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shape is a tuple that gives you an indication of the number of dimensions in the array. So in your case, since the index value of Y.shape[0] is 0, your are working along the first dimension of your array. 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). 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] 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 python - What does 'range (y.shape [1])' mean in "for i in range ... Toggle Shape Color For those new to PowerPoint, in the Insert tab under the Illustration group, there are Action Buttons at the bottom of the Shapes menu that can utilize (call) the following codes as they are. RuntimeError: Given normalized_shape= [256], expected input with shape [*, 256], but got input of size [640, 256, 14, 14] Asked 3 years, 2 months ago Modified 3 years, 2 months ago Viewed 5k times The requested array has an inhomogeneous shape after 2 dimensions. The detected shape was (200, 2) + inhomogeneous part. Concerned codelines are: ... Basically, the code iterates through a collections of images and performs some operations on such images that outputs the (1x4) vector. The image and its operated-output are then added to the dataset.
