The shape attribute for numpy arrays returns the dimensions of the array. If Y has n rows and m columns, then Y.shape is (n,m). So Y.shape[0] is n.
python - What does .shape [] do in "for i in range (Y.shape [0 ...
Shape n, expresses the shape of a 1D array with n items, and n, 1 the shape of a n-row x 1-column array. (R,) and (R,1) just add (useless) parentheses but still express respectively 1D and 2D array shapes, Parentheses around a tuple force the evaluation order and prevent it to be read as a list of values (e.g. in function calls).
On the other hand, x.shape is a 2-tuple which represents the shape of x, which in this case is (10, 1024). x.shape[0] gives the first element in that tuple, which is 10. Here's a demo with some smaller numbers, which should hopefully be easier to understand.
python - x.shape [0] vs x [0].shape in NumPy - Stack Overflow
In python shape[0] returns the dimension but in this code it is returning total number of set. Please can someone tell me work of shape[0] and shape[1]? Code: m_train = train_set_x_orig.shape[0]
python - What does 'range (y.shape [1])' mean in "for i in range ...
Why not start from the 2D shape and then switch the behavior to 1D which is what you need to do anyway after grouping the line. However, when switching the shape behavior to 1D it still doesn't act as a line connector, anymore. Instead, moving the endpoint simply changes the length of the shape. I'm using Visio 2302.