我一直在使用glmmTMB软件包在大型数据集上运行GLMM。下面的示例数据
structure(list(code = structure(c(1L, 1L, 1L, 1L), .Label = c("2388",
"2390", "12950", "12952", "12954", "12956", "12958", "12960",
"12962", "12964", "12966", "12968", "13573", "13574", "13575",
"13576", "13577", "14203", "19318", "19319", "19320", "19321",
"19322", "19515", "19517", "19523", "19524", "25534", "25535",
"25536", "25537", "25539", "25540", "25541", "25542", "25543",
"25544", "25545", "25546", "25547", "25548", "25549", "25550",
"25552", "25553", "27583", "27584", "27585", "27586", "27588",
"27589", "27590", "27591", "27592", "27593", "27594", "27595",
"27596", "27597", "27598", "27599", "27600", "27601", "27602",
"27603", "27604", "27605", "27606", "27607", "27608", "27609",
"27610", "27611", "27613", "27614", "27615", "27616", "27617",
"27618", "27619", "27620", "27621", "27622", "27624", "27625",
"27626", "27627", "27629", "27630", "27631", "27632", "34176",
"34177", "34178", "34179", "52975", "52977", "52978", "54814",
"54815", "54816", "54817", "54821", "54822", "54823", "54824",
"54825", "54835", "54837", "54838", "54839", "54840", "54841",
"54842", "54843", "54844", "54845", "54846", "54847", "54848",
"54849", "54851", "54852", "54853", "54856", "54858", "54859",
"54860", "54863", "54864", "54866", "54869", "54872", "54873",
"54874", "54875", "54876", "54877", "54878", "54880", "54882",
"54883", "54884", "54886", "54887", "54889", "54890", "54892",
"54893", "54895", "54896", "54898", "54899", "54900", "54901",
"54903", "54904", "54905", "54906", "54911", "54912", "54914",
"54915", "54931", "54933", "54934", "54935", "54937", "54939",
"54940", "54941", "54942", "54943", "54944", "54945", "54946",
"54947", "54948", "54950", "54952", "54954", "54955", "54957",
"54958", "54959", "54961", "54962"), class = "factor"), station =
c("PB14",
"PB14", "PB16", "PB16"), species = c("Silvertip Shark", "Silvertip Shark",
"Silvertip Shark", "Silvertip Shark"), sex = c("F", "F", "F",
"F"), size = c(112, 112, 112, 112), datetime = c("1466247120",
"1466247420", "1467026100", "1469621400"), year = c("2016", "2016",
"2016", "2016"), month = c(6, 6, 6, 7), hour = c("11", "11",
"12", "13"), season = c("dry season", "dry season", "dry season",
"dry season"), daynight = c("day", "day", "day", "day"), time_diff = c(4,
5, 5821, 43255), offshore = structure(c(2L, 2L, 1L, 1L), .Label =
c("offshore",
"onshore"), class = "factor"), rowN = 1:4), row.names = c(NA,
4L), class = "data.frame")
我希望对80％的数据运行模型，然后使用predict（）函数对其进行验证。使用下面的代码
Off_80 <- Off %>% sample_frac(.80)
Off_20 <- anti_join(Off, Off_80, by = 'rowN')
OffMod_80 <- glmmTMB(offshore ~ sex + log(size) + species*daynight + species*season + (1|code), family=binomial(), data=Off_80)
pred_Off20 <- as.data.frame(predict(OffMod_80, newdata=Off_20, type="response", allow.new.levels=TRUE))
然后我将比较预测结果和观察结果，以验证模型的强度。
但是使用此方法而不是得到“离岸”或“在岸”响应，我得到了一个数值。
predict()
1 0.2807461
2 0.2631816
3 0.2631816
4 0.2807461
反正有没有获得预测函数以吐出二项式响应的方法？或者如何将这些值解释为二项式？
最初，我的响应变量为1或0，但是遵循此post，我将值更改为factor。但仍然predict（）吐出一个数值。
任何帮助表示赞赏！
答案 0 :(得分：1)
predict()（对于二项式模型）返回成功或失败的概率，它不会返回1或0（因为您只能以某个 probability < / em>）。因此，如果要检查模型性能，可以尝试计算曲线下的面积：
library(glmmTMB)
library(pROC)
data(mtcars)
n <- nrow(mtcars)
train <- sample(1:n, n * .8, TRUE)
mtcars_train <- mtcars[train, ]
mtcars_test <- mtcars[-train, ]
m <- glmmTMB(formula = vs ~ hp + wt + (1 | gear), family = binomial, data = mtcars_train)
p <- predict(m, newdata = mtcars_test)
auc(roc(response = mtcars_test$vs, predictor = p))
#> Area under the curve: 0.9107
由reprex package（v0.3.0）于2019-05-29创建
另一种方法是比较完整模型与零模型的正确预测（PCP）的百分比（参见 Herron，M.（1999）。有限因变量模型中的后估计不确定性。政治分析，8， 83–98 ）。在这里，完整模型的PCP应该明显更高：
library(glmmTMB)
library(insight)
data(mtcars)
m <- glmmTMB(formula = vs ~ hp + wt + (1 | gear), family = binomial, data = mtcars)
m0 <- glmmTMB(formula = vs ~ 1 + (1 | gear), family = binomial, data = mtcars)
y <- insight::get_response(m)
y0 <- insight::get_response(m0)
n <- nobs(m)
n0 <- nobs(m0)
p <- predict(m, type = "response")
p0 <- predict(m0, type = "response")
pcp_full <- (sum(1 - p[y == 0]) + sum(p[y == 1])) / n
pcp_null <- (sum(1 - p0[y0 == 0]) + sum(p0[y0 == 1])) / n0
# percentage correct predictions full model
pcp_full
#> [1] 0.8374718
# percentage correct predictions null model
pcp_null
#> [1] 0.6614221
由reprex package（v0.3.0）于2019-05-29创建
最后，如果您确实想要0和1之类的东西，则可以使用simulate()，它返回的值与原始响应相同。比起您可以比较多少模拟响应与实际响应相符：
library(glmmTMB)
data(mtcars)
m <- glmmTMB(formula = vs ~ hp + wt + (1 | gear), family = binomial, data = mtcars)
# simulate response, first column = successes
s <- simulate(m)$sim_1[, 1]
# proportion of response values that equal simulated responses
mean(mtcars$vs == s)
#> [1] 0.875
由reprex package（v0.3.0）于2019-05-29创建
答案 1 :(得分：0)
预测函数可为您提供在模型中为测试数据集中表示的值提供的预测变量的情况下，离岸变量为“离岸”的概率。您可能需要查看以下问题的答案：Type parameter of the predict() function。它们突出显示了？predict.glm和？predict.rpart之间的区别。我无法从您提供的数据中复制您的分析，而是查看以下示例：https://www.theanalysisfactor.com/r-tutorial-glm1/。这显示了glm的工作原理（通常）以及predict.glm函数为您提供了什么。希望对您有所帮助。