[python] Transposing a 1D NumPy array

I use Python and NumPy and have some problems with "transpose":

import numpy as np
a = np.array([5,4])
print(a)
print(a.T)

Invoking a.T is not transposing the array. If a is for example [[],[]] then it transposes correctly, but I need the transpose of [...,...,...].

This question is related to python numpy transpose

The answer is


The transpose of

x = [[0 1],
     [2 3]]

is

xT = [[0 2],
      [1 3]]

well the code is:

x = array([[0, 1],[2, 3]]);
np.transpose(x)        

this a link for more information:

http://docs.scipy.org/doc/numpy/reference/generated/numpy.transpose.html


You can only transpose a 2D array. You can use numpy.matrix to create a 2D array. This is three years late, but I am just adding to the possible set of solutions:

import numpy as np
m = np.matrix([2, 3])
m.T

For 1D arrays:

a = np.array([1, 2, 3, 4])
a = a.reshape((-1, 1)) # <--- THIS IS IT

print a
array([[1],
       [2],
       [3],
       [4]])

Once you understand that -1 here means "as many rows as needed", I find this to be the most readable way of "transposing" an array. If your array is of higher dimensionality simply use a.T.


instead use arr[:,None] to create column vector


numpy 1D array --> column/row matrix:

>>> a=np.array([1,2,4])
>>> a[:, None]    # col
array([[1],
       [2],
       [4]])
>>> a[None, :]    # row, or faster `a[None]`
array([[1, 2, 4]])

And as @joe-kington said, you can replace None with np.newaxis for readability.


There is a method not described in the answers but described in the documentation for the numpy.ndarray.transpose method:

For a 1-D array this has no effect, as a transposed vector is simply the same vector. To convert a 1-D array into a 2D column vector, an additional dimension must be added. np.atleast2d(a).T achieves this, as does a[:, np.newaxis].

One can do:

import numpy as np
a = np.array([5,4])
print(a)
print(np.atleast_2d(a).T)

Which (imo) is nicer than using newaxis.


Another solution.... :-)

import numpy as np

a = [1,2,4]

[1, 2, 4]

b = np.array([a]).T

array([[1], [2], [4]])


Basically what the transpose function does is to swap the shape and strides of the array:

>>> a = np.ones((1,2,3))

>>> a.shape
(1, 2, 3)

>>> a.T.shape
(3, 2, 1)

>>> a.strides
(48, 24, 8)

>>> a.T.strides
(8, 24, 48)

In case of 1D numpy array (rank-1 array) the shape and strides are 1-element tuples and cannot be swapped, and the transpose of such an 1D array returns it unchanged. Instead, you can transpose a "row-vector" (numpy array of shape (1, n)) into a "column-vector" (numpy array of shape (n, 1)). To achieve this you have to first convert your 1D numpy array into row-vector and then swap the shape and strides (transpose it). Below is a function that does it:

from numpy.lib.stride_tricks import as_strided

def transpose(a):
    a = np.atleast_2d(a)
    return as_strided(a, shape=a.shape[::-1], strides=a.strides[::-1])

Example:

>>> a = np.arange(3)
>>> a
array([0, 1, 2])

>>> transpose(a)
array([[0],
       [1],
       [2]])

>>> a = np.arange(1, 7).reshape(2,3)
>>> a     
array([[1, 2, 3],
       [4, 5, 6]])

>>> transpose(a)
array([[1, 4],
       [2, 5],
       [3, 6]])

Of course you don't have to do it this way since you have a 1D array and you can directly reshape it into (n, 1) array by a.reshape((-1, 1)) or a[:, None]. I just wanted to demonstrate how transposing an array works.


As some of the comments above mentioned, the transpose of 1D arrays are 1D arrays, so one way to transpose a 1D array would be to convert the array to a matrix like so:

np.transpose(a.reshape(len(a), 1))

To 'transpose' a 1d array to a 2d column, you can use numpy.vstack:

>>> numpy.vstack(numpy.array([1,2,3]))
array([[1],
       [2],
       [3]])

It also works for vanilla lists:

>>> numpy.vstack([1,2,3])
array([[1],
       [2],
       [3]])

I am just consolidating the above post, hope it will help others to save some time:

The below array has (2, )dimension, it's a 1-D array,

b_new = np.array([2j, 3j])  

There are two ways to transpose a 1-D array:


slice it with "np.newaxis" or none.!

print(b_new[np.newaxis].T.shape)
print(b_new[None].T.shape)

other way of writing, the above without T operation.!

print(b_new[:, np.newaxis].shape)
print(b_new[:, None].shape)

Wrapping [ ] or using np.matrix, means adding a new dimension.!

print(np.array([b_new]).T.shape)
print(np.matrix(b_new).T.shape)

The name of the function in numpy is column_stack.

>>>a=np.array([5,4])
>>>np.column_stack(a)
array([[5, 4]])

Use two bracket pairs instead of one. This creates a 2D array, which can be transposed, unlike the 1D array you create if you use one bracket pair.

import numpy as np    
a = np.array([[5, 4]])
a.T

More thorough example:

>>> a = [3,6,9]
>>> b = np.array(a)
>>> b.T
array([3, 6, 9])         #Here it didn't transpose because 'a' is 1 dimensional
>>> b = np.array([a])
>>> b.T
array([[3],              #Here it did transpose because a is 2 dimensional
       [6],
       [9]])

Use numpy's shape method to see what is going on here:

>>> b = np.array([10,20,30])
>>> b.shape
(3,)
>>> b = np.array([[10,20,30]])
>>> b.shape
(1, 3)

The way I've learned to implement this in a compact and readable manner for 1-D arrays, so far:

h = np.array([1,2,3,4,5])

v1 = np.vstack(h)
v2 = np.c_[h]

h1 = np.hstack(v1)
h2 = np.r_[v2[:,0]]

numpy.r_ and numpy.c_ translate slice objects to concatenation along the first and second axis, respectively. Therefore the slicing v2[:,0] in transposing back the vertical array v2 into the horizontal array h2

numpy.vstack is equivalent to concatenation along the first axis after 1-D arrays of shape (N,) have been reshaped to (1,N). Rebuilds arrays divided by vsplit.


You can convert an existing vector into a matrix by wrapping it in an extra set of square brackets...

from numpy import *
v=array([5,4]) ## create a numpy vector
array([v]).T ## transpose a vector into a matrix

numpy also has a matrix class (see array vs. matrix)...

matrix(v).T ## transpose a vector into a matrix

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