我实现了这一点是为了在文件夹名称采用"NAME_(ID("格式的文件夹中查找重复项。该程序查找重复项，其中 NAME 相同且 ID 不同，并将它们放入具有完整路径的元组列表中以供以后合并。
虽然这种方法效果很好(文件夹计数永远不会是>1000(，但我忍不住想知道是否有一种更pythonic、可能更快的方法。
def getFolderNumber(f_name):
return f_name.split("_")[-1]
def discoverDuplicates(path):
'''
Discovers duplicate folders
'''
#Gets the list of folder ID numbers
folder_nums = [getFolderNumber(f) for f in os.listdir(path)]
#Gets the list of full folder names
folders = [f for f in os.listdir(path)]
#Returns folder ID numbers if there are more than 1
dupes = [item for item, count in collections.Counter(folder_nums).items() if count > 1]
#Gets the indices for the duplicates in folder_nums, which is of equal length and ordering to folders
f_indices = [i for i, x in enumerate(folder_nums) if x in dupes]
#Gets all items in folders that match the indices from above. Sorts this list using the ID number
pre_zip = sorted([folders[x] for x in f_indices],key=lambda x: getFolderNumber(x).lstrip("(").rstrip(")"))
#Packages together duplicates into a list of lists
iterator = iter(pre_zip)
pre_join = list(map(list,zip(iterator,iterator)))
#Lambda for joining paths
path_builder = lambda x: os.path.join(path,x)
#Joins the final paths within each sublist within the final lists of lists
return [final.append(list(map(path_builder,x))) for x in pre_join]
下面是一个输入示例：
输入： ListOfFolder(List( 包含 [NAME1_(459(，NAME2(459(，NameN(ID_N(...]
输出： [('路径/到/NAME1_(459('，'路径/到/NAME2_(459'(，(重复对2(,...]
这样的东西可以完成这项工作吗？
from collections import defaultdict
def group_duplicates(folders_names):
dups_dict = defaultdict(list)
for folder_name in folders_names:
folder_name_prefix = folder_name.split('_')[:-1]
dups_dicts[folder_name_prefix].append(folder_name)
return [tuple(dups) for dups in dups_dict.values()]
它最终可以被视为更"pythonic"，因为它非常简单明了(简单比复杂好(.
人们可以用一些嵌套的推导来一行，但它最终会变得不那么可读。
至于速度，它会执行得相当快，所以只要它不是你程序的瓶颈，你就不应该太担心它。