Let's say that we have the following 2d numpy array:
arr = np.array([[1,1,0,1,1],
[0,0,0,1,0],
[1,0,0,0,0],
[0,0,1,0,0],
[0,1,0,0,0]])
and the following indices for rows and columns:
rows = np.array([0,2,4])
cols = np.array([1,2])
The objective is to slice arr using rows and cols to take the following expected result:
arr_sliced = np.array([[1,0],
[0,0],
[1,0]])
Using directly the arrays as indices like arr[rows, cols] leads to:
IndexError: shape mismatch: indexing arrays could not be broadcast together with shapes (3,) (2,)
So what is the straightforward way to achieve this kind of slicing?
Update: useful information about the solution
So the solution was simple enough and it demands a basic comprehension about numpy's broadcasting. Someone could read these nice but not so representative examples from numpy. Also, the general broadcasting rules explains why there is no shape mismatch in:
arr[rows[:, np.newaxis], cols]
# rows[:, np.newaxis].shape == (3,1)
# cols.shape == (2,)
You can use:
arr[rows[:,None], cols[None]]
Output:
array([[1, 0],
[0, 0],
[1, 0]])
It looks like it is much quicker than indexing for large arrays.
arr[np.ix_([0,2,4],[1,2])]
array([[1, 0],
[0, 0],
[1, 0]])
document: https://docs.scipy.org/doc/numpy-1.10.0/reference/generated/numpy.ix_.html
This function takes N 1-D sequences and returns N outputs with N dimensions each, such that the shape is 1 in all but one dimension and the dimension with the non-unit shape value cycles through all N dimensions.
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 [...,...,...].
It's working exactly as it's supposed to. The transpose of a 1D array is still a 1D array! (If you're used to matlab, it fundamentally doesn't have a concept of a 1D array. Matlab's "1D" arrays are 2D.)
If you want to turn your 1D vector into a 2D array and then transpose it, just slice it with np.newaxis (or None, they're the same, newaxis is just more readable).
import numpy as np
a = np.array([5,4])[np.newaxis]
print(a)
print(a.T)
Generally speaking though, you don't ever need to worry about this. Adding the extra dimension is usually not what you want, if you're just doing it out of habit. Numpy will automatically broadcast a 1D array when doing various calculations. There's usually no need to distinguish between a row vector and a column vector (neither of which are vectors. They're both 2D!) when you just want a vector.
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)
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.
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
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.
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]])
instead use arr[:,None] to create column vector
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
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.
Another solution.... :-)
import numpy as np
a = [1,2,4]
[1, 2, 4]
b = np.array([a]).T
array([[1],
[2],
[4]])
The name of the function in numpy is column_stack.
>>>a=np.array([5,4])
>>>np.column_stack(a)
array([[5, 4]])
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)
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.
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 1-D array (flat array) as you have in your example, you can use the np.expand_dims() function:
>>> a = np.expand_dims(np.array([5, 4]), axis=1)
array([[5],
[4]])
np.expand_dims() will add a dimension to the chosen axis. In this case, we use axis=1, which adds a column dimension, effectively transposing your original flat array.
I am trying to get the dotproduct of two arrays in python using the numpy package. I get as output an array of size (n,). It says that my array has no column while I do see the results when I print it. Why does my array have no column and how do I fix this?
My goal is to calculate y - np.dot(x,b). The issue is that y is (124, 1) while np.dot(x,b) is (124,)
Thanks
It seems that you are trying to subtract two arrays of a different shape. Fortunately, it is off by a single additional axis, so there are two ways of handling it.
(1) You slice the y array to match the shape of the dot(x,b) array:
y = y[:,0]
print(y-np.dot(x,b))
(2) You add an additional axis on the np.dot(x,b) array:
dot = np.dot(x,b)
dot = dot[:,None]
print(y-dot)
Hope this helps
it may depends on the dimension of your array
For example :
a = [1, 0]
b = [[4, 1], [2, 2]]
c = np.dot(a,b)
gives
array([4, 1])
and its shape is (2,)
but if you change a like :
a = [[1, 0],[1,1]]
then result is :
array([[4, 1],
[6, 3]])
and its shape is (2,2)
I am using the following example to understand the working of axis in 3D arrays in Numpy.
a = np.array([[9],[9],[8]])
b = np.array([[1],[4],[6]])
print(np.stack([a,b],axis=0)
>>>
array([[[9],
[9],
[8]],
[[1],
[4],
[6]]])
print(np.stack([a,b],axis=1)
>>>
array([[[9],
[1]],
[[9],
[4]],
[[8],
[6]]])
print(np.stack([a,b],axis=2)
>>>
array([[[9, 1]],
[[9, 4]],
[[8, 6]]])
I am able to understand how axis=0 and axis=1 work. Can anyone explain how axis=2 works with pictorial representation as it is done for 2D arrays?
For reference
print(np.stack([a,b],axis=0.shape) #(2,3,1)
print(np.stack([a,b],axis=1.shape) #(3,2,1)
print(np.stack([a,b],axis=2.shape) #(3,1,2)
You can see 3D numpy arrays as data cube.
Let's suppose we have an np.array A.
(z, y, x) = np.shape(A). You notice that the z dimensions corresponds to the indices 0.
Your array A is simply z 2d array of dimensions (y, x) that you stack together.
It explains why A[0,:,:] is a 2d array.
Axis = 2 simply points out that you consider 2d arrays stacked on the z direction!
I have this np array and trying to add a number to one of just one of lines(trying to make asymmetric array if possible and if not a 100*3 array is also ok)
a=np.arange(100*2).reshape(-1,2)
a[40]=np.append(a[40],6)
note that a=np.arange(100*2).reshape(-1,2) is just simplified example and not real code that I wanna manipulate.
and I receive this error
ValueError: could not broadcast input array from shape (3) into shape (2)
is there any simple solution(except making new array and filling it with loop with previous value then adding 6)?
Would that be a solution to your problem?
import numpy as np
a = np.zeros((100, 3))
a[:,0:2] = np.arange(100*2).reshape(-1,2)
a[40,2]=6
The closest thing in numpy to a ragged array is an object dtype array:
In [475]: a = np.empty(2, object)
In [476]: a
Out[476]: array([None, None], dtype=object)
If an element is a list, you can use its append to add a value:
In [477]: a[0] = [1,2]
In [478]: a[1] = [2,3]
In [479]: a[1].append(4)
In [480]: a
Out[480]: array([list([1, 2]), list([2, 3, 4])], dtype=object)
But it's questionable whether such an array is any better that a list
In [481]: a.tolist()
Out[481]: [[1, 2], [2, 3, 4]]