Python - Split array into multiple arrays

Question

I have a array contains file names like below:

['001_1.png', '001_2.png', '001_3.png', '002_1.png','002_2.png', '003_1.png', '003_2.png', '003_3.png', '003_4.png', ....]

I want to quickly group these files into multiple arrays like this:

[['001_1.png', '001_2.png', '001_3.png'], ['002_1.png', '002_2.png'], ['003_1.png', '003_2.png', '003_3.png', '003_4.png'], ...]

Could anyone tell me how to do it in few lines in python?

• tda: In your desired output, should the third element be 001_3.png?
• BcK: Is it always like this, I mean ordered ?
• Vafliik: The third one should be 001_3.png, right?

Answer 1

If your data is already sorted by the file name, you can use itertools.groupby:

files = ['001_1.png', '001_2.png', '001_3.png', '002_1.png','002_2.png',
 '003_1.png', '003_2.png', '003_3.png']
import itertools
keyfunc = lambda filename: filename[:3]
# this creates an iterator that yields `(group, filenames)` tuples,
# but `filenames` is another iterator
grouper = itertools.groupby(files, keyfunc)
# to get the result as a nested list, we iterate over the grouper to
# discard the groups and turn the `filenames` iterators into lists
result = [list(files) for _, files in grouper]
print(list(result))
# [['001_1.png', '001_2.png', '001_3.png'],
# ['002_1.png', '002_2.png'],
# ['003_1.png', '003_2.png', '003_3.png']]

---

Otherwise, you can base your code on this recipe, which is more efficient than sorting the list and then using groupby.

• Input: Your input is a flat list, so use a regular ol' loop to iterate over it:

  for filename in files:
  

• Group identifier: The files are grouped by the first 3 letters:

  group = filename[:3]
  

• Output: The output should be a nested list rather than a dict, which can be done with

  result = list(groupdict.values())
  

Putting it together:

files = ['001_1.png', '001_2.png', '001_3.png', '002_1.png','002_2.png',
 '003_1.png', '003_2.png', '003_3.png']
import collections
groupdict = collections.defaultdict(list)
for filename in files:
 group = filename[:3]
 groupdict[group].append(filename)
result = list(groupdict.values())
print(result)
# [['001_1.png', '001_2.png', '001_3.png'],
# ['002_1.png', '002_2.png'],
# ['003_1.png', '003_2.png', '003_3.png']]

Read the recipe answer for more details.

• Mihai Alexandru-Ionut: Beatiful answer and upvote. I know the feeling when write a long and correct answer and nobody upvote it in order to recognize the invested time and effort into answering it.
• Aran-Fey: @MihaiAlexandru-Ionut Thanks :) I couldn't stand the low text/code ratio in this thread, so I decided to do something about it. Your answer was the only one that had a significant amount of explanation.

Answer 2

Something like that should work:

import itertools
mylist = [...]
[list(v) for k,v in itertools.groupby(mylist, key=lambda x: x[:3])]

If input list isn't sorted, than use something like that:

import itertools
mylist = [...]
keyfunc = lambda x:x[:3]
mylist = sorted(mylist, key=keyfunc)
[list(v) for k,v in itertools.groupby(mylist, key=keyfunc)]

Answer 3

You can do it using a dictionary.

list = ['001_1.png', '001_2.png', '003_3.png', '002_1.png', '002_2.png', '003_1.png', '003_2.png', '003_3.png', '003_4.png']
dict = {}
for item in list:
 if item[:3] not in dict:
 dict[item[:3]] = []
 dict[item[:3]].append(item)

Then you have to sort the dictionary by key value.

dict = {k:v for k,v in sorted(dict.items())}

The last step is to use a list comprehension in order to achieve your requirement.

list = [v for k,v in dict.items()]
print(list)

Output

[['001_1.png', '001_2.png'], ['002_1.png', '002_2.png'], ['003_3.png', '003_1.png', '003_2.png', '003_3.png', '003_4.png']]

Answer 4

Using a simple iteration and dictionary.

Ex:

l = ['001_1.png', ' 001_2.png', ' 003_3.png', ' 002_1.png', ' 002_2.png', ' 003_1.png', ' 003_2.png', ' 003_3.png', ' 003_4.png']
r = {}
for i in l:
 v = i.split("_")[0][-1]
 if v not in r:
 r[v] = []
 r[v].append(i)
print(r.values())

Output:

[['001_1.png', ' 001_2.png'], [' 003_3.png', ' 003_1.png', ' 003_2.png', ' 003_3.png', ' 003_4.png'], [' 002_1.png', ' 002_2.png']]

Answer 5

If your list is ordered like this here is a short script for this task.

myList = []
for i in a:
 if i[:-4].endswith('1'):
 myList.append([i])
 else:
 myList[-1].append(i)
# [['001_1.png', '001_2.png', '003_3.png'], ['002_1.png', '002_2.png'], ...]

Answer 6

#IYN
mini_list = []
p = ['001_1.png', '001_2.png', '001_3.png', '002_1.png','002_2.png', '003_1.png', '003_2.png', '003_3.png', '003_4.png']
new_p = []
for index, element in enumerate(p):
 if index == len(p)-1:
 mini_list.append(element)
 new_p.append(mini_list)
 break
 if element[0:3]==p[index+1][0:3]:
 mini_list.append(element)
 else:
 mini_list.append(element)
 new_p.append(mini_list)
 mini_list = []
print (new_p)

The code above will cut the initial list into sub lists and append them as individual lists into a resulting, larger list. Note: not a few lines, but you can convert this to a function.

def list_cutter(ls):
 mini_list = []
 new_list = []
 for index, element in enumerate(ls):
 if index == len(ls)-1:
 mini_list.append(element)
 new_list.append(mini_list)
 break
 if element[0:3]==ls[index+1][0:3]:
 mini_list.append(element)
 else:
 mini_list.append(element)
 new_list.append(mini_list)
 mini_list = []
 return new_list