long list that has many lists of longs

Question

I have this variable of type long that is a list of longs:

print(list_id)
[6L]
[6L]
[6L]
[6L]
[7L]

How to convert this list to something like:

list_id = [6, 6, 6, 6, 7]

I did something like this:

list_orgs_id = []
for i in list_id[0]:
    list_orgs_id.append(i)
print(list_orgs_id)

but it says:

TypeError: 'long' object is not iterable

Possible duplicate: https://stackoverflow.com/q/1614236/2125723 — ryanwebjackson, Jun 22 '18 at 12:32
@ryanwebjackson it's not the same thing. I have a nested list. ( list of lists) — Souad, Jun 22 '18 at 12:35
The question you are refering is quite simple. My question is different!! — Souad, Jun 22 '18 at 12:44

Jim Wright · Accepted Answer · 2018-06-22T12:45:00.760

1

You can cast them to integers with int().

list_id = [[6L], [6L], [6L], [6L], [7L]]
int_id = [int(i[0]) for i in list_id]
print(int_id)

Output:

[6, 6, 6, 6, 7]

Terminal output:

Python 2.7.10 (default, Oct  6 2017, 22:29:07)
[GCC 4.2.1 Compatible Apple LLVM 9.0.0 (clang-900.0.31)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> list_id = [[6L], [6L], [6L], [6L], [7L]]
>>> int_id = [int(i[0]) for i in list_id]
>>> print(int_id)
[6, 6, 6, 6, 7]
>>>

edited Jun 22 '18 at 12:45

answered Jun 22 '18 at 12:29

Jim Wright

5,905
1
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the list you mentioned is not correct. This is the list : `list_id = [[6L], [6L], [6L], [6L], [7L]]` – Souad Jun 22 '18 at 12:37
I tried this and it gives this error: `TypeError: 'long' object is not iterable` – Souad Jun 22 '18 at 12:40
Sorry, small typo fixed again. – Jim Wright Jun 22 '18 at 12:41
I tried your update and now i have: `TypeError: 'long' object has no attribute '__getitem__'` – Souad Jun 22 '18 at 12:43
I just ran it in my terminal and works perfectly. Check that `list_id` is what you think it is. – Jim Wright Jun 22 '18 at 12:44
We are not using the same version – Souad Jun 22 '18 at 12:45

score 1 · Answer 2 · answered Jun 22 '18 at 12:31

1

Do you want something like this?Simply use list comprehension

main_list = [[1,2,3],[2,3,4]]
list_id=[j for i in main_list for j in i]
print(list_id)

Output:

[1, 2, 3, 2, 3, 4]

answered Jun 22 '18 at 12:31

Taohidul Islam

5,246
3
26
39

I tried `list_ids = [int(i) for i in list_id[0]]` and it says 'long' object is not iterable – Souad Jun 22 '18 at 12:36
Have you tried `list_ids = [int(i) for i in list_id[0].items()]` ? – Taohidul Islam Jun 22 '18 at 12:39
Yes just did and it said: `AttributeError: 'long' object has no attribute 'items' ` – Souad Jun 22 '18 at 12:39
Which version of Python you are using? – Taohidul Islam Jun 22 '18 at 12:41
I'm using Python 3.6 – Souad Jun 22 '18 at 12:41
When I write `list_id = [[6L], [6L], [6L], [6L], [7L]]` it shows `Python version 3.6 does not support tailing l or L`. How have you recovered this error? – Taohidul Islam Jun 22 '18 at 12:45
I get the list from SQLAlchemy query – Souad Jun 22 '18 at 12:46
1

@Somar `long` datatypes do not exist in python 3.x. The `int` itself is `long` & these don't have to be suffixed by `L`. – Melvin Abraham Jun 22 '18 at 13:08

Melvin Abraham · Answer 3 · 2018-06-22T13:13:23.040

Method 1

You could apply int() to all the elements in the list of list:

for x in range(len(list_id)):
    list_id[x] = int(list_id[x][0])

Method 2

You could also choose to write it in a fancy list comprehension:

list_id = [int(y) for x in list_id for y in x]

In this case you have to iterate over all the lists inside list_id and during each iteration you will iterate over the long integer inside those list. This is similar to:

temp = []

for x in list_id:
    for y in x:
        temp.append(int(x))

list_id = temp

Method 3

You could, if you want, write a user defined function and apply that using map() function on all the elements of list_id:

def func(list):
    return int(list[0])

list_id = map(func, list_id)

list_id = list(map)
# This line is not required if you just want to iterate over the values.

long list that has many lists of longs

3 Answers3

Method 1

Method 2

Method 3