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.

Approach #1

Using NumPy's numpy.split to have list of arrays as output -

import numpy as np

arr = np.array(a) # a is input list
out = np.split(arr,np.flatnonzero(arr[1:] < arr[:-1])+1)

Approach #2

Using loop comrehension to split the list directly and thus avoid numpy.split for efficiency purposes -

idx = np.r_[0, np.flatnonzero(np.diff(a)<0)+1, len(a)]
out = [a[idx[i]:idx[i+1]] for i in range(len(idx)-1)]

Output for given sample -

In [52]: idx = np.r_[0, np.flatnonzero(np.diff(a)<0)+1, len(a)]

In [53]: [a[idx[i]:idx[i+1]] for i in range(len(idx)-1)]
Out[53]: 
[[100, 564, 572, 578, 584, 590, 596, 602, 608, 614, 620, 625, 631],
 [70, 119, 125, 130, 134, 139, 144, 149, 154, 159, 614, 669],
 [100, 136, 144, 149, 153, 158, 163, 167, 173, 179],
 [62, 72, 78, 82, 87, 92, 97, 100, 107, 112, 117, 124, 426],
 [100, 129, 135, 140, 145, 151]]

We are using np.diff here, which feeds in a list in this case and then computes the differentiation. So, a better alternative would be with converting to array and then using comparison between shifted slices of it instead of actually computing the differentiation values. Thus, we could get idx like this as well -

arr = np.asarray(a)
idx = np.r_[0, np.flatnonzero(arr[1:] < arr[:-1])+1, len(arr)]

Let's time it and see if there's any improvement -

In [84]: a = np.random.randint(0,100,(1000,100)).cumsum(1).ravel().tolist()

In [85]: %timeit np.r_[0, np.flatnonzero(np.diff(a)<0)+1, len(a)]
100 loops, best of 3: 3.24 ms per loop

In [86]: arr = np.asarray(a)

In [87]: %timeit np.asarray(a)
100 loops, best of 3: 3.05 ms per loop

In [88]: %timeit np.r_[0, np.flatnonzero(arr[1:] < arr[:-1])+1, len(arr)]
10000 loops, best of 3: 77 µs per loop

In [89]: 3.05+0.077
Out[89]: 3.127

So, a marginal improvement there with the shifting and comparing method with the conversion : np.asarray(a) eating-up most of the runtime.

Try this below:

list_data = ['7072624', '7072631', '7072672', '7072687', '7072752', '7072759', '7072768', '7072783']
    array1 = [int(list_data[i]) for i in range(len(list_data)) if i % 2 == 0]
    array2 = [int(list_data[i]) for i in range(len(list_data)) if i % 2 != 0]