Why np.zeros(3) and np.zeros((3,1)) are of different dimensions in python?-CodePudding

I'm trying to generate N data points for three random variables that are jointly normal in python. If I use the following code:

import numpy as np
import scipy
import pandas
import sys
from scipy.linalg import block_diag
from pandas import *
N=100
Sigma=np.identity(3)
Mu=np.zeros((3,1))
Z=np.random.multivariate_normal(Mu, Sigma, N)

I got the following error message:

in <module>
    Z=np.random.multivariate_normal(Mu, Sigma, N)
  File "mtrand.pyx", line 4067, in numpy.random.mtrand.RandomState.multivariate_normal
ValueError: mean must be 1 dimensional

This means that the dimension of np.zeros((3,1)) is not 1. After changing the line Mu=np.zeros((3,1)) to Mu=np.zeros(3), it works. This implies that np.zeros(3) is 1 dimensional.

As np.zeros(3) and np.zeros((3,1)) are both an array of three zeros, I guess naturally both should be 1 dimensional. Using Mu.ndim in each case, I found that the dimension of np.zeros(3) is one and the dimension of np.zeros((3,1)) is two. My question is:

Why does Python make a distinction between np.zeros((3,1)) and np.zeros(3) regarding their dimensions (why is this distinction useful)?

CodePudding user response：

It's normal for them to have different dimensions. The first one only has 1 array made of 3 zeros and the second one has 3 arrays each one made of 1 zero.

If you print Mu[0] in your example, you will get a list [0.] while if you print Mu[0] after using np.zeros(3) to define it, you will get 0.0

I can think of cases where this is distinction is useful especially when working with features in machine learning. If I have a sequence of features of size 1, I would want to use a dimension [n,1] and not [n] because that helps the model (let's say LSTM) make a difference between the sequence size and the feature size.