So i'm trying to append 2 np array together but it gives me this error ValueError: all the input arrays must have same number of dimensions, but the array at index 0 has 2 dimension(s) and the array at index 1 has 1 dimension(s)I know that this mean the shape of the array are not the same but I don't understand why and how to fix it.
arr1 = np.array([
[10.24217065 5.63381577]
[ 2.71521988 -3.33068004]
[-3.43022486 16.40921457]
[ 1.4461307 12.59851726]
[12.34829023 29.67531647]
[16.65382971 9.8915765 ]])
arr2 = np.array([4.62643996 5.14587112])
arr3 = np.append(arr1,arr2,axis=0)
Simply make them the same dimension:
arr3 = np.append(arr1, [arr2], axis=0)
arr2 has only a single dimension, since its shape is (2,). arr1 on the other hand has two dimensions, since its shape is (6, 2). These aren't compatible for np.append, as it says.
You can make arr2 have the required number of dimensions in many ways. One of them is reshaping:
arr3 = np.append(arr1, arr2.reshape(1, 2), axis=0)
At this point, the arrays have shape (6, 2) and (1, 2), which np.append knows how to deal with. The output will have shape (7, 2).
The error message tells you exactly what is the problem. The first array has two dimensions and the second array has one dimension. Another pair of [ ] in the second array will do the job.
arr2 = np.array([[4.62643996 5.14587112]])
arr3 = np.vstack((arr1, arr2))
or if you really want to use append, my favorite is
arr3 = np.append(arr1, arr2[np.newaxis, :])
Related
I have a numpy ndarray train_data of length 200, where every row is another ndarray of length 10304.
However when I print np.shape(train_data), I get (200, 1), and when I print np.shape(train_data[0]) I get (1, ), and when I print np.shape(train_data[0][0]) I get (10304, ).
I am quite confused with this behavior as I supposed the first np.shape(train_data) should return (200, 10304).
Can someone explains to me why this is happening, and how could I get the array to be in shape of (200, 10304)?
This is because the arrays are constructed to be arrays of objects. Basically each element in the array is pointing to another array of size (1, ) which points to another array of size (10304, ). This is not equivalent to a normal ndarray in numpy so the shape is not recognized correctly. You can check this by looking at the dtypes.
To replicate what you see:
import numpy as np
arr = np.empty(200, dtype='object')
for i in range(200):
temp_arr = np.empty(1, dtype='object')
temp_arr[0] = np.zeros(10304)
arr[i] = temp_arr
print(arr.shape)
print(arr[0].shape)
print(arr[0][0].shape)
(200,)
(1,)
(10304,)
To get the (200, 10304) array back you need to "unpack" them:
new_arr = np.array([x[0] for x in arr])
#(200, 10304)
I'm not sure why that's happening, try reshaping the array:
B = np.reshape(A, (-1, 2))
I am following a tutorial to implement the K-nearest Neighbor algorithm on a dataset.
I have an array of shape (6003,) and I want to do this:
data = data.reshape((data.shape[0], 3072))
However, I am getting this error:
cannot reshape array of size 6003 into shape (6003,3072)
Any help on this, please? Thanks!
when you reshape a numpy array the total number elements shouldn't change.
e.g. a =[2,3,4,5,1,7] if you want to reshape this to a 2Darray then the dimensions multiplied should be equal to the total number elements in the original array a.
this means you can reshape array a in to dimension of (1,6) (2,3),(6,1),(3,2).
the title of your question does give away the error by the way.
Reshaping array of shape (x,) into an array of shape (x,y)
is impossible because you are trying to add more elements into your original data.
an array of shape (x,) can only be reshaped into an array of shape (x/y,y)
I hope this helps.
You are trying to reshape into an incompatible shape. Now, what do I mean by that? Look at this example:
a = np.array([[1, 2, 3],
[4, 5, 6],
])
The shape of this array is:
a.shape
>> (2, 3)
Array a has 2 x 3 = 6 elements. Let's try to reshape it into a (2, 6) array
a.reshape(2, 6)
This raises
>> ValueError: cannot reshape array of size 6 into shape (2,6)
Notice that we were trying to make an array that has 2 x 3 = 6 elements into an array that would have 2 x 6 = 12 elements. But NumPy cannot add those extra elements into your original array and give that your desired shape. So it raises ValueError.
In your case, you are trying to make an array with 6003 elements into an array that will have 6003 x 3072 = 18441216 elements!
I have a numpy array of images with the shape of (5879,). Inside every index of the numpy array, I have the Pixels of the image with a shape of (640,640,3).
I want to reshape the complete array in such a way that the shape of the numpy array becomes (5879,640,640,3).
please check, whether below code works for you or not
import numpy as np
b = np.array([5879])
b.shape
output (1,)
a = np.array([[640],[640],[3]])
a = a.reshape((a.shape[0], 1))
a.shape
output (3, 1)
c = np.concatenate((a,b[:,None]),axis=0)
c.shape
Output:
(4, 1)
np.concatenate((a,b[:,None]),axis=0)
output
array([[ 640],
[ 640],
[ 3],
[5879]])
You want to stack your images along the first axis, into a 4D array. However, your images are all 3D.
So, first you need to add a leading singleton dimension to all images, and then to concatenate them along this axis:
imgs = [i_[None, ...] for i_ in orig_images] # add singleton dim to all images
x = np.concatenate(imgs, axis=0) # stack along the first axis
Edit:
Based on Mad Phyiscist's comment, it seems like using np.stack is more appropriate here: np.stack takes care of adding the leading singleton dimension for you:
x = np.stack(orig_images, axis=0)
import numpy as np
a = np.array([1,2,3,4])
print a.shape[0]
Why it will output 4?
The array [1,2,3,4], it's rows should be 1, I think , so who can explain the reason for me?
because
print(a.shape) # -> (4,)
what you think (or want?) to have is
a = np.array([[1],[2],[3],[4]])
print(a.shape) # -> (4, 1)
or rather (?)
a = np.array([[1, 2 , 3 , 4]])
print(a.shape) # -> (1, 4)
If you'll print a.ndim you'll get 1. That means that a is a one-dimensional array (has rank 1 in numpy terminology), with axis length = 4. It's different from 2D matrix with a single row or column (rank 2).
More on ranks
Related questions:
numpy: 1D array with various shape
Python: Differentiating between row and column vectors
The shape attribute for numpy arrays returns the dimensions of the array. If a has n rows and m columns, then a.shape is (n,m). So a.shape[0] is n and a.shape[1] is m.
numpy arrays returns the dimensions of the array. So, when you create an array using,
a = np.array([1,2,3,4])
you get an array with 4 dimensions. You can check it by printing the shape,
print(a.shape) #(4,)
So, what you get is NOT a 1x4 matrix. If you want that do,
a = numpy.array([1,2,3,4]).reshape((1,4))
print(a.shape)
Or even better,
a = numpy.array([[1,2,3,4]])
a = np.array([1, 2, 3, 4])
by doing this, you get a a as a ndarray, and it is a one-dimension array. Here, the shape (4,) means the array is indexed by a single index which runs from 0 to 3. You can access the elements by the index 0~3. It is different from multi-dimensional arrays.
You can refer to more help from this link Difference between numpy.array shape (R, 1) and (R,).
If I create an array X = np.random.rand(D, 1) it has shape (3,1):
[[ 0.31215124]
[ 0.84270715]
[ 0.41846041]]
If I create my own array A = np.array([0,1,2]) then it has shape (1,3) and looks like
[0 1 2]
How can I force the shape (3, 1) on my array A?
You can assign a shape tuple directly to numpy.ndarray.shape.
A.shape = (3,1)
As of 2022, the docs state:
Setting arr.shape is discouraged and may be deprecated in the future.
Using ndarray.reshape is the preferred approach.
The current best solution would be
A = np.reshape(A, (3,1))
A=np.array([0,1,2])
A.shape=(3,1)
or
A=np.array([0,1,2]).reshape((3,1)) #reshape takes the tuple shape as input
The numpy module has a reshape function and the ndarray has a reshape method, either of these should work to create an array with the shape you want:
import numpy as np
A = np.reshape([1, 2, 3, 4], (4, 1))
# Now change the shape to (2, 2)
A = A.reshape(2, 2)
Numpy will check that the size of the array does not change, ie prod(old_shape) == prod(new_shape). Because of this relation, you're allowed to replace one of the values in shape with -1 and numpy will figure it out for you:
A = A.reshape([1, 2, 3, 4], (-1, 1))
You can set the shape directy i.e.
A.shape = (3L, 1L)
or you can use the resize function:
A.resize((3L, 1L))
or during creation with reshape
A = np.array([0,1,2]).reshape((3L, 1L))
Your 1-D array has the shape (3,):
>>>A = np.array([0,1,2]) # create 1-D array
>>>print(A.shape) # print array shape
(3,)
If you create an array with shape (1,3), you can use the numpy.reshape mentioned in other answers or numpy.swapaxes:
>>>A = np.array([[0,1,2]]) # create 2-D array
>>>print(A.shape) # print array shape
>>>A = np.swapaxes(A,0,1) # swap 0th and 1st axes
>>>A # display array with swapped axes
(1, 3)
array([[0],
[1],
[2]])