Reputation: 24084
import numpy as np
matrix1 = np.array([[1,2,3],[4,5,6]])
vector1 = matrix1[:,0] # This should have shape (2,1) but actually has (2,)
matrix2 = np.array([[2,3],[5,6]])
np.hstack((vector1, matrix2))
ValueError: all the input arrays must have same number of dimensions
The problem is that when I select the first column of matrix1 and put it in vector1, it gets converted to a row vector, so when I try to concatenate with matrix2, I get a dimension error. I could do this.
np.hstack((vector1.reshape(matrix2.shape[0],1), matrix2))
But this looks too ugly for me to do every time I have to concatenate a matrix and a vector. Is there a simpler way to do this?
Upvotes: 21
Views: 48730
Reputation: 1925
The even simpler way is to subset the matrix.
>>> matrix1
[[1 2 3]
[4 5 6]]
>>> matrix1[:, [0]] # Subsetting
[[1]
[4]]
>>> matrix1[:, 0] # Indexing
[1 4]
>>> matrix1[:, 0:1] # Slicing
[[1]
[4]]
I also mentioned this in a similar question.
It works somewhat similarly to a Pandas dataframe. If you index the dataframe, it gives you a Series. If you subset or slice the dataframe, it gives you a dataframe.
Your approach uses indexing, David Z's approach uses slicing, and my approach uses subsetting.
Upvotes: 0
Reputation: 74154
Here are three other options:
You can tidy up your solution a bit by allowing the row dimension of the vector to be set implicitly:
np.hstack((vector1.reshape(-1, 1), matrix2))
You can index with np.newaxis
(or equivalently, None
) to insert a new axis of size 1:
np.hstack((vector1[:, np.newaxis], matrix2))
np.hstack((vector1[:, None], matrix2))
You can use np.matrix
, for which indexing a column with an integer always returns a column vector:
matrix1 = np.matrix([[1, 2, 3],[4, 5, 6]])
vector1 = matrix1[:, 0]
matrix2 = np.matrix([[2, 3], [5, 6]])
np.hstack((vector1, matrix2))
Upvotes: 16
Reputation: 131550
The easier way is
vector1 = matrix1[:,0:1]
For the reason, let me refer you to another answer of mine:
When you write something like
a[4]
, that's accessing the fifth element of the array, not giving you a view of some section of the original array. So for instance, if a is an array of numbers, thena[4]
will be just a number. Ifa
is a two-dimensional array, i.e. effectively an array of arrays, thena[4]
would be a one-dimensional array. Basically, the operation of accessing an array element returns something with a dimensionality of one less than the original array.
Upvotes: 28