Locating sublists of one list in another list in Python

Question

I have two lists G3, G333. I want to locate the sublists of G333 in G3 and print the indices i. For example, [0, 4, 5, 9] occurs at i=0 in G3, [10, 14, 15, 19] occurs at i=5 in G3. But I am getting an error. I present the expected output.

G3=[[0, 4, 5, 9], [1, 5, 6, 10], [2, 6, 7, 11], [3, 7, 8, 12], [9, 13, 14, 18], [10, 14, 15, 19]]

G333=[[0, 4, 5, 9], [1, 5, 6, 10], [10, 14, 15, 19]] 


G3s = set(map(tuple, G3))
G333s = set(map(tuple, G333)) 

for i in range(0,len(G3s)): 
    if (G3s[i] & G333s[i]): 
        print(i)

The error is

in <module>
    if (G3s[i] & G333s[i]):

TypeError: 'set' object is not subscriptable

The expected output is

i=[0,1,5]

Asked By: Wiz123

||

Source

Answer 1

You can directly use list.index to get the index.

for l in G333:
    print(G3.index(l))

Or, with a list comprehension:

i = [G3.index(l) for l in G333] # [0, 1, 5]

Answered By: Unmitigated

Answer 2

Here’s another way to do it:

G3 = [
    [0, 4, 5, 9],
    [1, 5, 6, 10],
    [2, 6, 7, 11],
    [3, 7, 8, 12],
    [9, 13, 14, 18],
    [10, 14, 15, 19],
]

G333 = [[0, 4, 5, 9], [1, 5, 6, 10], [10, 14, 15, 19]]


for idx, sublist in enumerate(G3):
    if sublist in G333:
        print(idx)

The above solution uses enumerate to iterate through G3 members, while also retrieving each element from G3 to compare with the sub-lists from
G333.

If you want to use the same approach as your current solution, you could
use something like:

G3 = [
    [0, 4, 5, 9],
    [1, 5, 6, 10],
    [2, 6, 7, 11],
    [3, 7, 8, 12],
    [9, 13, 14, 18],
    [10, 14, 15, 19],
]

G333 = [[0, 4, 5, 9], [1, 5, 6, 10], [10, 14, 15, 19]]

G3s = set(map(tuple, G3))
G333s = set(map(tuple, G333))

for element in G333s & G3s:
    print(G3.index(list(element)))

However, I personally find the first option to be easier to understand and use.

Side-note: when using for...loops, avoid using i as name for the elements of your iterator. Rather, try names that better explain what each element represents. For iterating through indexes, for example, I prefer using the name idx (it’s a common shorthand for index).

Answered By: Ingwersen_erik

Answer 3

You can use index function and if any item is missing that can be handled like below:

G3=[[0, 4, 5, 9], [1, 5, 6, 10], [2, 6, 7, 11], [3, 7, 8, 12], [9, 13, 14, 18], [10, 14, 15, 19]]

G333=[[0, 4, 5, 9], [1, 5, 6, 10], [10, 14, 15, 19]] 

i = []

for item in G333:
    try:
        i.append(G3.index(item))
    except ValueError:
        pass

print(i)

Answered By: Prakash S

Locating sublists of one list in another list in Python

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