This question already has answers here:
Numpy reshape 1d to 2d array with 1 column
(7 answers)
Closed 4 years ago.
I can't turn a numpy array into a vector. What I want to do is transform this:
>> x.shape
(784,)
into this:
>> x.shape
(784,1)
x was created in the normal numpy.array() way.
Note: I want to change, not create a new array.
You can add a new axis to a vector in multiple ways: e.g. using np.reshape
x = np.zeros(784)
x0 = x.reshape(784, 1) # shape: (784, 1)
or using np.newaxis while slicing:
x = np.zeros(784)
x0 = x[:,None] # short-hand for x[:,np.newaxis], shape: (784, 1)
Vectors, arrays and lists are totally different data structures, especially in Python. In your question you mentioned I can't turn a numpy array into a vector. Actually what you are trying to do is exactly the opposite, means you want to turn a vector of shape (784,) to an array of shape(784, 1). Anyways #cheersmate gave you the correct answer.
Related
I created two numpy 1D arrays
x1 = np.linspace(0, 1, 5)
x2 = np.linspace(0, 10, 5)
I wrote a function
def myfoo(x1,x2):
return x1**2+x1*x2+x2**2
To get a 2D numpy array, I use the following code :
y=np.empty((x1.size,x2.size))
for a in range(0,x2.size):
y[a]=myfoo(x1,x2[a])
I would like to know if is it possible to write a function that outputs this 2D array DIRECTLY. I simply wonder if is possible to write y=myfoo2(x1,x2) instead of three code lines as above.
I know I can insert these lines into the function as suggested in the comment. But, I wonder if it exists in Numpy or Python "something" (function, operators, ...) like the mathematicals dyadic product of two vectors (i.e. from two 1D vectors of size m,n, this operation gives a matrix of size m x n)
Thanks for answer
myfoo(x1[:,None], x2). x1[:,None]*x2
produces a (5,5) array.
This question already has answers here:
How do I add an extra column to a NumPy array?
(17 answers)
Closed 2 years ago.
I met with a problem when doing appending in NumPy. array_1 will throw an error: ValueError: could not broadcast input array from shape (4) into shape (3) but the bottom one does not, where am I doing it wrong? I need to write a loop to append arrays to each array in array_1.
The blunt way is to convert my 2d-arrays to a 2d-list, but as a keen learner, I am really curious about how to do it properly.
array_1 = np.array([[1,2,3], [4,5,6]])
array_1[0] = np.append(array_1[0], 1)
array_2 = np.array([[1,2,3]])
array_2 = np.append(array_2, 1)
Change it to this:
array_1 = np.array([[1,2,3], [4,5,6]])
array_1 = [np.append(array_1[0], 1), array_1[1]]
This question already has answers here:
How to get the element-wise mean of an ndarray
(2 answers)
Closed 3 years ago.
I have 'n' numpy arrays each with shape (128,)
How to get an average numpy array of shape (128,) for the list of numpy arrays.
I have seen the documentation of numpy's average() and mean() which describes that the average is calculated for all the elements in a single numpy array rather than multiple or list of numpy arrays.
Example
numpyArrayList = [ar1,ar2,ar3,ar4...arn]
avgNumpyArray = avg(numpyArrayList)
avgNumpyArray.shape
should give result as (128,)
and this array should contain the average of all the numpy arrays
Thanks in advance
I would use np.mean([ar1,ar2,ar3,ar4...arn], axis=0).
You can achieve this by using the following code
ar = [ar1,ar2,ar3,...,arn]
r = np.mean(ar)
for axis=0 use following
r = np.mean(ar, axis=0)
for axis=1 use following
r = np.mean(ar, axis=1)
something like?
mean=0
n=len(numpyArrayList)
for i in numpyArrayList:
mean += i.sum()/(128.*n)
Edit: misunderstood the question, sry
My goal is to to turn a row vector into a column vector and vice versa. The documentation for numpy.ndarray.transpose says:
For a 1-D array, this has no effect. (To change between column and row vectors, first cast the 1-D array into a matrix object.)
However, when I try this:
my_array = np.array([1,2,3])
my_array_T = np.transpose(np.matrix(myArray))
I do get the wanted result, albeit in matrix form (matrix([[66],[640],[44]])), but I also get this warning:
PendingDeprecationWarning: the matrix subclass is not the recommended way to represent matrices or deal with linear algebra (see https://docs.scipy.org/doc/numpy/user/numpy-for-matlab-users.html). Please adjust your code to use regular ndarray.
my_array_T = np.transpose(np.matrix(my_array))
How can I properly transpose an ndarray then?
A 1D array is itself once transposed, contrary to Matlab where a 1D array doesn't exist and is at least 2D.
What you want is to reshape it:
my_array.reshape(-1, 1)
Or:
my_array.reshape(1, -1)
Depending on what kind of vector you want (column or row vector).
The -1 is a broadcast-like, using all possible elements, and the 1 creates the second required dimension.
If your array is my_array and you want to convert it to a column vector you can do:
my_array.reshape(-1, 1)
For a row vector you can use
my_array.reshape(1, -1)
Both of these can also be transposed and that would work as expected.
IIUC, use reshape
my_array.reshape(my_array.size, -1)
Consider the following simple example:
X = numpy.zeros([10, 4]) # 2D array
x = numpy.arange(0,10) # 1D array
X[:,0] = x # WORKS
X[:,0:1] = x # returns ERROR:
# ValueError: could not broadcast input array from shape (10) into shape (10,1)
X[:,0:1] = (x.reshape(-1, 1)) # WORKS
Can someone explain why numpy has vectors of shape (N,) rather than (N,1) ?
What is the best way to do the casting from 1D array into 2D array?
Why do I need this?
Because I have a code which inserts result x into a 2D array X and the size of x changes from time to time so I have X[:, idx1:idx2] = x which works if x is 2D too but not if x is 1D.
Do you really need to be able to handle both 1D and 2D inputs with the same function? If you know the input is going to be 1D, use
X[:, i] = x
If you know the input is going to be 2D, use
X[:, start:end] = x
If you don't know the input dimensions, I recommend switching between one line or the other with an if, though there might be some indexing trick I'm not aware of that would handle both identically.
Your x has shape (N,) rather than shape (N, 1) (or (1, N)) because numpy isn't built for just matrix math. ndarrays are n-dimensional; they support efficient, consistent vectorized operations for any non-negative number of dimensions (including 0). While this may occasionally make matrix operations a bit less concise (especially in the case of dot for matrix multiplication), it produces more generally applicable code for when your data is naturally 1-dimensional or 3-, 4-, or n-dimensional.
I think you have the answer already included in your question. Numpy allows the arrays be of any dimensionality (while afaik Matlab prefers two dimensions where possible), so you need to be correct with this (and always distinguish between (n,) and (n,1)). By giving one number as one of the indices (like 0 in 3rd row), you reduce the dimensionality by one. By giving a range as one of the indices (like 0:1 in 4th row), you don't reduce the dimensionality.
Line 3 makes perfect sense for me and I would assign to the 2-D array this way.
Here are two tricks that make the code a little shorter.
X = numpy.zeros([10, 4]) # 2D array
x = numpy.arange(0,10) # 1D array
X.T[:1, :] = x
X[:, 2:3] = x[:, None]