R - 在data.table中使用glm [英] R - using glm inside a data.table

问题描述

我一直在做这个成功的方法：

我试图在data.table中使用一些glm来生成由关键因素分割的模型结果。 / p>

• 高级glm

  
  
  

  glm（modellingDF，formula = Outcome〜 IntCol + DecCol，family = binomial（link = logit））




• 具有单列的范围glm

  
  
  

  modellingDF [，list（Outcome，
  fitted = glm（x，formula = Outcome〜IntCol，family = binomial（link = logit））$ fits），
  by = variable]

  / li>
  

• 具有两个整数列的Scoped glm

  
  
  

  modellingDF [，list（Outcome，
  fitted = glm ，formula = Outcome〜IntCol + IntCol2，family = binomial（link = logit））$ fitted），
  by = variable]

但是，当我尝试在范围内使用我的十进制列执行高级glm时，会产生此错误。

  model.frame.default中的错误（公式= Outcome〜IntCol + DecCol，data = x，：
变量长度不同（对于'DecCol'）
  
 
 
 
我想也许是因为分区的长度可变，所以我用一个可重现的例子测试：
  library（data.table）
 
 testing <-data.table（letters = sample（rep（LETTERS，5000），5000） 
 letters2 = sample（rep（LETTERS [1：5]，10000），5000），
 cont.var = rnorm（5000），
 cont.var2 = round * 1000,0），
 outcome = rbinom（5000,1,0.8）
，key =letters）
 testing.glm< -testing [，list（outcome，
 fit = glm（x，formula = outcome_cont.var + cont.var2，family = binomial（link = logit））$ fitted）
），by = list code> 
但这没有错误。我认为也许是由于NAs或东西，但data.table modellingDF的总结没有指示应该有任何问题：
  DecCol 
最小。 ：0.0416 
第一次查询：0.6122 
中值：0.7220 
平均值：0.6794 
第三个查询：0.7840 
最大。 ：0.9495 
 
 nrow（modellingDF [is.na（DecCol），]）＃results in 0 
 
 modellingDF [，list（len = .N，DecCollen = length ），IntCollen = length 
（IntCol），Outcomelen = length（Outcome）），by = Bracket] 
 
 Bracket len DecCollen IntCollen Outcomelen 
 1：3-6 39184 39184 39184 39184 
 2：1-2 19909 19909 19909 19909 
 3：0 9912 9912 9912 9912 
  
 
 $ b b 
也许我有一个舒适的一天，但任何人都可以提出解决方案或手段进一步挖掘这个问题。
解决方案
您需要在 glm 中正确指定数据参数。在 data.table （使用 [）中引用 .SD 。 （请参阅在data.table环境中创建公式R 的相关问题）
 
 
 
因此
  modellingDF [，list（Outcome，fitted = glm（data = .SD，
 formula = Outcome〜IntCol，family = binomial（link = logit））$ fits），
 by = variable] 
 
 
 
 
在这种情况下（简单提取这个方法是合理的，使用 data.table 和 .SD 可以得到一个如果您保存整个模型，然后尝试更新（参见）
 
I'm trying to do some glm's inside a data.table to produce modelled results split by key factors.

I've been doing this sucessfully for:


High level glm 

glm(modellingDF,formula=Outcome~IntCol + DecCol,family=binomial(link=logit))
Scoped glm with single columns

modellingDF[,list(Outcome,
                  fitted=glm(x,formula=Outcome~IntCol ,family=binomial(link=logit))$fitted   ),
                  by=variable]
Scoped glm with two integer columns

modellingDF[,list(Outcome,
                  fitted=glm(x,formula=Outcome~IntCol + IntCol2 ,family=binomial(link=logit))$fitted   ),
                  by=variable]


But, when I try and do the high level glm inside the scope with my decimal column, it produces this error
Error in model.frame.default(formula = Outcome ~ IntCol + DecCol, data = x,  : 
  variable lengths differ (found for 'DecCol')
I thought perhaps it was due to variable lengths of the partitions, so I tested with a reproducible example:
library("data.table")

testing<-data.table(letters=sample(rep(LETTERS,5000),5000),
                    letters2=sample(rep(LETTERS[1:5],10000),5000), 
                    cont.var=rnorm(5000),
                    cont.var2=round(rnorm(5000)*1000,0),
                    outcome=rbinom(5000,1,0.8)
                    ,key="letters")
testing.glm<-testing[,list(outcome,
                  fitted=glm(x,formula=outcome~cont.var+cont.var2,family=binomial(link=logit))$fitted)
        ),by=list(letters)]
But this did not have the error.  I thought maybe it was due to NAs or something but a summary of the data.table modellingDF gives no indication that there should be any issues:
DecCol
Min.   :0.0416
1st Qu.:0.6122
Median :0.7220
Mean   :0.6794
3rd Qu.:0.7840
Max.   :0.9495

nrow(modellingDF[is.na(DecCol),])   # results in 0

modellingDF[,list(len=.N,DecCollen=length(DecCol),IntCollen=length
(IntCol ),Outcomelen=length(Outcome)),by=Bracket]

  Bracket  len DecCollen IntCollen Outcomelen
1:     3-6 39184  39184       39184      39184
2:     1-2 19909  19909       19909      19909
3:       0  9912   9912        9912       9912
Perhaps I'm having a dozy day, but could anyone suggest a solution or a means for digging into this issue further?
解决方案
You need to correctly specify the data argument within glm. Inside a data.table (using [), this is referenced by .SD.  (see create a formula in a data.table environment in R for related question)

So 
modellingDF[,list(Outcome, fitted = glm(data = .SD, 
  formula = Outcome ~ IntCol ,family = binomial(link = logit))$fitted),
 by=variable]
will work.

While in this case (simply extracting the fitted values and moving on), this approach is sound, using data.table and .SD can get in a mess of environments if you are saving the whole model and then attempting to update it (see Why is using update on a lm inside a grouped data.table losing its model data?)

                        
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