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The shape attribute for numpy arrays returns the dimensions of the array The actual relation between the two is size = np.prod(shape) so the distinction should indeed be a bit more obvious in the arguments names. If y has n rows and m columns, then y.shape is (n,m)
(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 Shape (in the numpy context) seems to me the better option for an argument name Yourarray.shape or np.shape() or np.ma.shape() returns the shape of your ndarray as a tuple
And you can get the (number of) dimensions of your array using yourarray.ndim or np.ndim()
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. You can think of a placeholder in tensorflow as an operation specifying the shape and type of data that will be fed into the graph.placeholder x defines that an unspecified number of rows of shape (128, 128, 3) of type float32 will be fed into the graph A placeholder does not hold state and merely defines the type and shape of the data to flow.
So in line with the previous answers, df.shape is good if you need both dimensions, for a single dimension, len() seems more appropriate conceptually Looking at property vs method answers, it all points to usability and readability of code. For any keras layer (layer class), can someone explain how to understand the difference between input_shape, units, dim, etc. For example the doc says units specify the output shape of a layer.
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] 8 list object in python does not have 'shape' attribute because 'shape' implies that all the columns (or rows) have equal length along certain dimension Let's say list variable a has following properties:
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