Compare a list to a list of tuple to get another list

Question

I am new to python so apologies for the naive question. I have a list

l1 = [2, 4, 6, 7, 8] 

and another list of tuples

l2 = [(4,6), (6,8), (8,10)]

I want to output a list l3 of size l1 that compares the value of l1 to the first co-ordinates of l2 and stores the second co-ordinate if the first co-ordinate is found in l1, else stores 0.

output :

l3 = [0, 6, 8, 0, 10]

I tired to do a for loop like:

l3 = []
for i in range(len(l1)):
   if l1[i] == l2[i][0]:
      l3.append(l2[i][1])
   else:
      l3.append(0)  

but this doesn't work. It gives the error

IndexError: list index out of range

which is obvious as l2 is shorter than l1.

Answer 1

You can create a dictionary from l2:

l1 = [2,4,6,7,8] 
l2 =[(4,6),(6,8),(8,10)]
new_l2 = dict(l2)
l3 = [new_l2.get(i, 0) for i in l1]

Output:

l3 = [0,6,8,0,10]

Answer 2

I would always use Ajax1234's solution instead, but I wanted to illustrate how I would approach it using a for-loop, as you intended:

l3 = []
for elem in l1:
    pairs = list(filter(lambda x: x[0] == elem, l2))
    l3.append(pairs[0][1] if pairs else 0)

---

An alternate approach would be using next() and a list comprehension instead of filter() and a for-loop. This one is far more efficient and readable:

l3 = [next((u[1] for u in l2 if u[0] == elem), 0) for elem in l1]