如何根据特定表达式在子列表中保存字符串列表?

Question

我正在做一个脚本来保存子列表中的读取路径。假设我在一个列表中保存了 400 个文件路径，每个路径都有特定的语法Ci_whaterver.csv，那么在我的路径列表中我有这样的东西：

pathlist=[C1_01.csv,C1_02.csv,...,Cn_01.csv,Cn_02.csv] 

我想最终对这种情况进行排序或排序pathlistf：

pathlistf=[[C1_01.csv,C1_02.csv,...],[C2_01.csv,C2_02.csv,...],...,[Cn_01.csv,Cn_02.csv,...]]

我不知道以这种方式重新排序路径。

再次您好，我遇到了这个问题： 我想问一些与过去案例非常相似的事情，假设我有下一条路：path=[case1_Qxxx_cap1_whatever.csv,case1_Qxxx_cap1_whatever2.csv,case1_Qxxx_cap1_whatever3.csv,....,case1_Qxxx_cap2_whatever.csv,case1_Qxxx_cap2_whatever2.csv,case1_Qxxx_cap2_whatever3.csv,case2_Qxxx_cap1_whatever.csv,case2_Qxxx_cap1_whatever2.csv,...,case2_Qxxx_cap2_whatever.csv,case2_Qxxx_cap2_whatever2.csv]

我想要这个：

pathf=[[[case1_Qxxx_cap1_whatever.csv,case1_Qxxx_cap1_whatever2.csv,...],[case1_Qxxx_cap2_whatever.csv,Qxxx_cap2_whatever2.csv,...]],[[case2_Qxxx_cap1_whatever.csv,case2_Qxxx_cap1_whatever2.csv,...],[case2_Qxxx_cap2_whatever.csv,case2_Qxxx_cap2_whatever2.csv,...]]]

Answer 1

如果pathlist是预排序的，则可以根据itertools.groupby使用以下代码。

from itertools import groupby
pathlist=['Cn_01.csv', 'C1_02.csv', 'C9_01.csv', 'C9_02.csv', 'Ca_01.csv', 'C9_03.csv', 'Ca_02.csv', 'C1_01.csv', 'Cn_02.csv']
pathlist.sort()
groupedfilenames = (list(g) for _, g in groupby(pathlist, key=lambda a: a[:2]))
print(list(groupedfilenames))

输出：

[['C1_01.csv', 'C1_02.csv'], ['C9_01.csv', 'C9_02.csv', 'C9_03.csv'], ['Ca_01.csv', 'Ca_02.csv'], ['Cn_01.csv', 'Cn_02.csv']]

Answer 2

一种方法是创建一个字典，并使用部分Ci作为键，并以Ci开头的文件名列表将是值。例如，取pathlist = ['C1_01.csv','C1_02.csv', 'C2_01.csv' , 'C3_01.csv', 'C2_02.csv']，然后我们将创建一个字典，该字典将存储

{'C1': ['C1_01.csv', 'C1_02.csv'], 'C2': ['C2_01.csv', 'C2_02.csv'], 'C3': ['C3_01.csv']}

这是代码：

pathlist = ['C1_01.csv','C1_02.csv', 'C2_01.csv' , 'C3_01.csv', 'C2_02.csv']
d = {}
for path in pathlist:
if path[:2] not in d:
d[path[:2]] = [path]
else:
d[path[:2]].append(path)
pathlistf = []
for key in d:
pathlistf.append(d[key])
print(pathlistf)
# Output: [['C1_01.csv', 'C1_02.csv'], ['C3_01.csv'], ['C2_01.csv', 'C2_02.csv']]

希望这能解决问题。请随时提出任何问题。

Answer 3

这个函数对你有帮助吗？

最好！

import traceback
import pandas as pd
from typing import List

def organize_paths(path_lst:List[str],sep:str):
"""
Method to create a DataFrame from csv_path items
Args:
path_lst (List[str]): lits with paths in string format
sep (str): separator for split method
Returns:
df (DataFrame): matrix with the splitting info
"""
lst = list()
print(f'(SUCCESS organize_paths) -> starting process -> input: ({path_lst},{sep})')
try:
for item in path_lst:
calc_data = item.split(sep)
lst.append({calc_data[0]:calc_data[1]})
df = pd.DataFrame(lst)
print(f'(SUCCESS organize_paths) -> finishing process -> input: ({path_lst},{sep}) -> output sample: {df.head().to_dict()}')
return df
except BaseException as exc:
print(f'(ERROR organize_paths) -> finishing process -> input: ({path_lst},{sep}) -> exception: {traceback.format_exc()}')

pathlist=['C1_01.csv','C1_02.csv','C2_01.csv','C2_02.csv']
df = organize_paths(path_lst=pathlist,sep='_') 
# display(df) to validate calculations