使用列表输出而不是元组进行压缩

Question

What is the fastest and most elegant way of doing list of lists from two lists?

I have

In [1]: a=[1,2,3,4,5,6]

In [2]: b=[7,8,9,10,11,12]

In [3]: zip(a,b)
Out[3]: [(1, 7), (2, 8), (3, 9), (4, 10), (5, 11), (6, 12)]

And I'd like to have

In [3]: some_method(a,b)
Out[3]: [[1, 7], [2, 8], [3, 9], [4, 10], [5, 11], [6, 12]]

I was thinking about using map instead of zip, but I don't know if there is some standard library method to put as a first argument.

I can def my own function for this, and use map, my question is if there is already implemented something. No is also an answer.

Answer 1

If you are zipping more than 2 lists (or even only 2, for that matter), a readable way would be:

[list(a) for a in zip([1,2,3], [4,5,6], [7,8,9])]

This uses list comprehensions and converts each element in the list (tuples) into lists.

Answer 2

You almost had the answer yourself. Don't use map instead of zip . Use map AND zip .

You can use map along with zip for an elegant, functional approach:

list(map(list, zip(a, b)))

zip returns a list of tuples. map(list, [...]) calls list on each tuple in the list. list(map([...]) turns the map object into a readable list.

Answer 3

I love the elegance of the zip function, but using the itemgetter() function in the operator module appears to be much faster. I wrote a simple script to test this:

import time
from operator import itemgetter

list1 = list()
list2 = list()
origlist = list()
for i in range (1,5000000):
        t = (i, 2*i)
        origlist.append(t)

print "Using zip"
starttime = time.time()
list1, list2 = map(list, zip(*origlist))
elapsed = time.time()-starttime
print elapsed

print "Using itemgetter"
starttime = time.time()
list1 = map(itemgetter(0),origlist)
list2 = map(itemgetter(1),origlist)
elapsed = time.time()-starttime
print elapsed

I expected zip to be faster, but the itemgetter method wins by a long shot:

Using zip
6.1550450325
Using itemgetter
0.768098831177

Answer 4

How about this?

>>> def list_(*args): return list(args)

>>> map(list_, range(5), range(9,4,-1))
[[0, 9], [1, 8], [2, 7], [3, 6], [4, 5]]

Or even better:

>>> def zip_(*args): return map(list_, *args)
>>> zip_(range(5), range(9,4,-1))
[[0, 9], [1, 8], [2, 7], [3, 6], [4, 5]]

Answer 5

I generally don't like using lambda, but...

>>> a = [1, 2, 3, 4, 5]
>>> b = [6, 7, 8, 9, 10]
>>> c = lambda a, b: [list(c) for c in zip(a, b)]
>>> c(a, b)
[[1, 6], [2, 7], [3, 8], [4, 9], [5, 10]]

If you need the extra speed, map is slightly faster:

>>> d = lambda a, b: map(list, zip(a, b))
>>> d(a, b)
[[1, 6], [2, 7], [3, 8], [4, 9], [5, 10]]

However, map is considered unpythonic and should only be used for performance tuning.

Answer 6

Using numpy

The definition of elegance can be quite questionable but if you are working with numpy the creation of an array and its conversion to list (if needed...) could be very practical even though not so efficient compared using the map function or the list comprehension.

import numpy as np 
a = b = range(10)
zipped = zip(a,b)
result = np.array(zipped).tolist()
Out: [[0, 0],
 [1, 1],
 [2, 2],
 [3, 3],
 [4, 4],
 [5, 5],
 [6, 6],
 [7, 7],
 [8, 8],
 [9, 9]]

Otherwise skipping the zip function you can use directly np.dstack :

np.dstack((a,b))[0].tolist()

Answer 7

List comprehension would be very simple solution I guess.

a=[1,2,3,4,5,6]

b=[7,8,9,10,11,12]

x = [[i, j] for i, j in zip(a,b)]

print(x)

output : [[1, 7], [2, 8], [3, 9], [4, 10], [5, 11], [6, 12]]