python 按某個 id 切列表

Question

我有一個包含日期和 ID 的列表，例如：

olist = ['20191101_01.csv','20191101_02.csv','20191101_03.csv','20191101_04.csv','20191102_01.csv','20191102_02.csv','20191102_03.csv','20191102_04.csv','20191103_01.csv','20191103_02.csv','20191103_03.csv','20191103_04.csv']

我想通過 id 來剪切它們，例如：

nlist = [['20191101_01.csv','20191102_01.csv','20191103_01.csv','20191104_01.csv'],['20191101_02.csv','20191102_02.csv','20191103_02.csv','20191104_02.csv']......]

有沒有一種簡單而干凈的方法來做到這一點？

Answer 1

我建議使用字典。 然后，您可以在 o(n) 時間內實現它

olist = ['20191101_01.csv','20191101_02.csv','20191101_03.csv','20191101_04.csv','20191102_01.csv','20191102_02.csv','20191102_03.csv','20191102_04.csv','20191103_01.csv','20191103_02.csv','20191103_03.csv','20191103_04.csv']
parsed_dict = {}
for el in olist:
  key = el.split('_')[1]
  if parsed_dict.get(key) is None:
    parsed_dict[key] = [el]
  else:
    parsed_dict[key].append(el)

print(parsed_dict)

編輯，根據二戰的評論更新：

from collections import defaultdict

olist = ['20191101_01.csv','20191101_02.csv','20191101_03.csv','20191101_04.csv','20191102_01.csv','20191102_02.csv','20191102_03.csv','20191102_04.csv','20191103_01.csv','20191103_02.csv','20191103_03.csv','20191103_04.csv']
parsed_dict = defaultdict(list)
for el in olist:
  key = el.split('_')[1]
  parsed_dict[key].append(el)

print(parsed_dict)

Answer 2

我會使用collections.defaultdict和列表壓縮，即：

from collections import defaultdict
olist = ['20191101_01.csv','20191101_02.csv','20191101_03.csv','20191101_04.csv','20191102_01.csv','20191102_02.csv','20191102_03.csv','20191102_04.csv','20191103_01.csv','20191103_02.csv','20191103_03.csv','20191103_04.csv']
d = defaultdict(list)
[d[x.split("_")[1].split(".")[0]].append(x) for x in olist]
print(dict(d))

---

{'01': ['20191101_01.csv', '20191102_01.csv', '20191103_01.csv'], '02': ['20191101_02.csv', '20191102_02.csv', '20191103_02.csv'], '03': ['20191101_03.csv', '20191102_03.csv', '20191103_03.csv'], '04': ['20191101_04.csv', '20191102_04.csv', '20191103_04.csv']}

---

演示

Answer 3

也可以為此使用 pandas ：

import pandas as pd
df = pd.DataFrame({'files':olist})

df['grouper'] = df['files'].str.split('_',expand=True)[1]
nlist = df.groupby('grouper')['files'].agg(list).tolist()

output：

[['20191101_01.csv', '20191102_01.csv', '20191103_01.csv'], ['20191101_02.csv', '20191102_02.csv', '20191103_02.csv'], ['20191101_03.csv', '20191102_03.csv', '20191103_03.csv'], ['20191101_04.csv', '20191102_04.csv', '20191103_04.csv']]

Answer 4

您可以使用兩個字符 id 對列表進行排序，然后使用itertools.groupby對其進行分組。

from itertools import groupby

olist = ['20191101_01.csv','20191101_02.csv','20191101_03.csv','20191101_04.csv',
         '20191102_01.csv','20191102_02.csv','20191102_03.csv','20191102_04.csv',
         '20191103_01.csv','20191103_02.csv','20191103_03.csv','20191103_04.csv']

file_id = lambda filename: filename[-6:-4]

slist = sorted(olist, key=file_id)

result = [list(value) for key, value in groupby(slist, key=file_id)]

print(result)

output：

[['20191101_01.csv', '20191102_01.csv', '20191103_01.csv'],
 ['20191101_02.csv', '20191102_02.csv', '20191103_02.csv'],
 ['20191101_03.csv', '20191102_03.csv', '20191103_03.csv'],
 ['20191101_04.csv', '20191102_04.csv', '20191103_04.csv']]