How do you extract a column from a multi-dimensional array?

Question

Does anybody know how to extract a column from a multi-dimensional array in Python?

Answer 1

>>> import numpy as np
>>> A = np.array([[1,2,3,4],[5,6,7,8]])

>>> A
array([[1, 2, 3, 4],
    [5, 6, 7, 8]])

>>> A[:,2] # returns the third columm
array([3, 7])

See also: "numpy.arange" and "reshape" to allocate memory

Example: (Allocating a array with shaping of matrix (3x4))

nrows = 3
ncols = 4
my_array = numpy.arange(nrows*ncols, dtype='double')
my_array = my_array.reshape(nrows, ncols)

Answer 2

Could it be that you're using a NumPy array? Python has the array module, but that does not support multi-dimensional arrays. Normal Python lists are single-dimensional too.

However, if you have a simple two-dimensional list like this:

A = [[1,2,3,4],
     [5,6,7,8]]

then you can extract a column like this:

def column(matrix, i):
    return [row[i] for row in matrix]

Extracting the second column (index 1):

>>> column(A, 1)
[2, 6]

Or alternatively, simply:

>>> [row[1] for row in A]
[2, 6]

Answer 3

If you have an array like

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

Then you extract the first column like that:

[row[0] for row in a]

So the result looks like this:

[1, 2, 3]

Answer 4

check it out!

a = [[1, 2], [2, 3], [3, 4]]
a2 = zip(*a)
a2[0]

it is the same thing as above except somehow it is neater the zip does the work but requires single arrays as arguments, the *a syntax unpacks the multidimensional array into single array arguments

Answer 5

def get_col(arr, col):
    return map(lambda x : x[col], arr)

a = [[1,2,3,4], [5,6,7,8], [9,10,11,12],[13,14,15,16]]

print get_col(a, 3)

map function in Python is another way to go.

Answer 6

>>> x = arange(20).reshape(4,5)
>>> x array([[ 0,  1,  2,  3,  4],
        [ 5,  6,  7,  8,  9],
        [10, 11, 12, 13, 14],
        [15, 16, 17, 18, 19]])

if you want the second column you can use

>>> x[:, 1]
array([ 1,  6, 11, 16])

Answer 7

[matrix[i][column] for i in range(len(matrix))]

Answer 8

The itemgetter operator can help too, if you like map-reduce style python, rather than list comprehensions, for a little variety!

# tested in 2.4
from operator import itemgetter
def column(matrix,i):
    f = itemgetter(i)
    return map(f,matrix)

M = [range(x,x+5) for x in range(10)]
assert column(M,1) == range(1,11)

Answer 9

You can use this as well:

values = np.array([[1,2,3],[4,5,6]])
values[...,0] # first column
#[1,4]

Note: This is not working for built-in array and not aligned (e.g. np.array([[1,2,3],[4,5,6,7]]) )

Answer 10

I think you want to extract a column from an array such as an array below

import numpy as np
A = np.array([[1,2,3,4],[5,6,7,8],[9,10,11,12]])

Now if you want to get the third column in the format

D=array[[3],
[7],
[11]]

Then you need to first make the array a matrix

B=np.asmatrix(A)
C=B[:,2]
D=asarray(C)

And now you can do element wise calculations much like you would do in excel.

Answer 11

let's say we have n X m matrix(n rows and m columns) say 5 rows and 4 columns

matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16],[17,18,19,20]]

To extract the columns in python, we can use list comprehension like this

[ [row[i] for row in matrix] for in range(4) ]

You can replace 4 by whatever number of columns your matrix has. The result is

[ [1,5,9,13,17],[2,10,14,18],[3,7,11,15,19],[4,8,12,16,20] ]

Answer 12

One more way using matrices

>>> from numpy import matrix
>>> a = [ [1,2,3],[4,5,6],[7,8,9] ]
>>> matrix(a).transpose()[1].getA()[0]
array([2, 5, 8])
>>> matrix(a).transpose()[0].getA()[0]
array([1, 4, 7])

Answer 13

Well a 'bit' late ...

In case performance matters and your data is shaped rectangular, you might also store it in one dimension and access the columns by regular slicing e.g. ...

A = [[1,2,3,4],[5,6,7,8]]     #< assume this 4x2-matrix
B = reduce( operator.add, A ) #< get it one-dimensional

def column1d( matrix, dimX, colIdx ):
  return matrix[colIdx::dimX]

def row1d( matrix, dimX, rowIdx ):
  return matrix[rowIdx:rowIdx+dimX] 

>>> column1d( B, 4, 1 )
[2, 6]
>>> row1d( B, 4, 1 )
[2, 3, 4, 5]

The neat thing is this is really fast. However, negative indexes don't work here! So you can't access the last column or row by index -1.

If you need negative indexing you can tune the accessor-functions a bit, e.g.

def column1d( matrix, dimX, colIdx ):
  return matrix[colIdx % dimX::dimX]

def row1d( matrix, dimX, dimY, rowIdx ):
  rowIdx = (rowIdx % dimY) * dimX
  return matrix[rowIdx:rowIdx+dimX]

Answer 14

I prefer the next hint: having the matrix named matrix_a and use column_number, for example:

import numpy as np
matrix_a = np.array([[1,2,3,4],[5,6,7,8],[9,10,11,12]])
column_number=2

# you can get the row from transposed matrix - it will be a column:
col=matrix_a.transpose()[column_number]

Answer 15

Despite using zip(*iterable) to transpose a nested list, you can also use the following if the nested lists vary in length:

map(None, *[(1,2,3,), (4,5,), (6,)])

results in:

[(1, 4, 6), (2, 5, None), (3, None, None)]

The first column is thus:

map(None, *[(1,2,3,), (4,5,), (6,)])[0]
#>(1, 4, 6)

Answer 16

All columns from a matrix into a new list:

N = len(matrix) 
column_list = [ [matrix[row][column] for row in range(N)] for column in range(N) ]

Answer 17

If you want to grab more than just one column just use slice:

 a = np.array([[1, 2, 3],[4, 5, 6],[7, 8, 9]])
    print(a[:, [1, 2]])
[[2 3]
[5 6]
[8 9]]