如何对多个物种分别lapply或申请GLM？

Question

我正在尝试运行我的数据集中多个不同物种的 GLM。目前我一直在为每个物种子设置我的数据并复制这段代码，结果变得一团糟。我知道必须有更好的方法来执行此操作（也许使用 lapply 函数？）但我不确定如何开始。

我运行在一个物种的 CPUE（每单位努力量的捕获量）上建立模型，并使用年份、盐度、流量和降雨量作为我的解释变量。

我的数据在这里：https://drive.google.com/file/d/1_ylbMoqevvsuucwZn2VMA_KMNaykDItk/view?usp=sharing

这是我试过的代码。它完成了工作，但我一直在复制这段代码并每次都改变物种。我希望找到一种方法来简化这个过程并稍微清理一下我的代码。

fish_df$pinfishCPUE <- ifelse(fish_df$Commonname == "Pinfish", fish_all$CPUE, 0)
#create binomial column
fish_df$binom <- ifelse(fish_df$pinfishCPUE > 0, 1,0)


glm.full.bin = glm(binom~Year+Salinity+Discharge +Rainfall,data=fish_df,family=binomial)
glm.base.bin = glm(binom~Year,data=fish_df,family=binomial)

#step to simplify model and get appropriate order
glm.step.bin = step(glm.base.bin,scope=list(upper=glm.full.bin,lower=~Year),direction='forward',
                    trace=1,k=log(nrow(fish_df)))

#final model - may choose to reduce based on deviance and cutoff in above step
glm.final.bin  = glm.step.bin
print(summary(glm.final.bin))

#calculate the LSMeans for the proportion of positive trips
lsm.b.glm = emmeans(glm.final.bin,"Year",data=fish_df)
LSMeansProp = summary(lsm.b.glm)

输出：

Call:
glm(formula = log.CPUE ~ Month + Salinity + Temperature, family = gaussian, 
    data = fish_B_pos)

Deviance Residuals: 
    Min       1Q   Median       3Q      Max  
-3.8927  -0.7852   0.1038   0.8974   3.5887  

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  2.38530    0.72009   3.313  0.00098 ***
Month        0.10333    0.03433   3.010  0.00272 ** 
Salinity    -0.13530    0.01241 -10.900  < 2e-16 ***
Temperature  0.06901    0.01434   4.811  1.9e-06 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for gaussian family taken to be 1.679401)

    Null deviance: 1286.4  on 603  degrees of freedom
Residual deviance: 1007.6  on 600  degrees of freedom
AIC: 2033.2

Number of Fisher Scoring iterations: 2

Answer 1

我会建议下一个方法，为模型创建一个函数，然后在列表上使用 lapply，该列表是通过变量 Commonname:[=17 将 split() 应用于数据框而产生的=]

library(emmeans)
#Load data
fish_df <- read.csv('fish_df.csv',stringsAsFactors = F)
#Code
List <- split(fish_df,fish_df$Commonname)
#Function for models
mymodelfun <- function(x)
{
  #Create binomial column
  x$binom <- ifelse(x$pinfishCPUE > 0, 1,0)
  
  
  glm.full.bin = glm(binom~Year+Salinity+Discharge +Rainfall,data=x,family=binomial)
  glm.base.bin = glm(binom~Year,data=x,family=binomial)
  
  #step to simplify model and get appropriate order
  glm.step.bin = step(glm.base.bin,scope=list(upper=glm.full.bin,lower=~Year),direction='forward',
                      trace=1,k=log(nrow(x)))
  
  #final model - may choose to reduce based on deviance and cutoff in above step
  glm.final.bin  = glm.step.bin
  print(summary(glm.final.bin))
  
  #calculate the LSMeans for the proportion of positive trips
  lsm.b.glm = emmeans(glm.final.bin,"Year",data=x)
  LSMeansProp = summary(lsm.b.glm)
  return(LSMeansProp)
}
#Apply function
Lmods <- lapply(List,mymodelfun)

在Lmods里面会有模型的结果，这里举个例子：

Lmods$`Atlantic Stingray`

输出：

 Year emmean    SE  df asymp.LCL asymp.UCL
 2009  -22.6 48196 Inf    -94485     94440

Results are given on the logit (not the response) scale. 
Confidence level used: 0.95

如何对多个物种分别lapply或申请GLM？

How to us lapply or sapply for GLM on multiple species separately?

r

glm