"Cloning" row or column vectors

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Sometimes it is useful to "clone" a row or column vector to a matrix. By cloning I mean converting a row vector such as

[1, 2, 3]

Into a matrix

[[1, 2, 3],
 [1, 2, 3],
 [1, 2, 3]]

or a column vector such as

[[1],
 [2],
 [3]]

into

[[1, 1, 1]
 [2, 2, 2]
 [3, 3, 3]]

In MATLAB or octave this is done pretty easily:

 x = [1, 2, 3]
 a = ones(3, 1) * x
 a =

    1   2   3
    1   2   3
    1   2   3
    
 b = (x') * ones(1, 3)
 b =

    1   1   1
    2   2   2
    3   3   3

I want to repeat this in numpy, but unsuccessfully

In [14]: x = array([1, 2, 3])
In [14]: ones((3, 1)) * x
Out[14]:
array([[ 1.,  2.,  3.],
       [ 1.,  2.,  3.],
       [ 1.,  2.,  3.]])
# so far so good
In [16]: x.transpose() * ones((1, 3))
Out[16]: array([[ 1.,  2.,  3.]])
# DAMN
# I end up with 
In [17]: (ones((3, 1)) * x).transpose()
Out[17]:
array([[ 1.,  1.,  1.],
       [ 2.,  2.,  2.],
       [ 3.,  3.,  3.]])

Why wasn't the first method (In [16]) working? Is there a way to achieve this task in python in a more elegant way?

linear-algebra numpy python
12 Answers

One clean solution is to use NumPy's outer-product function with a vector of ones:

np.outer(np.ones(n), x)

gives n repeating rows. Switch the argument order to get repeating columns. To get an equal number of rows and columns you might do

np.outer(np.ones_like(x), x)
user_62633170

Returning to the original question

In MATLAB or octave this is done pretty easily:

x = [1, 2, 3]

a = ones(3, 1) * x ...

In numpy it's pretty much the same (and easy to memorize too):

x = [1, 2, 3]
a = np.tile(x, (3, 1))
user_69325100

If you have a pandas dataframe and want to preserve the dtypes, even the categoricals, this is a fast way to do it:

import numpy as np
import pandas as pd
df = pd.DataFrame({1: [1, 2, 3], 2: [4, 5, 6]})
number_repeats = 50
new_df = df.reindex(np.tile(df.index, number_repeats))
user_9926870

Another solution

>> x = np.array([1,2,3])
>> y = x[None, :] * np.ones((3,))[:, None]
>> y
array([[ 1.,  2.,  3.],
       [ 1.,  2.,  3.],
       [ 1.,  2.,  3.]])

Why? Sure, repeat and tile are the correct way to do this. But None indexing is a powerful tool that has many times let me quickly vectorize an operation (though it can quickly be very memory expensive!).

An example from my own code:

# trajectory is a sequence of xy coordinates [n_points, 2]
# xy_obstacles is a list of obstacles' xy coordinates [n_obstacles, 2]
# to compute dx, dy distance between every obstacle and every pose in the trajectory
deltas = trajectory[:, None, :2] - xy_obstacles[None, :, :2]
# we can easily convert x-y distance to a norm
distances = np.linalg.norm(deltas, axis=-1)
# distances is now [timesteps, obstacles]. Now we can for example find the closest obstacle at every point in the trajectory by doing
closest_obstacles = np.argmin(distances, axis=1)
# we could also find how safe the trajectory is, by finding the smallest distance over the entire trajectory
danger = np.min(distances)
user_40114070

To answer the actual question, now that nearly a dozen approaches to working around a solution have been posted: x.transpose reverses the shape of x. One of the interesting side-effects is that if x.ndim == 1, the transpose does nothing.

This is especially confusing for people coming from MATLAB, where all arrays implicitly have at least two dimensions. The correct way to transpose a 1D numpy array is not x.transpose() or x.T, but rather

x[:, None]

or

x.reshape(-1, 1)

From here, you can multiply by a matrix of ones, or use any of the other suggested approaches, as long as you respect the (subtle) differences between MATLAB and numpy.

user_29887300
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