python 2D列表的索引

Question

In order to access the second dimension of a 2D numpy array, we could use eg

A[:,0:9]

How could we do this to a 2D list?

Answer 1

Wasteful but should work:

list(zip(*(list(zip(*A))[0:9])))

Slightly more economical using itertools.isclice :

list(zip(*(itertools.islice(zip(*A), 0, 9))))

Or one could use map and operator.itemgetter :

list(map(operator.itemgetter(slice(0,9)), A))

Answer 2

An array and nested list version:

In [163]: A=np.arange(12).reshape(3,4)
In [164]: Al = A.tolist()

For sliced indexing, a list comprehension (or mapping equivalent) works fine:

In [165]: A[:,1:3]
Out[165]: 
array([[ 1,  2],
       [ 5,  6],
       [ 9, 10]])
In [166]: [l[1:3] for l in Al]
Out[166]: [[1, 2], [5, 6], [9, 10]]

For advanced indexing, the list requires a further level of iteration:

In [167]: A[:,[0,2,3]]
Out[167]: 
array([[ 0,  2,  3],
       [ 4,  6,  7],
       [ 8, 10, 11]])

In [169]: [[l[i] for i in [0,2,3]] for l in Al]
Out[169]: [[0, 2, 3], [4, 6, 7], [8, 10, 11]]

Again there are various mapping alternatives.

In [171]: [operator.itemgetter(0,2,3)(l) for l in Al]
Out[171]: [(0, 2, 3), (4, 6, 7), (8, 10, 11)]

itemgetter uses tuple(obj[i] for i in items) to generate those tuples.

Curiously, itemgetter returns tuples for the list index, and lists for slices:

In [176]: [operator.itemgetter(slice(1,3))(l) for l in Al]
Out[176]: [[1, 2], [5, 6], [9, 10]]