I have read about concatenate. But, did not see the function taking a single list as input.It must have two lists as input.
Consider the following statement in a program that I want to execute
row = np.concatenate(row, 1)
What is concatenate doing here? It is taking only one list named row.
Probably you have seen it most often used like this:
c = np.concatenate([a, b])
but you can of course also do:
ab = [a, b]
c = np.concatenate(ab)
Look at row before and after concatenating to see what is going on.
The first argument of np.concatenate is supposed to be a sequence of objects (think vectors or matrices). The second argument is the axis along which the concatenation is to be performed. See help(np.concatenate) for the full docstring.
For your command to be valid, the objects in the row sequence must have at least a 0th and a 1st dimension. This would typically be a matrix, but the name row is suggestive of a set of row vectors that have dimension [0, d].
If you concatenate n vectors of shape [0, d] along the 1st dimension, this will result in an object of shape [0, n*d]. Which is a very long row vector.
Related
In order to modify the value of a particular vertical line in the image, i want to extract the 3-band vertical line of the image.
When I was a two dimensional array, I extracted the vertical lines with the following code:
vertical_line = array[:,[1]]
I tried to use this code in a 3d array image but it failed.
#image shape(b,g,r) (100,100,3)
# my try
vertical_line_try_1 = image[:,[1]][3]#result => [[255,255,255]]
vertical_line_try_2 = image[:,[1][3]]#result => IndexError: list index out of range
vertical_line_try_3 = image[:,[1],[3]]#result => IndexError: index 3 is out of bounds for axis 2 with size 3
How can I extract the vertical lines of three bands at once without writing a loop?
When you index with a slice the corresponding dimension is kept (since you get a possible stridded range of values). On the other hand, when you index with a single number that dimension disappears.
Ex:
a = np.random.randn(4,5)
# Let's get the third column in the matrix
print(a[:, 2].shape) # prints `(4,)` -> we get a vector
print(a[:, 2:3].shape) # prints `(4,1)` -> we keep the result as a column matrix
From your two dimensional example it seems you are using advanced indexing just as a way to keep the dimension. If that is true, using a slice containing a single element as I have shown is cleaner (my opinion) and faster (tested with %timeit in IPython).
Now, back to your question. It looks like you want to extract all values whose index of the second dimension is equal to 1, similar to the red part in the figure below.
If that is the case, then just use either image[:,1, :] (or just image[:,1]) if you to get the values as a matrix, or image[:,1:2, :] if you want a 3D array.
It is also useful to understand what your failed attempts were actually doing.
image[:,[1]][3]: This is actually two indexing operations in the same line.
First, image[:,[1]] would return a 100x1x3 array, and then the second indexing with [3] would take the fourth line. Since this second indexing is a regular one (not a fancy and not a slice), then the indexed dimension disappears and you get a 1x3 array in the end
image[:,[1][3]]: This one I'm not really sure what it is doing
image[:,[1],[3]]: This means "take all values from the first dimension, but only the ones from the second index of the first dimension and the fourth index of the third dimension. Since the third dimension only has "size" 3, then trying to get the fourth index results in the out of bound error.
I have a dataset of one-line sentences and each sentence belongs to a class based on the context. I have created a lexicon of important words and converted my input data into a list of features, where each feature is a vector of the length of the lexicon.
I want to input this data to a dynamic LSTM cell, but can't figure out how to reshape it.
Consider my batch_size = 100, length_lexicon = 64, nRows_Input = 1000
Why not use numpy.reshape?
Check out this documentation: https://docs.scipy.org/doc/numpy/reference/generated/numpy.reshape.html
For example:
>>> a = np.arange(6).reshape((3, 2))
>>> a
array([[0, 1],
[2, 3],
[4, 5]])
numpy.reshape¶
numpy.reshape(a, newshape, order='C')
Gives a new shape to an array without changing its data.
Parameters:
a : array_like
Array to be reshaped.
newshape : int or tuple of ints
The new shape should be compatible with the original shape. If an integer, then the result will be a 1-D array of that length. One
shape dimension can be -1. In this case, the value is inferred from
the length of the array and remaining dimensions.
order : {‘C’, ‘F’, ‘A’}, optional
Read the elements of a using this index order, and place the elements into the reshaped array using this index order. ‘C’ means to
read / write the elements using C-like index order, with the last axis
index changing fastest, back to the first axis index changing slowest.
‘F’ means to read / write the elements using Fortran-like index order,
with the first index changing fastest, and the last index changing
slowest. Note that the ‘C’ and ‘F’ options take no account of the
memory layout of the underlying array, and only refer to the order of
indexing. ‘A’ means to read / write the elements in Fortran-like index
order if a is Fortran contiguous in memory, C-like order otherwise.
Returns:
reshaped_array : ndarray
This will be a new view object if possible; otherwise, it will be a copy. Note there is no guarantee of the memory layout (C- or
Fortran- contiguous) of the returned array.
I have a 2-d numpy array that I convert into a 3-d array by using:
trainSet = numpy.reshape(trainSet, (trainSet.shape[0], 1, trainSet.shape[1]))
My problem is I'm trying to adjust values in the 1st axis (the new one). However, no matter what I try I can not actually figure out how to access the 1's in the new axis. I've tried a multitude of print statements such as
print(trainSet[0][0][0])
print(trainSet[1][0][1])
but no matter what I try I can not print out any 1's, just the contents of the 2-d array. I know it's a 3d array because when I use
print(trainSet.shape)
I get
(12, 1, 793)
Any help would be immensely appreciated!
The one refers to the size of that particular dimension, not the content of the array. The array [[5 5]] has shape (1, 2) but, still, you don't really expect its values must include 1 for that reason, do you?
What inserting a dimension of length one does is increasing the nesting level by one, so a 2d array [[a b] [c d]] will become [[[a b] [c d]]] or [[[a b]] [[c d]]] or [[[a] [b]] [[c] [d]]] depending on where you insert the new axis.
In particular, there are no new elements.
If I didn't mistake your question, what you want is
trainSet[:,0,:]
the amount of values remains the same, you're just creating a 3D structure instead of the 2D structure.
I am trying to translate a piece of 'for' loop code from Matlab to Python. And there is one statement in this block: A[B]=C. All those three A, B and C are matrices. In python, I need to write as A[B-1]=C, because of the difference of index criteria between Matlab and Python.
When B is non-empty, this statement goes well in python. However, if B is empty, this statement goes like this:
A11 = np.copy(A[:,B-1]) #Remind that B is an empty matrix, like B=np.array([0])
IndexError:arrays used as indices must be of integer (or boolean) type
Actually, if B is empty, what I want to have for matrix A11 is just another empty matrix. Definitely I can use a if block to define what matrix A11 should be when B is an empty matrix. But it will be too fussy because I have another 5 statement like this kind of using matrix as an index. Could you give me an example that shows me how to fix this problem? Thanks a lot!
B = np.array([0]) does not generate an empty matrix, it just converts the list [0] into a numpy array.
I suppose you meant something like B = np.zeros(0) (where the argument is a shape). Numpy's default is dtype =float64 but in order to use an array for indexing integer or boolean type is required. For a non-empty array with values that are in fact integers numpy figures out that it can just change the dtype.
To fix your problem you can simply specify the dtype (to int or boolean) when you initialize it, i.e. B = np.zeros(0, dtype=np.int) works fine. A will then be an 'empty matrix' in the sense that one of its shape dimensions is 0 - the others however do not change.
I have an array of 3 dimensional vectors. The dimension of the array is arbitrary: it could be a single (N×3), double (M×N×3), triple (K×M×N×3) etc. I need to operate on two components of the vector while preserving the other dimensions.
For example, if I know it is three dimensionsional, I could do the following:
R = numpy.arctan2(A[:,:,:,1], A[:,:,:,0])
which gives me a three dimensional array of scalar values.
Now, to be able to do this on arbitrary number of dimensions. I need to slice over all other dimensions except the the last. So far, I'm able to do it with this:
s = [numpy.s_[:]] * (len(A.shape)-1)
R = numpy.arctan2(A[s+[1]], A[s+[0]])
which works even for single vectors. Is there a more numpythonic way of achieving the above?
I found an even nicer way. This here works for me
R = numpy.arctan2(A[...,1],A[...,0])