python 按某个 id 切列表

Question

I have a list contains dates and id, for example:

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']

and I want to cut them by ids, for example:

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']......]

is there a simple and clean way to do it?

Answer 1

I would suggest using a dict. You can then achieve it o(n) time

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)

edit, updated according to wwii's comment:

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

I'd use collections.defaultdict and a list compreension , ie:

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']}

---

Demo

Answer 3

could also use pandas for this:

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

You could sort the list using the two character id and then group it using itertools.groupby .

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)

The 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']]