'list' object has no attribute 'shape'

Question

how to create an array to numpy array?

def test(X, N):
    [n,T] = X.shape
    print \"n : \", n
    print \"T : \", T



if __name__==\"__main__\":

    X =         


        

Answer 1

Use numpy.array to use shape attribute.

>>> import numpy as np
>>> X = np.array([
...     [[-9.035250067710876], [7.453250169754028], [33.34074878692627]],
...     [[-6.63700008392334], [5.132999956607819], [31.66075038909912]],
...     [[-5.1272499561309814], [8.251499891281128], [30.925999641418457]]
... ])
>>> X.shape
(3L, 3L, 1L)

NOTE X.shape returns 3-items tuple for the given array; [n, T] = X.shape raises ValueError.

Answer 2

firstly u have to import numpy library (refer code for making a numpy array) shape only gives the output only if the variable is attribute of numpy library .in other words it must be a np.array or any other data structure of numpy. Eg.

`>>> import numpy
>>> a=numpy.array([[1,1],[1,1]])
>>> a.shape
(2, 2)`

Answer 3

import numpy
X = numpy.array(the_big_nested_list_you_had)

It's still not going to do what you want; you have more bugs, like trying to unpack a 3-dimensional shape into two target variables in test.

Answer 4

if the type is list, use len(list) and len(list[0]) to get the row and column.

l = [[1,2,3,4], [0,1,3,4]]

len(l) will be 2 len(l[0]) will be 4

Answer 5

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: a = [[2, 3, 4] [0, 1] [87, 8, 1]]

it is impossible to define 'shape' for variable 'a'. That is why 'shape' might be determined only with 'arrays' e.g.

b = numpy.array([[2, 3, 4]
                [0, 1, 22]
                [87, 8, 1]])

I hope this explanation clarifies well this question.

Answer 6

İf you have list, you can print its shape as if it is converted to array

import numpy as np
print(np.asarray(X).shape)