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这个是regress的使用说明，用来进行多元线性回归。 - ^1 e, E* ~$ w- _第一个问题：regress的第三个参数为置信水平，可填可不填，但是不管我填写与否，都会有一个warning：R-square and the F statistic are not well-defined unless X has a column of ones. 3 _- k4 h0 A! d% L0 PType "help regress" for more information. |! U& p; X* R% @: R
- o0 |( t6 U5 f r- q( G/ p/ @( D第二个问题：r是预测值和真实值的差，r'*r应该是残差平方和吗？它能够用来评价回归模型的好坏吗？ 2 i6 x, G2 A& G2 n* D: L4 y6 `1 f' A) m. I. G4 v* X
第三个问题：stats是一个数组，The vector stats contains the R2 statistic along with the F and p values for the regression# x" [6 l# ?) h0 |% y% o) P! Q. G
很多网上的使用说明，包括matlab的help都只提到了stats数组的3个成员，但是我使用regress函数后stats有4个成员，请问另外一个是代表什么问题+ R3 u; O( n. V) H, G$ w# f
]. Y j: H1 D& f. Y; B; K- @8 D 作者: ExxNEN 时间: 2020-4-14 18:28
REGRESS Multiple linear regression using least squares. % P( {2 e" k* T: y4 eB = REGRESS (Y,X) & k' d0 n9 w4 k, Y! [returns the vector B of regression coefficients in the 3 r0 u2 T( F [- ^8 m% {- S/ e' k+ ylinear model Y = X*B. % ?6 Q; R+ r: ^3 ] ?' b, R2 h h9 A- r
X is an n-by-p design matrix, with rows # j# \- F! |8 Gcorresponding to observations and columns to predictor variables. ( b$ L0 d* I- ~2 N8 g2 m / ]3 ~, |% R) q; z6 }, X5 ]% wY is an n-by-1 vector of response observations.0 m. W7 G" W' ^% O& h! x {" c
REGRESS ; {+ `! t2 M$ W. B多元线性回归——用最小二乘估计法* @6 c" K" ^ W" v
B = REGRESS (Y,X) ，& m& K- R8 [& ^, W# a% }
/ T' {! A' t1 m; a' Y5 |0 V返回值为线性模型Y = X*B的回归系数向量7 {9 ]: R3 X& T
X ，n-by-p 矩阵，行对应于观测值，列对应于预测变量 ' Q* E* I8 X+ |' P2 P$ U5 I Y ，n-by-1 向量，观测值的响应（即因变量）2 P5 p( G) }7 }
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[B,BINT] = REGRESS (Y,X) # N% V% u/ {8 l- x2 ereturns a matrix BINT of 95% confidence intervals for B.; G, u( q/ T' A: n1 J, j; ~4 u! D
BINT，B的95%的置信区间矩阵 D4 `: \ j; }& k/ B; ^
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[B,BINT,R] = REGRESS (Y,X) & g* W/ A) U& i/ {9 creturns a vector R of residuals.% j& f0 ~/ y% C0 y V6 Z/ X) t
R，残差向量 & B2 l! l( L, [9 h1 o4 \( E2 x3 ]6 A" }; a% Q
[B,BINT,R,RINT] = REGRESS (Y,X) 9 `& o. P, g0 g4 K( D- a
returns a matrix RINT of intervals that 4 i2 q6 u" G! a- V# ycan be used to diagnose outliers.5 r% R0 D' E% w" v
9 K ~9 p! `, t* CIf RINT(i,: ) does not contain zero, & V9 [, U2 o/ q* f' x- M ! H: ^5 _& H5 }8 ~5 H9 wthen the i-th residual is larger than would be expected, at the 5%6 Z3 a% }1 G- B$ C! T I
significance level. 6 u* r9 g1 W9 @& |: P& c5 V: J ~, r3 z) u% q
This is evidence that the I-th observation is an outlier. 0 ^, ]9 l \6 a/ O& \& w. I, \/ B! ]
RINT，区间矩阵，该矩阵可以用来诊断异常（即发现奇异观测值，译者注）。9 z- D$ I5 `& v) o2 T
如果RINT(i，：)所定区间没有包含0，则第i个残差在默认的5%的显著性水平比我们所预期的要大，这可说明第i个观测值是个奇异点（即说明该点可能是错误而无意义的，如记录错误等，译者注）; @/ ^9 @5 v+ V. y
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[B,BINT,R,RINT,STATS] = REGRESS (Y,X) 8 O" D, ^6 A5 x9 V8 _6 {: A0 {
returns a vector STATS containing 0 m- w6 N8 T% X- L. m; I! O8 Pthe R-square statistic, the F statistic and p value for the full model,and an estimate of the error variance.) J) {8 p% y% @
7 e0 ~( L% V# y4 o3 o K[...] = REGRESS (Y,X,ALPHA) ' P) a& e {5 @+ z
uses a 100*(1-ALPHA)% confidence level to compute BINT, and a (100*ALPHA)% significance level to compute RINT.$ c# d6 l* p3 U3 y, D: j! n
用100*(1-ALPHA)%的置信水平来计算BINT， 2 m5 Q+ ~- Z- Z6 D用(100*ALPHA)%的显著性水平来计算RINT 2 F1 t. Z8 Z' k% [9 b6 |0 ` 8 W8 t( D! ^9 p) c9 f- [4 tX should include a column of ones so that the model contains a constant / q/ ?; D' i! \$ G$ kterm. 4 s( o, x: B9 _. X lThe F statistic and p value are computed under the assumption ( l# w9 G5 H* m2 x8 ythat the model contains a constant term, and they are not correct for& p0 Q: g: M4 {' l8 _$ ^
models without a constant. ( E* u) M: [3 n% w8 H {% ~ @The R-square value is one minus the ratio of , L2 x* X% |. p# Bthe error sum of squares to the total sum of squares.' Q2 S, R3 C, X$ a2 Z
This value can+ n0 X+ a3 O$ f( x; t. x$ P
be negative for models without a constant, which indicates that the model is not appropriate for the data.$ a- Z% Y Y& P
X应该包含一个全“1”的列，这样则该模型包含常数项。F统计量和p值是在模型有常数项的假设下计算的，如果模型没有常数项，则计算得的F统计量和p值是不正确的。The R-square value is one minus the ratio of the error sum of squares to the total sum of squares.（此句无法把握，请高手帮忙~~!）若模型没有常数项，则这个值可以为负值，这也表明这个模型对数据是不合适的。（即数据不适合用多元线性模型，译者注）* `; l4 H6 Q. U
5 z5 i$ F" L9 AIf columns of X are linearly dependent, REGRESS sets the maximum ) K7 x+ _1 P. c' o8 A/ m+ B) spossible number of elements of B to zero to obtain a "basic solution", 2 U, z4 C8 Z2 b/ d0 S0 [% z" Tand returns zeros in elements of BINT corresponding to the zero elements of B.1 K8 X* M, w3 q) v5 h% g, l2 G9 g5 v
如果X的列是线性相关的，则REGRESS将使B的元素中“0”的数量尽量多，以此获得一个“基本解”，并且使B中元素“0”所对应的BINT元素为“0”。4 g* v- t8 R: O% Y& s/ k& K0 }- _
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REGRESS treats NaNs in X or Y as missing values, and removes them. REGRESS" G& \: F* j, C. m% H2 E" T5 u
将X或者Y中的NaNs当作缺失值处理，并且移除它们。