Match.arg（类型）中的错误：＆＃x27; arg＆＃x27;试图运行方差分析时必须长度为1

Question

我正在尝试将两个GLM模型与R中的ANOVA函数进行比较，以测试它们是否变得更好或更复杂，但是当我尝试运行ANOVA时，我总是会收到此错误消息：

Error in match.arg(type) : 'arg' must be of length 1

我的代码看起来像这样：

model1 <- glm(Tempus~Typ, data=tempusB, family=binomial)

model2 <- glm(Tempus~Typ+Semantik, data=tempusB, family=binomial)

Anova(model1, model2, test="LRT")

我已经看过在ANOVA功能的“帮助”部分中，如果我正确理解，似乎使用GLM模型，我只能使用一个参数？这是这里的问题吗？这是否意味着在这种情况下不可能使用ANOVA功能？

很抱歉，如果这是一个愚蠢的问题，我是R的绝对初学者，我的教授告诉我使用ANOVA功能，所以我尝试了。

编辑：这是我数据的示例。

dput(head(tempusB)
structure(list(Tempus = structure(c(2L, 2L, 1L, 1L, 2L, 2L), .Label = c("kT", 
"T"), class = "factor"), Typ = structure(c(3L, 1L, 3L, 3L, 3L, 
3L), .Label = c("K", "M", "V"), class = "factor"), Person = structure(c(3L, 
3L, 3L, 3L, 3L, 1L), .Label = c("1", "2", "3", "U"), class = "factor"), 
    Numerus = structure(c(2L, 1L, 1L, 1L, 1L, 1L), .Label = c("PL", 
    "SG", "U"), class = "factor"), Aktionsart = structure(c(1L, 
    1L, 2L, 2L, 2L, 1L), .Label = c("dur", "perf"), class = "factor"), 
    Semantik = structure(c(3L, 3L, 1L, 1L, 1L, 1L), .Label = c("Tat", 
    "Vor", "Zust"), class = "factor"),  
    Satz = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = c("HS", 
    "NS"), class = "factor"), Klasse = structure(c(3L, 2L, 1L, 
    1L, 1L, 1L), .Label = c("sch", "st", "unr"), class = "factor"), 
    Thema = structure(c(2L, 2L, 5L, 5L, 2L, 1L), .Label = c("Arbeit", 
    "Familie", "Gem", "Leben", "Mission", "Pers", "PNG"), class = "factor"), 
    Speaker = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = c("AC", 
    "AH", "BC", "BG", "CW", "DH", "DK", "EHA", "EHI", "EMH", 
    "EP", "ESE", "EUG", "EWH", "EWO", "HT", "JC", "JE", "JHI", 
    "JL", "MCA", "MG", "MT", "PK", "RM", "VK", "VR", "WR"), class = "factor"), 
    Interviewer = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = c("1", 
    "2", "12", "13"), class = "factor"), Sprache = structure(c(2L, 
    2L, 2L, 2L, 2L, 2L), .Label = c("SD", "UD", "UDSD"), class = "factor"), 
    Anzahl_P = c(4, 4, 4, 4, 4, 4), Geschlecht = structure(c(2L, 
    2L, 2L, 2L, 2L, 2L), .Label = c("m", "w"), class = "factor"), 
    Alter = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = c("2", 
    "3"), class = "factor"), Bildung = structure(c(3L, 3L, 3L, 
    3L, 3L, 3L), .Label = c("C", "HS", "NE", "U"), class = "factor"), 
    Stolz = structure(c(2L, 2L, 2L, 2L, 2L, 2L), .Label = c("a", 
    "b", "U"), class = "factor"), Identität = structure(c(3L, 
    3L, 3L, 3L, 3L, 3L), .Label = c("a", "b", "c", "U"), class = "factor"), 
    Erstsprache = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = c("L1", 
    "L2", "U"), class = "factor"), Englisch = c(10.56, 10.56, 
    10.56, 10.56, 10.56, 10.56)), row.names = c(NA, -6L), class = c("tbl_df", 
"tbl", "data.frame"), na.action = structure(c(`12861` = 12861L), class = "omit"))

Answer 1

因此，我发现了自己的错误，这是一个愚蠢的错误，但是如果其他任何人都跌跌撞撞：我已经将ANOVA和ANOVA功能互相困惑。
a nova函数（来自统计数据）与我尝试的语法一起使用：

anova（model1，model2，test =“ lrt”），

但 a nova函数（来自汽车包装）只需要一个模型（这就是为什么它说它只需要1个参数），因此语法将是：

anova（model1，test =“ lrt”）代码>，

就是这样。