[英]How to pass "weights" column name as a variable in R's lm?
The code below creates a linear model with R's lm, then a weighted model with a weights column.下面的代码创建一个带有 R 的 lm 的线性模型,然后是一个带有权重列的加权模型。 Finally, I try to pass in the weight column name with a variable
weight_col
and that fails.最后,我尝试使用变量
weight_col
传入权重列名称,但失败了。 I'm pretty sure it's looking for "weight_col" in df, then the caller's environment, finds a variable of length 1, and the lengths don't match.我很确定它在 df 中寻找“weight_col”,然后调用者的环境找到一个长度为 1 的变量,并且长度不匹配。
How do I get it to use weight_col as a name for the weights column in df?我如何让它使用 weight_col 作为 df 中权重列的名称?
I've tried several combinations of things without success.我已经尝试了几种组合,但都没有成功。
> df <- data.frame(
x=c(1,2,3),
y=c(4,5,7),
w=c(1,3,5)
)
> lm(y ~ x, data=df)
Call:
lm(formula = y ~ x, data = df)
Coefficients:
(Intercept) x
2.333 1.500
> lm(y ~ x, data=df, weights=w)
Call:
lm(formula = y ~ x, data = df, weights = w)
Coefficients:
(Intercept) x
1.947 1.658
> weight_col <- 'w'
> lm(y ~ x, data=df, weights=weight_col)
Error in model.frame.default(formula = y ~ x, data = df, weights = weight_col, :
variable lengths differ (found for '(weights)')
> R.version.string
[1] "R version 3.6.3 (2020-02-29)"
You can use the data frame name with extractor operator:您可以将数据框名称与提取器运算符一起使用:
lm(y ~ x, data = df, weights = df[[weight_col]])
Or you can use function get
:或者您可以使用函数
get
:
lm(y ~ x, data = df, weights = get(weight_col))
We can use [[
to extract the value of the column我们可以使用
[[
来提取列的值
lm(y ~ x, data=df, weights=df[[weight_col]])
Or with tidyverse
或者用
tidyverse
library(dplyr)
df %>%
summarise(model = list(y ~ x, weights = .data[[weight_col]]))
Your first example if weights = w
, which is using non-standard evaluation to find w
in the context of df
.您的第一个示例 if
weights = w
,它使用非标准评估在df
的上下文中查找w
。 So far, this is normal for interactive use.到目前为止,这对于交互式使用来说是正常的。
Your second set is weights = weight_col
which resolves to weights = "w"
, which is very different.你的第二组是
weights = weight_col
解析为weights = "w"
,这是非常不同的。 There is nothing in R's non-standard (or standard) evaluation in which that makes sense. R 的非标准(或标准)评估中没有任何内容是有意义的。
As I said in my comment, use the standard-evaluation form with [[
.正如我在评论中所说,使用带有
[[
的标准评估形式。
lm(y ~ x, data=df, weights=df[[weight_col]])
# Call:
# lm(formula = y ~ x, data = df, weights = df[[weight_col]])
# Coefficients:
# (Intercept) x
# 1.947 1.658
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