data.table和分层的手段

时间:2012-11-18 16:10:22

标签: r data.table

我有一些代码可以生成分层加权平均值 我确信这几个月前就有效了。但是,但我不确定目前的问题是什么。 (我道歉 - 这一定是非常基本的东西):

dp=
structure(list(seqn = c(1L, 2L, 3L, 4L, 6L, 7L, 8L, 9L, 10L, 
11L, 12L, 13L, 3L, 4L, 9L, 10L, 11L, 14L, 8L, 11L, 12L, 10L, 
5L, 13L, 2L, 14L, 3L, 9L, 6L, 7L), sex = c(2L, 1L, 2L, 2L, 1L, 
2L, 2L, 1L, 2L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L), bmi = c(22.8935608711259, 
27.0944623781918, 40.4637162938634, 23.7649712675423, 15.3193372705538, 
31.1280302540991, 21.4866354393239, 20.3200254374398, 32.331092513536, 
25.3679771839413, 33.9400508162971, 14.7048592172926, 25.5243757788688, 
23.4331882363495, 27.6428134168995, 29.3923629426172, 24.9547209666314, 
17.0522203606383, 15.51, 22, 30.62, 30.94, 29.1, 25.57, 24.9, 
27.33, 17.63, 18.48, 22.56, 29.39), tc = c(273L, 181L, 150L, 
201L, 142L, 165L, 235L, 219L, 298L, 222L, 143L, 134L, 268L, 160L, 
236L, 225L, 260L, 140L, 162L, 132L, 156L, 140L, 279L, 314L, 215L, 
174L, 129L, 148L, 153L, 245L), swt = c(1645, 3318, 2280, 1574, 
4062, 1627, 14604, 24675, 975, 975, 2697, 1559, 1737.58, 1730.23, 
19521.36, 28080.57, 1248.43, 13745.77, 5251.76464426326, 6497.194885522, 
15915.7023420765, 3740.96809540218, 16574.177622509, 307.32513798849, 
4720.89748295751, 3247.78896499604, 7698.70949077031, 1262.6450411464, 
6609.43340735515, 4254.23723479882)), .Names = c("seqn", "sex", 
"bmi", "tc", "swt"), row.names = c(20560L, 20561L, 20562L, 20563L, 
20565L, 20566L, 20567L, 20568L, 20569L, 20570L, 20571L, 20572L, 
61335L, 61336L, 61338L, 61339L, 61340L, 61341L, 95465L, 96890L, 
104613L, 105988L, 107581L, 112267L, 113403L, 114292L, 119979L, 
120271L, 125939L, 135699L), class = "data.frame")

dt=data.table(dp, key='sex')

sapply(df,function(x)weighted.mean(x,df$swt))  #this works to weighted mean
dt[,lapply(.SD, mean, na.rm=T), .SDcols=c('bmi','tc','swt')]  
     #this also works for overall unweighted mean

dt[,lapply(.SD, function(x)weighted.mean(x,swt, na.rm=TRUE)), by=key(dt), .SDcols=c('bmi','tc','swt')] 

但是这会给出错误: Error in weighted.mean.default(x, swt, na.rm = TRUE) : object 'swt' not found

sessionInfo()
R version 2.15.2 (2012-10-26)
Platform: i386-w64-mingw32/i386 (32-bit)

locale:
[1] LC_COLLATE=English_United States.1252  LC_CTYPE=English_United States.1252   
[3] LC_MONETARY=English_United States.1252 LC_NUMERIC=C                          
[5] LC_TIME=English_United States.1252    

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
[1] data.table_1.8.6

loaded via a namespace (and not attached):
[1] tools_2.15.2

2 个答案:

答案 0 :(得分:8)

UPDATE(来自Arun):现在已在v1.8.11中修复。来自NEWS

  当优化“开启”时,

o DT[, lapply(.SD, function(), by=]没有看到DT的列。现在已修复此问题#2381。测试成功添加和测试。感谢David F报告SO:        data.table and stratified means


这确实是介于1.8.2和1.8.6之间的错误。

dt[,lapply(.SD, function(x) weighted.mean(x,swt, na.rm=TRUE)), by=key(dt),
    .SDcols=c('bmi','tc','swt')] 
Error in weighted.mean.default(x, swt, na.rm = TRUE) : 
    object 'swt' not found

要解决此问题,请关闭优化:

options(datatable.optimize=FALSE)
dt[,lapply(.SD, function(x)weighted.mean(x,swt, na.rm=TRUE)), by=key(dt),    
    .SDcols=c('bmi','tc','swt')]
   sex      bmi       tc      swt
1:   1 25.64376 206.0115 17171.20
2:   2 23.73566 193.8727 11467.47

或者,不要用function()包裹:

options(datatable.optimize=TRUE)
dt[,lapply(.SD, weighted.mean, swt, na.rm=TRUE), by=key(dt),    
    .SDcols=c('bmi','tc','swt')] 
   sex      bmi       tc      swt
1:   1 25.64376 206.0115 17171.20
2:   2 23.73566 193.8727 11467.47

我们现在正在更多地使用优化,但是这个案例在测试套件中滑落:测试825.1,825.2和825.3没有在匿名function()内覆盖函数作为另一列的参数。如果没有给出功能,那将是一个问题;即,与这种情况不同,function()可以省略weighted.mean,因为verbose=TRUE已经给出,可以按原样应用。

您可以通过设置{{1}}(每个查询或使用全局选项)来查看优化如何修改j。在这种情况下,没有任何东西可以通过冗长的输出显示为错误,而只是将其视为一边。

现在以#2381: Optimization of lapply(.SD, function() ...) no longer sees columns inside ...提交。将修复并添加测试,这样就不会再次退化。

谢谢!

答案 1 :(得分:3)

我建议保持简单:

dt[,list(bmi_m=weighted.mean(bmi,swt,na.rm=TRUE),
         tc_m=weighted.mean(tc,swt,na.rm=TRUE),
         swt_m=weighted.mean(swt,swt,na.rm=TRUE)),by=key(dt)]

我认为这也相当快。