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Re: Linear Regression


From: Przemek Klosowski
Subject: Re: Linear Regression
Date: Fri, 7 Dec 2018 13:29:36 -0500
User-agent: Mozilla/5.0 (X11; Linux x86_64; rv:60.0) Gecko/20100101 Thunderbird/60.2.1

On 12/7/18 12:52 PM, LucaLuca wrote:
hi,
look this pics

https://na01.safelinks.protection.outlook.com/?url=https%3A%2F%2Fgyazo.com%2F1f61457f939308195d72bba5f4266bc5&data=02%7C01%7Cprzemek.klosowski%40nist.gov%7Cc39f54f956684fe06ae208d65c6ceccb%7C2ab5d82fd8fa4797a93e054655c61dec%7C1%7C0%7C636798020232877101&sdata=d%2B1VRCbGZzh435lokc8nOMNqo2aUVdPvVfHgcU2VjNk%3D&reserved=0

i need to find linea regression

i use this formula:

A=[4 3 2 0 1 8]
B=[1 2 3 4 5 6]

regress(A,B)

is write :error: regress: y must be a column vector

regress(A',B') resolves that issue, but your B/A data doesn't look like your picture, so I am not sure if that's what you need.

By the way, for simple linear regression, you can do polyfit(x,y,1)


i don't understand :|

i read example but i don't understand that..

i need to find a series of points that approximate my range A

Do you mean some sort of linear approximation of A based on fitted data?

polyval(polyfit(x,y,1),x)





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