That should do for your particular case since you include a string within list. Therefore you need to flatten it:

new_list = [int(item) for sublist in a for item in sublist if item.isdigit()]

>>> int(data[9][-4:-1])+1
103

Use iterable[start:end] to get a subset of them.

Another thing you can do is, if you want to get a single integer from a string, is something along the lines of (python2.x):

s = "test123\n"
x = int(filter(lambda x: x.isdigit(), s))

In Python3, filter returns a filter object which is an iterable. You can turn it into a list and join them for the same functional result.

s = "test123\n"
x = int(''.join(list(filter(lambda x: x.isdigit(), s))))

Note that this does not take into account separated numbers. For example, s = "123test456" will return 123456. If you want them separated, you will need to use something a bit more clever.

>>> s = "\n123test456\n"
>>> nums = map(int, ''.join([x if x.isdigit() else ' ' for x in s]).split())
>>> print(nums)
[123, 456]

Again, in python3, map returns a map object, so you just have to convert it to a list.

>>> s = "\n123test456\n"
>>> nums = list(map(int, ''.join([x if x.isdigit() else ' ' for x in s]).split()))
>>> print(nums)
[123, 456]

Assuming you want the numbers as floats:

miles = [' 86 miles\n', ' 43 miles\n', ' MV\n', ' 0.0 miles\n', ' 43 miles\n', '  15.0 miles\n', ' 0.0 miles\n', ' 0.0 miles\n', ' 0.0  miles  \n', ' 86 miles\n']

total = 0.0
for s in miles:
    for t in s.split():
        try:
            total += float(t)
        except ValueError:
            pass

From this question:

l=[[100], [500], [300]]
result=[item for sublist in l for item in sublist]

From wikibooks:

def flatten(seq, list = None):
    """flatten(seq, list = None) -> list

    Return a flat version of the iterator `seq` appended to `list`
    """
    if list == None:
        list = []
    try:                          # Can `seq` be iterated over?
        for item in seq:          # If so then iterate over `seq`
            flatten(item, list)      # and make the same check on each item.
    except TypeError:             # If seq isn't iterable
        list.append(seq)             # append it to the new list.
    return list

Google is your friend ...

Here is what you can do:

my_list=[['word:', 'house', 'garden', '0,2%'],
         ['word:', 'house', 'garden', '0,2%'],
         ['house', 'garden', '0,2%'],
         ['house', 'garden', '0,2%'],
         ['garden', '0,2%', '0,125%'],
         ['house', '0,2%', '?????'],
         ['house', 'garden', '0,02%'],
         ['house', 'garden', '0,02%'],
         ['garden', '0,02%'],
         ['house', 'garden', '0,2%'],
         ['garden', '0,2%'],
         ['house', '0,2'],
         ['house', '0,2', '%'],
         ['house', 'garden', 'kids', '0,2%'],
         ['house', 'garden', 'kids', '0,2%'],
         ['house', '0,2%', 'boy'],
         ['house', '0,12%'],
         ['house', '4%.'],
         ['house', '4%.', '4.'],
         ['house', '0,2%”.']]


my_dict = { 'garden':[], 'house':[]}
for lst in my_list:
    for s in lst:
        if any([n in s for n in '1234567890']):
            if 'house' in lst:
                my_dict['house'].append(s.replace('%',''))
            if 'garden' in lst:
                my_dict['garden'].append(s.replace('%',''))
print(my_dict)

Output:

{'garden': ['0,2', '0,2', '0,2', '0,2', '0,2', '0,125', '0,02', '0,02', '0,02', '0,2', '0,2', '0,2', '0,2'], 'house': ['0,2', '0,2', '0,2', '0,2', '0,2', '0,02', '0,02', '0,2', '0,2', '0,2', '0,2', '0,2', '0,2', '0,12', '4.', '4.', '4.', '0,2”.']}

Based on @alecxe solution you can also do it without any imports.

If your id numbers are always after id and have a fixed (7) number of digits I would probably just use .split('id ') to separate it and get the 7 digits from the second block onwards.

You can put them together in the desired format by using '; '.join()

Putting everything together:

pattern = ['; '.join([value[:7] for value in valueList.split('id ')[1:]]) for valueList in lst1]

Which prints out:

['3999595; 3999999', '3998895; 5555456; 3998899']