Is there an equivalent to the MATLAB size() command in Numpy?
In MATLAB,
>>> a = zeros(2,5)
0 0 0 0 0
0 0 0 0 0
>>> size(a)
2 5
In Python,
>>> a = zeros((2,5))
>>> a
array([[ 0., 0., 0., 0., 0.],
[ 0., 0., 0., 0., 0.]])
>>> ?????
This is called the "shape" in NumPy, and can be requested via the .shape attribute:
>>> a = zeros((2, 5))
>>> a.shape
(2, 5)
If you prefer a function, you could also use numpy.shape(a).
np.method(x) and x.method when x is a numpy object?
shape isn't a method, it's a property. And for many methods there are corresponding module-level functions, but not for all of them. You need to consult the documentation for details.Yes numpy has a size function, and shape and size are not quite the same.
Input
import numpy as np
data = [[1, 2, 3, 4], [5, 6, 7, 8]]
arrData = np.array(data)
print(data)
print(arrData.size)
print(arrData.shape)
Output
[[1, 2, 3, 4], [5, 6, 7, 8]]
8 # size
(2, 4) # shape
[w,k] = a.shape will give you access to individual sizes if you want to use it for loops like in matlab
shapeis an attribute of arrays and a function in the numpy model but not a method of array objects. Is there an obvious answer? Does it feel like it merits a separate SO question, or is it too potentially opinion-based?a.shape = (10,)