Example:
a = [[1,2,3],[4,5,6],[7,8,9]]
By doing something like a[:][0], I expect it to give me 1st elements in each row [1,4,7]
However, I get [1,2,3]. I can always make a function to do it.
def column(array,index):
result = []
for row in array:
result.append(row[index])
return result
But I was wondering: is there a smarter way of doing it?
The de facto standard way of manipulating arrays in such a way is to use the dedicated NumPy package.
Things work the way you want, with NumPy:
>>> import numpy
>>> a = numpy.array([[1,2,3],[4,5,6],[7,8,9]])
>>> a[:,0]
array([1, 4, 7])
In fact, the purpose of NumPy is to provide powerful array manipulations.
a[:][0] doesn't work because a[:] gives a copy of a. [0] then gets the first value from that copy, which is [1,2,3]
Instead use a list comprehension:
a = [[1,2,3],[4,5,6],[7,8,9]]
col = [row[0] for row in a] # => [1,4,7]
map version: map(operator.itemgetter(0), a).
import this): "There should be one-- and preferably only one --obvious way to do it."Use zip():
>>> a = [[1,2,3],[4,5,6],[7,8,9]]
>>> zip(*a)
[(1, 4, 7), (2, 5, 8), (3, 6, 9)]
>>> zip(*a)[0]
(1, 4, 7)
But note that zip builds tuples, which may require you to convert back to lists depending on your application.
zip builds a list of tuples (as opposed to a tuple of tuples), or an iterator of tuples on Python 3.zip(), which implies for most Python programmers that the program really needs to keep all the original data, but in a different structure; this is actually a waste of memory and execution time. Furthermore, this does not work in Python 3, because the result of zip() is a generator, in Python 3, so you would need list(zip(*a))[0]. The list comprehension approach is both extremely explicit and standard.
zip() for transposition might be handy.A bit of list comprehension will do the job:
a = [[1,2,3],[4,5,6],[7,8,9]]
col = [c[0] for c in a]
No idea of speed implications.
