元组列表中的2个项目

Question

I have a list of tuples:

Items = [(4, 2), (1, 1), (2, 4), (8, 6), (11, 4), (10, 2), (7, 3), (6, 1)]

and I want to get it like so in a for loop:

NewItems = [[(4, 2), (1, 1)], 
            [(2, 4), (8, 6)],
            [(11, 4), (10, 2)],
            [(7, 3), (6, 1)]]

I did this:

NewItems = []
while len(Items) > 0:
    NewItems.append([Items[0], Items[1]])
    del Items[0:2]
print NewItems

I don't think it's the best way, since I'm deleting Items variables. Then I tried to do this:

newList = iter(Items)
NewItems = []
for a, b in zip(newList, newList):
    NewItems.append([a, b])
print NewItems

but that's merging the tuples.

Is there any better solution that can do the same thing?

Answer 1

If you don't mind tuples of tuples instead of the inner lists, just zip it up:

>>> zip(Items[0::2], Items[1::2])
[((4, 2), (1, 1)), ((2, 4), (8, 6)), ((11, 4), (10, 2)), ((7, 3), (6, 1))]

Answer 2

You can use a list comprehension where you zip the items in pairs.

Items = [(4, 2), (1, 1), (2, 4), (8, 6), (11, 4), (10, 2), (7, 3), (6, 1)]
New_Items = [list(pair) for pair in zip(Items[::2], Items[1::2])]

>>> New_Items
[[(4, 2), (1, 1)], [(2, 4), (8, 6)], [(11, 4), (10, 2)], [(7, 3), (6, 1)]]

Answer 3

>>> items = [(4,2),(1,1),(2,4),(8,6),(11,4),(10,2),(7,3),(6,1)]
>>> new_items = [items[i:i+2] for i in range(0, len(items), 2)] 
>>> new_items
[[(4, 2), (1, 1)], [(2, 4), (8, 6)], [(11, 4), (10, 2)], [(7, 3), (6, 1)]]

You can do it with the while loop like this.

>>> new_items = []
>>> while items:
...     new_items.append((items.pop(0), items.pop(0)))
... 
>>> new_items
[((4, 2), (1, 1)), ((2, 4), (8, 6)), ((11, 4), (10, 2)), ((7, 3), (6, 1))]

However this is destructive to items and isn't very efficient due to using pop(0) which is O(n)

Answer 4

How about

Items = [(4,2),(1,1),(2,4),(8,6),(11,4),(10,2),(7,3),(6,1)]

NewItems = []
j =0   
for i in xrange(1,len(Items)-1): 
    if j+1 > len(Items):
        break
    NewItems.append([Items[j],Items[j+1]])
    j += 2

print NewItems

Answer 5

from time import time

Items = [(4,2),(1,1),(2,4),(8,6),(11,4),(10,2),(7,3),(6,1)]
NewItems = []
start_time = time()
while len(Items) > 0 :
    NewItems.append([Items[0],Items[1]])
    del Items[0:2]
print NewItems
print time() - start_time

Items = [(4,2),(1,1),(2,4),(8,6),(11,4),(10,2),(7,3),(6,1)]
NewItems = []
start_time = time()
for i in range(len(Items) / 2):
    NewItems.append([Items[2*i],Items[2*i + 1]])
print NewItems
print time() - start_time

With some timing so you can verify that the second solution is faster

Answer 6

Yet another approach, but probably not the prettiest.

x = []
i = iter(Items)

while True:
    try:
        x.append([i.next(), i.next()])
    except StopIteration:
        break