Sonuçlar lm () den bir denkleme nasıl çevrilir?

lm()Bir değeri tahmin etmek için kullanabiliriz , ancak bazı durumlarda sonuç formülünün denklemine hala ihtiyacımız var. Örneğin, denklemi parsellere ekleyin.

Consider this example:

set.seed(5)            # this line will allow you to run these commands on your
                       # own computer & get *exactly* the same output
x = rnorm(50)
y = rnorm(50)

fit = lm(y~x)
summary(fit)
# Call:
# lm(formula = y ~ x)
# 
# Residuals:
#      Min       1Q   Median       3Q      Max 
# -2.04003 -0.43414 -0.04609  0.50807  2.48728 
# 
# Coefficients:
#             Estimate Std. Error t value Pr(>|t|)
# (Intercept) -0.00761    0.11554  -0.066    0.948
# x            0.09156    0.10901   0.840    0.405
# 
# Residual standard error: 0.8155 on 48 degrees of freedom
# Multiple R-squared: 0.01449,  Adjusted R-squared: -0.006046 
# F-statistic: 0.7055 on 1 and 48 DF,  p-value: 0.4051 

The question, I'm guessing, is how to figure out the regression equation from R's summary output. Algebraically, the equation for a simple regression model is:

$$\hat y_i = \hat\beta_0 + \hat\beta_1 x_i + \hat\varepsilon_i \\ \text{where } \varepsilon\sim\mathcal N(0,~\hat\sigma^2)$$ We just need to map the summary.lm() output to these terms. To wit:

• $\hat\beta_0$ is the Estimate value in the (Intercept) row (specifically, -0.00761)
• $\hat\beta_1$ is the Estimate value in the x row (specifically, 0.09156)
• $\hat\sigma$ is the Residual standard error (specifically, 0.8155)

Plugging these in above yields:

$$\hat y_i = -0.00761~+~0.09156 x_i~+~\hat\varepsilon_i \\ \text{where } \varepsilon\sim\mathcal N(0,~0.8155^2)$$ For a more thorough overview, you may want to read this thread: Interpretation of R's lm() output.

If what you want is to predict scores using your resulting regression equation, you can construct the equation by hand by typing summary(fit) (if your regression analysis is stored in a variable called fit, for example), and looking at the estimates for each coefficient included in your model.

For example, if you have a simple regression of the type $y=\beta_0+\beta_1x+\epsilon$, and you get an estimate of the intercept ($\beta_0$) of +0.5 and an estimate of the effect of x on y ($\beta_1$) of +1.6, you would predict an individual's y score from their x score by computing: $\hat{y}=0.5+1.6x$.

However, this is the hard route. R has a built-in function, predict(), which you can use to automatically compute predicted values given a model for any dataset. For example: predict(fit, newdata=data), if the x scores you want to use to predict y scores are stored in the variable data. (Note that in order to see the predicted scores for the sample on which your regression was performed, you can simply type fit$fitted or fitted(fit); these will give you the predicted, a.k.a. fitted, values.)

If you want to show the equation, like to cut/paste into a doc, but don't want to fuss with putting the entire equation together:

R> library(MASS)
R> crime.lm <- lm(y~., UScrime)
R> cc <- crime.lm$coefficients
R> (eqn <- paste("Y =", paste(round(cc[1],2), paste(round(cc[-1],2), names(cc[-1]), sep=" * ", collapse=" + "), sep=" + "), "+ e"))
[1] "Y = -5984.29 + 8.78 * M + -3.8 * So + 18.83 * Ed + 19.28 * Po1 + -10.94 * Po2 + -0.66 * LF + 1.74 * M.F + -0.73 * Pop + 0.42 * NW + -5.83 * U1 + 16.78 * U2 + 0.96 * GDP + 7.07 * Ineq + -4855.27 * Prob + -3.48 * Time + e"

Building on keithpjolley's answer, this replaces the '+' signs used in the separator with the actual sign of the co-efficient.

modelcrime <- lm(y~., UScrime)
modelcrime_coeff <- modelcrime$coefficients
modelcrime_coeff_sign <- sign(modelcrime_coeff)
modelcrime_coeff_prefix <- case_when(modelcrime_coeff_sign == -1 ~ " - ",
                                     modelcrime_coeff_sign == 1 ~ " + ",
                                     modelcrime_coeff_sign == 0 ~ " + ")
modelcrime_eqn <- paste("y =", paste(if_else(modelcrime_coeff[1]<0, "- ", ""),
                                         abs(round(modelcrime_coeff[1],3)),
                                     paste(modelcrime_coeff_prefix[-1],
                                           abs(round(modelcrime_coeff[-1],3)),
                                           " * ",
                                           names(modelcrime_coeff[-1]),
                                           sep = "", collapse = ""),
                                     sep = ""))
modelcrime_eqn

produces the result

[1] "y = - 5984.288 + 8.783 * M - 3.803 * So + 18.832 * Ed + 19.28 * Po1 - 10.942 * Po2 - 0.664 * LF + 1.741 * M.F - 0.733 * Pop + 0.42 * NW - 5.827 * U1 + 16.78 * U2 + 0.962 * GDP + 7.067 * Ineq - 4855.266 * Prob - 3.479 * Time"