回为ti的值包含在所有子列表中连续重复395次(最长)的值.

Being the second tuple the one returned as ti contains the value that got repeated continuously 395 times (longest length) among all sub lists.

推荐答案

您只需使用 itertools.groupby():

In []:
import itertools as it

[(k, sum(1 for _ in g)) for k, g in it.groupby(L)]
# [(k, len(list(g)) for k, g in it.groupby(L)]  # alternative

Out[]:
[(1, 112), (0, 394), (1, 65), (2, 359), (1, 71)]

要获得最大值，可以将 max()与 key 一起使用，例如:

To get the maximum, you can use max() with a key, e.g.:

In []:
import operator as op

counts = [(k, sum(1 for _ in g)) for k, g in it.groupby(L)]
max(counts, key=op.itemgetter(1))

Out[]:
(0, 394)

但是，修复您的代码.

• 您在弄乱索引( counter )时，在 else:块中重置索引时，会从头开始.只需在 for 循环中为索引使用 range(1，len(l)).
• 您无需在 else:块中重置 sl (因此，它会以 111 的速度增长)，但是您确实不需要要创建 sl 列表，只需对项目进行计数
• 您错过了最后一个值的情况
• 处理最后一个值需要对逻辑进行一些重新排序

• You are confusing your indexing (counter), when you reset it in the else: block you start from the beginning again. Just use range(1, len(l)) in your for loop for the index.
• You don't reset sl in the else: block (hence it keeps growing by 111) but you really don't need to create the sl list just count the items
• You miss the case of the last value
• Dealing with the last value needs a little reordering of logic

如此固定，它看起来像:

So fixed, it would look like:

def longest_sequence(l):
    counter = 1
    sublists = []
    for i in range(1, len(l)):
        if l[i] != l[i-1]:
            sublists.append([l[i-1], counter])
            counter = 0
        counter += 1

    if counter > 0:
        sublists.append((l[i], counter))

    return sublists

In []:
longest_sequence(L)

Out[]:
[(1, 112), (0, 394), (1, 65), (2, 359), (1, 71)]

In []:
max(longest_sequence(L), key=op.itemgetter(1))

Out[]:
(0, 394)

这篇关于如何计算列表列表中相同值的最长序列，然后在元组中输出最大序列的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！