Equatiomatic 沒有從 logistic model 建立我的方程式

Question

我有以下數據，分類的響應變量是gleason_score ，而預測變量，連續變量是age 。 我已經使用glm function 擬合數據以進行邏輯回歸 model。

structure(list(age = c(69L, 60L, 78L, 73L, 80L, 78L, 89L, 75L, 
66L, 74L, 72L, 80L, 63L, 100L, 67L, 73L, 75L, 83L, 72L, 73L, 
50L, 75L, 70L, 56L, 75L, 70L, 90L, 65L, 70L, 80L, 73L, 70L, 70L, 
75L, 71L, 65L, 65L, 72L, 67L, 65L, 70L, 75L, 85L, 75L, 70L, 86L, 
74L, 78L, 64L, 70L, 65L, 65L, 70L, 74L, 77L, 75L, 65L, 80L, 70L, 
70L, 58L, 58L, 65L, 78L, 76L, 80L, 66L, 71L, 70L, 55L, 70L, 90L, 
78L, 67L, 65L, 60L, 69L, 80L, 72L, 76L, 68L, 77L, 88L, 69L, 79L, 
77L, 78L, 66L, 80L, 72L, 81L, 80L, 70L, 86L, 87L, 70L, 80L, 66L, 
60L, 50L, 69L, 63L, 75L, 68L, 68L, 75L, 63L, 74L, 54L, 81L, 72L, 
70L, 68L, 55L, 75L, 75L, 65L, 72L, 77L, 64L, 64L, 76L, 83L, 95L, 
85L, 70L, 75L, 75L, 61L, 95L, 72L, 81L, 87L, 70L, 77L, 70L, 65L, 
77L, 70L, 65L, 70L, 75L, 68L, 93L, 65L, 65L, 75L, 78L, 86L), 
    gleason_score = structure(c(2L, 2L, 1L, 2L, 2L, 2L, 2L, 1L, 
    2L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 1L, 2L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 
    2L, 1L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 2L, 1L, 
    2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
    2L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
    2L, 2L, 2L, 1L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 1L, 2L, 2L, 1L, 
    2L, 2L, 2L, 2L, 2L, 1L, 2L, 1L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 
    2L, 2L, 2L, 1L, 2L, 2L), .Label = c("0", "1"), class = "factor")), class = "data.frame", row.names = c(NA, 
-149L))

使用以下代碼：

attach(data2)
glm.fit=glm(gleason_score ~ age, family=binomial(link = "logit"))
plot(x=age, y=gleason_score)
lines(age, glm.fit$fitted.values)
summary(glm.fit)

我有這些結果：

Call:
glm(formula = gleason_score ~ age, family = binomial(link = "logit"))

Deviance Residuals: 
    Min       1Q   Median       3Q      Max  
-2.1515   0.4451   0.5806   0.6530   1.0105  

Coefficients:
            Estimate Std. Error z value Pr(>|z|)  
(Intercept) -2.17146    1.87588  -1.158   0.2470  
age          0.05155    0.02651   1.944   0.0518 .

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 141.02  on 148  degrees of freedom
Residual deviance: 137.00  on 147  degrees of freedom
AIC: 141

Number of Fisher Scoring iterations: 4

我使用以下代碼來構建方程式：

equatiomatic::extract_eq(glm.fit, wrap = FALSE,use_coefs = TRUE)

但我有這個錯誤信息：

Error in model$data[which(model$y == 1)[1], outcome_nm] : object of type 'environment' is not subsettable

請協助追蹤我的錯誤

Answer 1

不要使用attach(data2) .. 相反，將data2傳遞給glm()調用的data參數。

glm.fit=glm(gleason_score ~ age, data=data2, family=binomial(link = "logit"))
equatiomatic::extract_eq(glm.fit, wrap = FALSE,use_coefs = TRUE)

Output：

$$
\log\left[ \frac { \widehat{P( \operatorname{gleason\_score} = \operatorname{1} )} }{ 1 - \widehat{P( \operatorname{gleason\_score} = \operatorname{1} )} } \right] = -2.17 + 0.05(\operatorname{age})
$$

要了解原因，請在創建glm.fit時比較glm.fit$data

1. attach(data2)與
2. 不使用attach而是將data2傳遞給data arg

第二種方法是正確的。

在第一種方法（你的）下， glm.fit$data返回這個：

<environment: R_GlobalEnv>

在第二種方法（正確）下， glm.fit$data返回實際數據（注意這里只顯示前六行）

  age gleason_score
1  69             1
2  60             1
3  78             0
4  73             1
5  80             1
6  78             1