Question

所以我有这个数据 -

hours     A      B      id
    2.9      1      1      1
    2.3      1      1      2
    2.3      1      1      3
    2.5      1      1      4
    4.6      1      2      1

    attach(hayfever.data)
hayfever.lm = lm(hayfever.data$hours ~ hayfever.data$A * hayfever$B)

但是我收到了这个错误 -

Error: invalid type (NULL) for variable 'hayfever.data$A'

当我这样做时，我也是这样 -

   hayfever.lm = lm(hayfever.data$hours ~ A * B)

除变量名称更改外。

为什么会这样？

enter image description here

输出输出

## structure(list(hours = c(2.9, 2.3, 2.3, 2.5, 4.6, 4.4, 4.9, 4.5, 
## 4.8, 4.5, 4.4, 4.6, 5.8, 5.2, 5.2, 5.3, 8.4, 9.1, 8.7, 9, 9.1, 
## 9.4, 8.7, 9.4, 6.4, 5.7, 5.9, 7.2, 9.9, 10.5, 10.6, 12.1, 13.5, 
## 13, 13.3, 16.2), A = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
## 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 
## 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), B = c(1L, 1L, 1L, 1L, 
## 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 
## 3L, 3L, 3L, 3L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L
## ), id = c(1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L, 
## 2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L, 
## 2L, 3L, 4L, 1L, 2L, 3L, 4L)), .Names = c("hours", "A", "B", "id"
## ), class = "data.frame", row.names = c(NA, -36L))

Answer 1

此处无需使用attach。你应该怎么做：

lm(hours ~ A * B,data=hayfever.data)
Call:
lm(formula = hours ~ A * B, data = hayfever.data)

Coefficients:
(Intercept)            A            B          A:B  
        0.4           NA          2.1           NA

数据在哪里：

hayfever.data <- read.table(text='hours     A      B      id
+     2.9      1      1      1
+     2.3      1      1      2
+     2.3      1      1      3
+     2.5      1      1      4
+     4.6      1      2      1',header=TRUE)

Answer 2

如果我使用dput的输出将数据输入到R中，则代码可以正常工作。

hayfever.data <- structure(list(hours = c(2.9, 2.3, 2.3, 2.5, 4.6, 4.4, 4.9, 4.5, 
4.8, 4.5, 4.4, 4.6, 5.8, 5.2, 5.2, 5.3, 8.4, 9.1, 8.7, 9, 9.1, 
9.4, 8.7, 9.4, 6.4, 5.7, 5.9, 7.2, 9.9, 10.5, 10.6, 12.1, 13.5, 
13, 13.3, 16.2), A = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), B = c(1L, 1L, 1L, 1L, 
2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 
3L, 3L, 3L, 3L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L
), id = c(1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L, 
2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L, 
2L, 3L, 4L, 1L, 2L, 3L, 4L)), .Names = c("hours", "A", "B", "id"
), class = "data.frame", row.names = c(NA, -36L))

hayfever.lm <- lm(hours ~ A * B, data=hayfever.data)
hayfever.lm

# Call:
# lm(formula = hours ~ A * B, data = hayfever.data)
# 
# Coefficients:
# (Intercept)            A            B          A:B  
#      1.9833       0.4208      -0.5542       1.4062

你的R会话中可能存在一些挥之不去的问题，可能来自attach？ attach是邪恶的。不要使用它。说真的，这会引起很多麻烦。

尝试按照评论中的一些建议并清除工作区，然后重新启动R会话。请勿使用attach。

奇怪的是，您附加的屏幕截图与我从dput输出中读取的数据不符。

Answer 3

我在瑞士数据集中遇到了同样的错误。结果我的问题是因为区分大小写。 D'哦！基本上，R告诉我们引用的变量不存在，所以我们可能更好地查看我们的命名/拼写解决方案。我正在使用微软的R发行版，所以也许其他人不那么挑剔？

> lm(swiss$catholic~swiss$fertility, swiss)
Error in model.frame.default(formula = swiss$catholic ~ swiss$fertility,  : 
  invalid type (NULL) for variable 'swiss$catholic'
> lm(swiss$Catholic~swiss$Fertility)

Call:
lm(formula = swiss$Catholic ~ swiss$Fertility, swiss)

Coefficients:
    (Intercept)  swiss$Fertility  
        -67.441            1.548  

>

变量的类型（NULL）无效

3 个答案: