Question

我试图用glm估算足球队的优势。

# data is dataframe (structure on bottom). 
model <- glm(Goals ~ Home + Team + Opponent, family=poisson(link=log), data=data)

但得到错误：

Error in if (any(y < 0)) stop("negative values not allowed for the 'Poisson' family") : 
  missing value where TRUE/FALSE needed
In addition: Warning message:
In Ops.factor(y, 0) : ‘<’ not meaningful for factors

data：

> data
                      Team                 Opponent Goals Home
1 5a51f2589d39c31899cce9d9 5a51f2579d39c31899cce9ce     3    1
2 5a51f2579d39c31899cce9ce 5a51f2589d39c31899cce9d9     0    0
3 5a51f2589d39c31899cce9da 5a51f2579d39c31899cce9cd     3    1
4 5a51f2579d39c31899cce9cd 5a51f2589d39c31899cce9da     0    0

> is.factor(data$Goals)
[1] TRUE

Answer 1

来自glm()功能文档的“详细信息”部分：

典型的预测变量具有形式响应〜术语，其中响应是（数字）响应向量，术语是一系列术语，用于指定响应的线性预测变量。

因此，您需要确保目标列为数字：

df <- data.frame( Team= c("5a51f2589d39c31899cce9d9", "5a51f2579d39c31899cce9ce", "5a51f2589d39c31899cce9da", "5a51f2579d39c31899cce9cd"),
                  Opponent=c("5a51f2579d39c31899cce9ce", "5a51f2589d39c31899cce9d9", "5a51f2579d39c31899cce9cd", "5a51f2589d39c31899cce9da "),
                  Goals=c(3,0,3,0),
                  Home=c(1,0,1,0))

str(df)
#'data.frame':  4 obs. of  4 variables:
# $ Team    : Factor w/ 4 levels "5a51f2579d39c31899cce9cd",..: 3 2 4 1
# $ Opponent: Factor w/ 4 levels "5a51f2579d39c31899cce9cd",..: 2 3 1 4
# $ Goals   : num  3 0 3 0
# $ Home    : num  1 0 1 0


model <- glm(Goals ~ Home + Team + Opponent, family=poisson(link=log), data=df)

然后是输出：

> model


Call:  glm(formula = Goals ~ Home + Team + Opponent, family = poisson(link = log), 
    data = df)

Coefficients:
                      (Intercept)                               Home       Team5a51f2579d39c31899cce9ce  
                       -2.330e+01                          2.440e+01                         -3.089e-14  
     Team5a51f2589d39c31899cce9d9       Team5a51f2589d39c31899cce9da   Opponent5a51f2579d39c31899cce9ce  
                       -6.725e-15                                 NA                                 NA  
 Opponent5a51f2589d39c31899cce9d9  Opponent5a51f2589d39c31899cce9da   
                               NA                                 NA  

Degrees of Freedom: 3 Total (i.e. Null);  0 Residual
Null Deviance:      8.318 
Residual Deviance: 3.033e-10    AIC: 13.98

R：if（any（y <0））停止时出错（“Poisson'系列不允许负值”）

1 个答案: