CODEWITHSUNDEEP
python-3.xlist
Cat Hariss
Cat Hariss

Reputation: 143

Check if element exists in Nth position in list of lists

Consider having this list:

ini_list = [[1, 2, 5, 10, 7], 
            [4, 3, 4, 3, 21], 
            [45, 65, 8, 8, 9]]

I need to be able to find if 4 exists in any of the 3rd elements of the list of lists. So in this example, I see it does exist on the 2nd list [4, 3, 4, 3, 21], on positions 1 and 3 so it will return true.

I can do this:

elem_to_find = 4
res1 = any(elem_to_find in sublist for sublist in ini_list)

But by doing it this way I'm searching through all the list of list elements not only the 3rd of each list as expected. What would be the right approach?

Upvotes: 2

Views: 475

Answers (1)

M-Chen-3
M-Chen-3

Reputation: 2054

You can just do this:

res1 = any(elem_to_find == sublist[2] for sublist in ini_list if len(sublist) > 2)

This compares elem_to_find to sublist[2] instead of just sublist, i.e. the 3rd element instead of the whole list. It also makes sure the length of the sublist is greater than 2 (otherwise the 3rd element wouldn't exist).

Below is the entire code with 2 examples:

ini_list = [[1, 2, 5, 10, 7], 
            [4, 3, 4, 3, 21], 
            [45, 65, 8, 8, 9]]

elem_to_find = 4
res1 = any(elem_to_find == sublist[2] for sublist in ini_list if len(sublist) > 2)

print(res1)
# Prints True

ini_list = [[1, 2, 5, 10, 7], 
            [4, 3, 3, 3, 21],  # No longer has 4 in 3rd position
            [45, 65, 8, 8, 9]]

res1 = any(elem_to_find == sublist[2] for sublist in ini_list if len(sublist) > 2)
print(res1)
# Prints False

Upvotes: 2

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