How to flatten multiple lists with list comprehension in python

I currently have multiple lists that consists of inner lists. I've found out how to use list comprehension to flatten the lists but how can i do it without reusing the same line of code?

Here's an example code:

first = [[1,2,3], [4,5,6], [7,8,9]]
second = [[3,5,6], [0,3,4]]
third = [[2,5,0], [1,2,9]]

flat_first = [item for sublist in first for item in sublist]
flat_second = [item for sublist in second for item in sublist]
flat_third = [item for sublist in third for item in sublist]

Here is what i tried:

lists = [first, second, third]
for i in lists:
   flattened = [item for sublist in i for item in sublist]

Typically, if you don't want to "reuse" the same line of code, you would put it into a function. Assuming you want to "update" the variables first, second, and third, you cannot do so without repeating the variable name (and repeating something, like you have in your first example). In this case, might you prefer to use a dictionary? Also, itertools.chain.from_iterable is probably fastest:

import itertools

my_lists = {
    'first': [[1,2,3], [4,5,6], [7,8,9]],
    'second': [[3,5,6], [0,3,4]],
    'third': [[2,5,0], [1,2,9]],
}

# Put this into an update() function if you want a simple call!
for key, val in my_lists.items():
    my_lists[key] = list(itertools.chain.from_iterable(val))

Which updates the original dictionary:

>>> my_lists
{'first': [1, 2, 3, 4, 5, 6, 7, 8, 9], 'second': [3, 5, 6, 0, 3, 4], 'third': [2, 5, 0, 1, 2, 9]}

Otherwise, if you simply want to use a list comprehension to create new variables flat_first, flat_second, and flat_third, you've already done that in your OP, although if you want a one-liner to do that:

flat_first, flat_second, flat_third = (list(itertools.chain.from_iterable(val)) for val in [first, second, third])

Similar to what Ignatius asked, what exactly is your goal (intended output) here?

lists = [first, second, third]
flattened = [[item for sublist in some_list for item in sublist] for some_list in lists]

Could also be correct, as commented by Ignatius, which gives you:

[[1, 2, 3, 4, 5, 6, 7, 8, 9], [3, 5, 6, 0, 3, 4], [2, 5, 0, 1, 2, 9]]

Although you are creating new values, which I am not entirely sure is what you want. Unless it is.

I'm not 100% sure witch solution you want, so I came up with two options:

# Making a list of lists
first = [[1,2,3], [4,5,6], [7,8,9]]
second = [[3,5,6], [0,3,4]]
third = [[2,5,0], [1,2,9]]
lists = [first, second, third]

# Option 1 (a single flattened list)

flattened1 = [item for sublist in lists for subsublist in sublist for item in subsublist]
print(flattened1)
# Output:
# [1, 2, 3, 4, 5, 6, 7, 8, 9, 3, 5, 6, 0, 3, 4, 2, 5, 0, 1, 2, 9]

# Option 2 (a list of flattened lists)

flattened2 = [[item for subsublist in sublist for item in subsublist] for sublist in lists]
print(flattened2)
# Output:
# [[1, 2, 3, 4, 5, 6, 7, 8, 9], [3, 5, 6, 0, 3, 4], [2, 5, 0, 1, 2, 9]]

I checked your code too, flattened is being subscribed each interaction, an option to fix it would be:

flattened = []
for sublist in lists:
    flattened.append([item for subsublist in sublist for item in subsublist])

print(flattened)
# Output:
# [[1, 2, 3, 4, 5, 6, 7, 8, 9], [3, 5, 6, 0, 3, 4], [2, 5, 0, 1, 2, 9]]

Let me know if one of these options do help you,
have fun!