在python中合并1D和2D列表

Question

I'm trying to set up data to convert to a numpy array. I have three lists. Two are one dimensional, and one is two dimensional.

a = [1,2,3]
b = [4,5,6]
c = [ [7,8],[9,10],[11,12] ]

I want to end up with this:

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

I've tried using zip() , but it doesn't delve into the 2D array.

Answer 1

Assuming you don't mind if the use of NumPy in the conversion itself, the following should work.

from numpy import array

a = array([1, 2, 3])
b = array([4, 5, 6])
c = array([[7, 8], [9, 10], [11, 12]])

result = array(list(zip(a, b, c[:, 0], c[:, 1])))

Note that c[:, n] will only work with NumPy arrays, not standard Python lists.

Answer 2

A numpy solution (as per your tag) would be

In [398]: np.vstack([a,b,np.array(c).T]).T
Out[398]: 
array([[ 1,  4,  7,  8],
       [ 2,  5,  9, 10],
       [ 3,  6, 11, 12]])

this makes c into a 2x3 array:

In [399]: np.array(c).T
Out[399]: 
array([[ 7,  9, 11],
       [ 8, 10, 12]])

which can then be stacked (concatenated) vertically with a and b which also have 3 elements.

zip(*c) is a list form of transpose

In [412]: list(zip(a,b,c)) 
Out[412]: [(1, 4, [7, 8]), (2, 5, [9, 10]), (3, 6, [11, 12])]

In [418]: list(zip(a,b,*zip(*c)))
Out[418]: [(1, 4, 7, 8), (2, 5, 9, 10), (3, 6, 11, 12)]

Answer 3

With python3 it's as simple as

a = [1,2,3]
b = [4,5,6]
c = [ [7,8],[9,10],[11,12] ]
[[x, y, *z] for x, y, z in zip(a, b, c)]
[[1, 4, 7, 8], [2, 5, 9, 10], [3, 6, 11, 12]]

Answer 4

In case you want to operate with standard Python lists you can use zip and then process each row afterwards:

from itertools import chain

a = [1,2,3]
b = [4,5,6]
c = [[7,8], [9,10], [11,12]]

[list(chain.from_iterable(y if isinstance(y, list) else [y] for y in x)) for x in zip(a, b, c)]

Answer 5

For starters, zip yields tuples, not lists, so you'd have to convert its output into lists:

def lzip(*args):
    'Transforms the tuples yielded by zip into equivalent lists.'
    for tup in zip(*args):
        yield list(tup)
    return

# Later...
a = [1, 2, 3]
b = [4, 5, 6]
c = [
  [7, 8],
  [9, 10],
  [11, 12],
  ]
lzip(a, b)
# A generator that yields: [1, 4], [2, 5], [3, 6]

Second, zip won't return what you want no matter what, because the operation you're trying to perform on the sub-lists of c is concatenation , not zipping.

First, lzip the 1-d lists as in the example above, then concatenate each sublist that results with the matching sublist in c .

Below, I've used a list comprehension to build the combined list assign it to a variable, with two new local variables to make the zip operation more clear:

heads = lzip(a, b)
tails = c
zipped = [head + tail for head, tail in zip(heads, tails)]
# [[1, 4, 7, 8], [2, 5, 9, 10], [3, 6, 11, 12]]

The zip yields 2-tuples containing the n ⁠th head and n ⁠th tail:

# First iteration:
([1, 4], [7, 8])
# Next iteration:
([2, 5], [9, 10])
# Last iteration:
([3, 6], [11, 12])

These tuples are unpacked into two variables by the (invisible) assignment head, tail = ... inside the list comprehension. The tuples themselves are then thrown away.

Concatenating ("adding") lists with head + tail combines two lists into one. In this case, pairs of 2-element lists become individual 4-element lists.

Finally the [ and ] of the list comprehension collect the 4-element lists into a single list... of 4-element lists.