CODEWITHSUNDEEP
pythonarrays
eric2323223
eric2323223

Reputation: 3628

Python - Split array into multiple arrays

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?

Upvotes: 5

Views: 18103

Answers (6)

Michael Matta
Michael Matta

Reputation: 69

#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

Upvotes: 0

Aran-Fey
Aran-Fey

Reputation: 43126

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.

Upvotes: 6

oxyum
oxyum

Reputation: 6787

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)]

Upvotes: 4

Mihai Alexandru-Ionut
Mihai Alexandru-Ionut

Reputation: 48337

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']]

Upvotes: 1

BcK
BcK

Reputation: 2811

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'], ...]

Upvotes: 0

Rakesh
Rakesh

Reputation: 82755

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']]

Upvotes: 0

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