英文:
Split list in python when same values occurs into a list of sublists
问题
使用Python，我需要将my_list = ['1', '2', '2', '3', '3', '3', '4', '4', '5']拆分成一个避免相同值的子列表列表。正确的输出是[['1', '2', '3', '4', '5'], ['2', '3', '4'], ['3']]。
英文:
Using python, I need to split my_list = ['1','2','2','3','3','3','4','4','5'] into a list with sublists that avoid the same value. Correct output = [['1','2','3','4','5'],['2','3','4'],['3']]
答案1
得分: 3
这个代码的功能是将一个包含重复元素的列表 my_list 转换为一个包含子列表的列表 output，每个子列表包含独一无二的元素。以下是翻译后的代码部分和输出：
也许不是最高效的方法，但仍然有效：my_list = ['1', '2', '2', '3', '3', '3', '4', '4', '5']output = []for e in my_list:for f in output:if not e in f:f.append(e)breakelse:output.append([e])print(output)**输出：**[['1', '2', '3', '4', '5'], ['2', '3', '4'], ['3']]
英文:
Probably not the most efficient approach but effective nonetheless:
my_list = ['1','2','2','3','3','3','4','4','5']output = []for e in my_list:for f in output:if not e in f:f.append(e)breakelse:output.append([e])print(output)
Output:
[['1', '2', '3', '4', '5'], ['2', '3', '4'], ['3']]
答案2
得分: 2
以下是已翻译的代码部分：
我假设您正在对每个唯一元素及其出现次数进行索引，并将结果列表排序，以更好地满足您的期望输出。uniques = list(set(my_list))uniques.sort()unique_counts = {unique:my_list.count(unique) for unique in uniques}new_list = []for _ in range(max(unique_counts.values())):new_list.append([])for unique, count in unique_counts.items():for i in range(count):new_list[i].append(unique)new_list的输出为[['1', '2', '3', '4', '5'], ['2', '3', '4'], ['3']]
英文:
I assumed you are indexing every unique element with its occurrence and also sorted the result list to better suit your desired output.
uniques = list(set(my_list))uniques.sort()unique_counts = {unique:my_list.count(unique) for unique in uniques}new_list = []for _ in range(max(unique_counts.values())):new_list.append([])for unique,count in unique_counts.items():for i in range(count):new_list[i].append(unique)
The output for new_list is
[['1', '2', '3', '4', '5'], ['2', '3', '4'], ['3']]
答案3
得分: 2
使用 collections.Counter 来识别所需子列表的最大数量，然后根据它们的频率在子列表上分配连续的唯一键：
from collections import Countermy_list = ['1','2','2','3','3','3','4','4','5']cnts = Counter(my_list)res = [[] for i in range(cnts.most_common(1).pop()[1])]for k in cnts.keys():for j in range(cnts[k]):res[j].append(k)print(res)
输出结果：
[['1', '2', '3', '4', '5'], ['2', '3', '4'], ['3']]
英文:
By using collections.Counter for recognizing the maximum number of the needed sublists and then distributing consecutive unique keys on sublists according to their frequencies:
from collections import Countermy_list = ['1','2','2','3','3','3','4','4','5']cnts = Counter(my_list)res = [[] for i in range(cnts.most_common(1).pop()[1])]for k in cnts.keys():for j in range(cnts[k]):res[j].append(k)print(res)
[['1', '2', '3', '4', '5'], ['2', '3', '4'], ['3']]
答案4
得分: 0
这是一个基于列表推导的方法，用于获取唯一值和它们的计数。
my_list = ['1','2','2','3','3','3','4','4','5']unique = [val for i, val in enumerate(my_list) if val not in my_list[0:i]]counts = [my_list.count(val) for val in unique]output = [[val for val, ct in zip(unique, counts) if ct > i] for i in range(max(counts))]
希望这对你有帮助。
英文:
Here's a way to do it based on getting unique values and counts using list comprehension.
my_list = ['1','2','2','3','3','3','4','4','5']unique = [val for i,val in enumerate(my_list) if val not in my_list[0:i]]counts = [my_list.count(val) for val in unique]output = [[val for val,ct in zip(unique, counts) if ct > i] for i in range(max(counts))]
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