将列表与元组列表组合在一起

Question

l1 = [1, 2, 3, 4]
l2 = [(10, 20), (30, 40), (50, 60), (70, 80)]

>>> print(list(zip(l1, l2)))

[(1, (10, 20)), (2, (30, 40)), (3, (50, 60)), (4, (70, 80))]

However, I want it to be just a list of four tuples likes this:

[(1, 10, 20), (2, 30, 40), (3, 50, 60), (4, 70, 80)]

I've also tried:

>>> print(list(zip(l1, *l2)))

[(1, 10, 30, 50, 70), (2, 20, 40, 60, 80)]

So my question is:

How can I zip a list with a list of tuples or a list of lists?

Answer 1

A more generic way to approach this problem is to consider that both the lists should be scaled up/down to the same dimension as and when required

>>> [(a, ) + b for a, b in zip(l1, l2)]
[(1, 10, 20), (2, 30, 40), (3, 50, 60), (4, 70, 80)]

As in your example, the second list has an extra dimension in contrast to the first list, either

1. Convert the first list to the higher dimension matching the second list
2. Reduce the dimension of the second list

In this particular case, the first approach was easier and thus was the obvious choice

Answer 2

>>> l1=[1,2,3,4]
>>> l2=[(10,20),(30,40),(50,60),(70,80)]
>>> [tuple([x] + list(y)) for x,y in zip(l1,l2)]
[(1, 10, 20), (2, 30, 40), (3, 50, 60), (4, 70, 80)]

Answer 3

You need

l1 = [1, 2, 3, 4]
l2 = [(10, 20), (30, 40), (50, 60), (70, 80)]
print [(a, b[0], b[1]) for a, b in list(zip(l1, l2))]

This works like this: zip creates an array that combines the elements of l1 and l2

>>> zip (l1, l2)
[(1, (10, 20)), (2, (30, 40)), (3, (50, 60)), (4, (70, 80))]

for i, j in zip(l1, l2) will unpack the values into i and j sequentially picking up one element from each array in each loop iteration; ie, 1 and (10, 20) in the first iteration, 2 and (30, 40) in the second, etc.

You can then reshuffle the values to create the tuple you want. You can do this either with (a, b[0], b[1]) or (a, ) + b . The latter approach, which Abhijit used and is probably more efficient than mine, is based on the fact that you can add tuples (eg (1, 2) + (3, 4) + (5,) equals (1, 2, 3, 4, 5) ).

If what you want to do with the numbers isn't overly complex, you can do it at the beginning of the for expression instead of just creating the tuple. Maybe you can make your code more compact that way.

Answer 4

If let me do this, it will do as below:

l1 = [1, 2, 3, 4]
l2 = [(10, 20), (30, 40), (50, 60), (70, 80)]

print [ (a,b[0],b[1]) for a, b in zip(l1, l2)]

So you have to know zip() how to use:

This function returns a list of tuples, where the i-th tuple contains the i-th element from each of the argument sequences or iterables. The returned list is truncated in length to the length of the shortest argument sequence. When there are multiple arguments which are all of the same length, zip() is similar to map() with an initial argument of None. With a single sequence argument, it returns a list of 1-tuples. With no arguments, it returns an empty list.

Answer 5

Just putting down another way with explicit for loop, assuming len(l1) == len(l2)

for x in range(0,len(l1)):
    print (l1[x],l2[x][0],l2[x][1])