如何在多个模型上使用broom :: tidy？
我正在尝试使用扫帚总结19个多项式回归模型的结果。我已经关注了这个问题，并尝试与结合使用broom::tidy。我的脚本如下：
ALTER PROCEDURE [dbo].[spRegressionPeak]
@StudyID int
AS
BEGIN
Declare @sStudyID VARCHAR(50)
Set @sStudyID = CONVERT(VARCHAR(50),@StudyID)
--We are selecting the distinct StudyID, Productnumber, ResponseID and mean
values 1 thorugh 6 from the CodeMeans table.
--Note that spCodeMeans must be run before running this stored procedure to
ensure response data exists in the CodeMeans table.
--We use IsNull values to pass zeroes where an average wasn't calculated os that
the polynomial regression can be calculated.
DECLARE @inquery AS NVARCHAR(MAX) = '
Select
c.StudyID, c.RespID, c.LikingOrder, avg(isnull(C1,0)) as C1, avg(isnull(C2,0)) as C2, avg(isnull(C3,0)) as C3, avg(isnull(C4,0)) as C4,
avg(isnull(C5,0)) as C5, avg(isnull(C6,0)) as C6, avg(isnull(C7,0)) as C7, avg(isnull(C8,0)) as C8, avg(isnull(C9,0)) as C9,
avg(isnull(C10,0)) as C10, avg(isnull(C11,0)) as C11, avg(isnull(C12,0)) as C12, avg(isnull(C13,0)) as C13, avg(isnull(C14,0)) as C14,
avg(isnull(C15,0)) as C15, avg(isnull(C16,0)) as C16, avg(isnull(C17,0)) as C17, avg(isnull(C18,0)) as C18, avg(isnull(C19,0)) as C19
from ClosedStudyResponses c
where c.StudyID = @StudyID
group by StudyID, RespID, LikingOrder
order by RespID
'
--We are setting @inquery aka InputDataSet to be our initial dataset.
--R Services requires that a data.frame be passed to any calculations being
generated. As such, df is simply data framing the @inquery data.
--The res object holds the polynomial regression results by RespondentID and
LikingOrder for each of the averages in the @inquery resultset.
EXEC sp_execute_external_script @language = N'R'
, @script = N'
library(tidyr, broom)
studymeans <- InputDataSet
df <- data.frame(studymeans)
lin.mod.1 <- lm(df$LikingOrder ~ poly(df$C1,3, raw=TRUE))
lin.mod.2 <- lm(df$LikingOrder ~ poly(df$C2,3, raw=TRUE))
lin.mod.3 <- lm(df$LikingOrder ~ poly(df$C3,3, raw=TRUE))
lin.mod.4 <- lm(df$LikingOrder ~ poly(df$C4,3, raw=TRUE))
lin.mod.5 <- lm(df$LikingOrder ~ poly(df$C5,3, raw=TRUE))
lin.mod.6 <- lm(df$LikingOrder ~ poly(df$C6,3, raw=TRUE))
lin.mod.7 <- lm(df$LikingOrder ~ poly(df$C7,3, raw=TRUE))
lin.mod.8 <- lm(df$LikingOrder ~ poly(df$C8,3, raw=TRUE))
lin.mod.9 <- lm(df$LikingOrder ~ poly(df$C9,3, raw=TRUE))
lin.mod.10 <- lm(df$LikingOrder ~ poly(df$C10,3, raw=TRUE))
lin.mod.11 <- lm(df$LikingOrder ~ poly(df$C11,3, raw=TRUE))
lin.mod.12 <- lm(df$LikingOrder ~ poly(df$C12,3, raw=TRUE))
lin.mod.13 <- lm(df$LikingOrder ~ poly(df$C13,3, raw=TRUE))
lin.mod.14 <- lm(df$LikingOrder ~ poly(df$C14,3, raw=TRUE))
lin.mod.15 <- lm(df$LikingOrder ~ poly(df$C15,3, raw=TRUE))
lin.mod.16 <- lm(df$LikingOrder ~ poly(df$C16,3, raw=TRUE))
lin.mod.17 <- lm(df$LikingOrder ~ poly(df$C17,3, raw=TRUE))
lin.mod.18 <- lm(df$LikingOrder ~ poly(df$C18,3, raw=TRUE))
lin.mod.19 <- lm(df$LikingOrder ~ poly(df$C19,3, raw=TRUE))
lst <- lapply(ls(pattern="lin.mod"), get)
allmodels <- lapply(lst, summary)
res <- broom::tidy(allmodels)
'
, @input_data_1 = @inquery
, @output_data_1_name = N'res'
, @params = N'@StudyID int'
,@StudyID = @StudyID
--- Edit this line to handle the output data frame.
--WITH RESULT SETS ((StudyID int, RespID int, LikingOrder int, NewColumn int,
res varchar(max)));
END;
将有效的StudyID输入参数传递给上面的脚本时，它会引发以下错误：
Error in setNames(data.frame(data), value.name) :
'names' attribute [1] must be the same length as the vector [0]
Calls: source ... <Anonymous> -> <Anonymous> -> melt.default -> setNames
In addition: There were 50 or more warnings (use warnings() to see the first
50)
我的输入数据如下：
理想的结果是在data.frame中获得所有19个模型的摘要。如何解决错误并修改代码以完成最终结果？
没有您的工作环境，我不确定确切如何设置数据，但是似乎您正在尝试在多个预测变量列上使用相同的因变量来拟合模型。我认为缺失的部分是rowwise根据扫帚和dplyr小插图的，但不是完全确定。不过，这是mtcars数据集的有效示例。请注意，该结构将tidy在具有包含模型的列表列的行数据框上使用，而不是直接在列表上使用。您还可以通过映射包含预测变量列的数据框来直接创建模型，而不必在环境中存储模型且需要使用getand 的混乱局面ls。任何时候您发现自己使用ls，请考虑是否可以将元素放入列表中！
编辑：在再次看到此问题提示的小插图后，我意识到您实际上可以像现在显示的那样做一个快速管道（请参阅使用的方法的编辑历史记录enframe。通过gather将数据转换为适合分组模型拟合的格式，您可以整齐地获得预期的结果！
library(tidyverse)
library(broom)
mtcars %>%
gather(predictor, measure, -mpg) %>%
group_by(predictor) %>%
do(tidy(lm(mpg ~ measure, .)))
#> # A tibble: 20 x 6
#> # Groups: predictor [10]
#> predictor term estimate std.error statistic p.value
#> <chr> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 am (Intercept) 17.1 1.12 15.2 1.13e-15
#> 2 am measure 7.24 1.76 4.11 2.85e- 4
#> 3 carb (Intercept) 25.9 1.84 14.1 9.22e-15
#> 4 carb measure -2.06 0.569 -3.62 1.08e- 3
#> 5 cyl (Intercept) 37.9 2.07 18.3 8.37e-18
#> 6 cyl measure -2.88 0.322 -8.92 6.11e-10
#> 7 disp (Intercept) 29.6 1.23 24.1 3.58e-21
#> 8 disp measure -0.0412 0.00471 -8.75 9.38e-10
#> 9 drat (Intercept) -7.52 5.48 -1.37 1.80e- 1
#> 10 drat measure 7.68 1.51 5.10 1.78e- 5
#> 11 gear (Intercept) 5.62 4.92 1.14 2.62e- 1
#> 12 gear measure 3.92 1.31 3.00 5.40e- 3
#> 13 hp (Intercept) 30.1 1.63 18.4 6.64e-18
#> 14 hp measure -0.0682 0.0101 -6.74 1.79e- 7
#> 15 qsec (Intercept) -5.11 10.0 -0.510 6.14e- 1
#> 16 qsec measure 1.41 0.559 2.53 1.71e- 2
#> 17 vs (Intercept) 16.6 1.08 15.4 8.85e-16
#> 18 vs measure 7.94 1.63 4.86 3.42e- 5
#> 19 wt (Intercept) 37.3 1.88 19.9 8.24e-19
#> 20 wt measure -5.34 0.559 -9.56 1.29e-10
由reprex包（v0.2.0）创建于2018-07-10 。