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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)
Re: Linear Regression, Marco Atzeri, 2018/12/07