Difference between numpy.ndarray and list (dimension, size)

Since numpy's ndarray and Python's built-in list were messed up, I tried to summarize each dimension and size in my own way. This is a summary of beginners.

Summary

As a result of google, the major size acquisition method was to use the len () function, np.shape () function, and instance variable ndarray.shape.

In conclusion, I think it's cheap pie ** if you use ** np.shape (). Also, I think it is good to remember that the ** built-in list type realizes a two-dimensional array representation by using the list comprehension notation **.

Size acquisition function

How to get the size ndarray type Built-in list type
len() 1D only 1D only
np.shape() n dimensions n dimensions
ndarray.shape n dimensions - ※

Also, of course, the output representation did not change depending on the type and dimension.

Notation of size

dimension ndarray type Built-in list type
1D (3, ) (3, )
2D (3, 4) (3, 4) ※

numpy.ndarray type

The numpy.array () function seems to be a handy tool to easily create an ndarray.

code

Since it is a numpy.ndarray type, ** there is no , between numbers **.

import numpy as np 
A_np = np.array([5, 2], dtype=int)
B_np = np.array([10, 9], dtype=int)
C_np = np.array([1, 11], dtype=int)
D_np = np.array([9, 1], dtype=int)
#Result of substitution
A_np : [5 2]
B_np : [10  9]
C_np : [ 1 11]
D_np : [9 1]

Next, let's look at the size. Check with the instance variable ndarray.shape. A one-dimensional ndarray type array is represented by `(number of elements,)`.
#Array size
A_np.shape : (2,)
B_np.shape : (2,)
C_np.shape : (2,)
D_np.shape : (2,)

Built-in list type

Python built-in type

Code one-dimensional array

Since it is a built-in list format, **, `is inserted between numbers **.

A = [5, 2]
B = [10, 9]
C = [1, 11]
D = [9, 1]
#Result of substitution
A : [5, 2]
B : [10, 9]
C : [1, 11]
D : [9, 1]

#### result Similarly, let's look at the size next. Even in the built-in list, the array size is returned by numpy.shape (). There are two elements in a one-dimensional array.
#Size np.shape()Function version
np.shape(A) : (2,)
np.shape(B) : (2,)
np.shape(C) : (2,)
np.shape(D) : (2,)

In the case of one dimension, the size (= number of elements) can also be confirmed with the built-in variable len (). In the case of 2D, the size is not returned. It is just the number of elements in list.
#Size len()Function version
len(A) : 2
len(B) : 2
len(C) : 2
len(D) : 2

Code 2D array

The 2D of the built-in list uses "list comprehension notation".

A = [[12, -8, 4], [0, 6, -10]
#Result of substitution
[[12, -8, 4], [0, 6, -10]

The output of the built-in variable len () has 2 elements in list.

#size
len(A) : 2

There are two ways to find the 2D array size of the built-in list type. The first is to use the np.shape function. The second method is to convert to numpy.ndarray type and then use ndarray.shape.

First, the built-in list type remains

A = [[12, -8, 4], [0, 6, -10]]
A : 
[[12, -8, 4], [0, 6, -10]]

np.shape(A) : (2, 3)

Second, convert to ndarray type Since the built-in list type is changed to dumpy.ndarray type, , between numbers disappear.

A = [[12, -8, 4], [0, 6, -10]
A_np = np.array(A)
A_np : 
[[ 12  -8   4]
 [  0   6 -10]]

A_np.shape : (2, 3)

Caution

When you want to use a two-dimensional array with the built-in list type, it seems to be troublesome when the number of elements of each element is different. Specifically,

A = [[12, -8, 4], [0, 6, -10, 2]]

At this time, it is a built-in list type two-dimensional array with two list type elements inside. The lengths of those elements are different.
The result at this time is

len(A) : 2
np.shape(A) : (2,)

Even if you convert it to numpy.ndarray

A_np = np.array(A)
A_np : 
[[12, -8, 4] [0, 6, -10, 2]]

np.shape(A_np) : (2,)

The , between the numbers are not gone, and the elements are a list type ndarray type array (?). I'm sorry if I said something strange. I'm not confident. By the way, np.array ([[12, -8, 4] [0, 6, -10, 2]], dtype = int) resulted in an error.
In other words, list comprehensions with different numbers of elements return the length of the list because the size cannot be measured as a matrix. Well, is it natural?

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