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

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

``````>>> 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])
``````

Example: (Allocating a array with shaping of matrix (3×4))

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

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]
``````

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]
``````

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

``````>>> 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])
``````

``````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.

If you have a two-dimensional array in Python (not numpy), you can extract all the columns like so,

``````data = [
['a', 1, 2],
['b', 3, 4],
['c', 5, 6]
]

columns = list(zip(*data))

print("column[0] = {}".format(columns[0]))
print("column[1] = {}".format(columns[1]))
print("column[2] = {}".format(columns[2]))

``````

Executing this code will yield,

``````>>> print("column[0] = {}".format(columns[0]))
column[0] = ('a', 'b', 'c')

>>> print("column[1] = {}".format(columns[1]))
column[1] = (1, 3, 5)

>>> print("column[2] = {}".format(columns[2]))
column[2] = (2, 4, 6)
``````

Of course, you can extract a single column by index (e.g. `columns[0]`)

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

col1 = [val[1] for val in array]
col2 = [val[2] for val in array]
col3 = [val[3] for val in array]
col4 = [val[4] for val in array]
print(col1)
print(col2)
print(col3)
print(col4)

Output:
[1, 5, 9, 13]
[2, 6, 10, 14]
[3, 7, 11, 15]
[4, 8, 12, 16]
``````

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

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)
``````

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]]) )

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.

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])
``````

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] ]`

Just use transpose(), then you can get the columns as easy as you get rows

``````matrix=np.array(originalMatrix).transpose()
print matrix[NumberOfColumns]
``````

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]
``````

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]]
``````

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)
``````

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]
``````

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) ]
``````

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