CODEWITHSUNDEEP
python
Jan Erst
Jan Erst

Reputation: 89

How to create a sequence of two values?

I have a list with different combinations, i.e:

list1 = [(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)]

I also have another list, in my case one looking like:

list2 = [1,1]

What I would like to do is to take the two values of list2, put them together as (1,1), and compare them with the elements in list1, then returning the index. My current attempt is looking like this:

def return_index(comb):
    try:
         return comb_leaves.index(comb)
    except ValueError:
         print("no such value")

Unfortunately, it cant find it, because it's not a sequence. Anyone with any good idea of how to fix this?

Upvotes: 0

Views: 194

Answers (4)

Mad Physicist
Mad Physicist

Reputation: 114320

You are confusing "sequence" with "tuple". Lists and tuples are both sequences. Informally, a sequence is anything that has a length and supports direct indexing in addition to being iterable. A range object is considered to be a sequence too for example.

To create a two element tuple from any other sequence, use the constructor:

test_element = tuple(list_2)

Upvotes: 4

Jörg Beyer
Jörg Beyer

Reputation: 3671

list1 = [(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)]
list2 = [1,1]

tup2 = tuple(list2)

list1.append(tup2)
print('list1:',list1)

print('index:', list1.index(tup2))

will give this:

list1: [(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2), (1, 1)]
index: 4

Not sure if unconditionally adding tup2 is what you want.

Maybe you ask for the index, if the 2nd list is in list1:

list1 = [(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)]list2 = [1,1]

tup2 = tuple(list2)
if tup2 in list1:
    print('index:', list1.index(tup2))
else:
    print('not found')

That gives:

index: 4

the index function returns the first element that matches.

Upvotes: 1

iz_
iz_

Reputation: 16593

Try this:

list1 = [(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)]
list2 = [1, 1]

def return_index(comb):
    try:
        return list1.index(tuple(comb))
    except ValueError:
        print("Item not found")

print(return_index(list2)) # 4

With this line: 

list1.index(tuple(list2))

Convert list2 into a tuple from a list. list1's elements are tuples, so to make the comparison, list2 needs to be a tuple. tuple(list2) turns [1, 1] into (1, 1) (the same type as the elements of list1).

Upvotes: 1

Shariq
Shariq

Reputation: 506

list3 = tuple(list2)
print(list3 in list1) #Check if it exists.

Upvotes: 1

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