我很感激任何帮助从我的样本数据var.w_X
中随机选择5
的10
子集来自var.w_X
sampleDT
,同时保留所有其他不以var.w_
开头的变量。
下面是样本数据sampleDT
,其中包含其他变量(完全保留的变量),X
变量以其名称中的var.w_
开头(用于绘制随机样本的变量)。
在当前的例子中,X=10
,所以var.w_
包括var.w_1
到var.w_10
,我想从这些5
中随机抽取10
样本。然而,在我的实际数据中,X>1,000,000
and我可能想要从这些7,500
中抽取var.w_
X>1,000,000
变量的样本。
因此,在任何给定的解决方案中,效率的考虑是至关重要的,因为recently我遇到了mutate_at
的一些性能问题,其原因我仍然没有解释。
重要的是,要保持的其他变量(那些不以var.w_
开头的变量)不能保证保持任何预先指定的顺序,因为它们可能位于例如var.w_
变量之前和/或之间和/或之后。因此,依赖于列顺序的解决方案将无法工作。
#样本数据
sampleDT<-structure(list(n = c(62L, 96L, 17L, 41L, 212L, 143L, 143L, 143L,
73L, 73L), r = c(3L, 1L, 0L, 2L, 170L, 21L, 0L, 33L, 62L, 17L
), p = c(0.0483870967741935, 0.0104166666666667, 0, 0.0487804878048781,
0.80188679245283, 0.146853146853147, 0, 0.230769230769231, 0.849315068493151,
0.232876712328767), var.w_8 = c(1.94254385942857, 1.18801169942857,
3.16131123942857, 3.16131123942857, 1.13482609242857, 1.13042157942857,
2.13042157942857, 1.13042157942857, 1.12335579942857, 1.12335579942857
), var.w_9 = c(1.942365288, 1.187833128, 3.161132668, 3.161132668,
1.134647521, 1.130243008, 2.130243008, 1.130243008, 1.123177228,
1.123177228), var.w_10 = c(1.94222639911111, 1.18769423911111,
3.16099377911111, 3.16099377911111, 1.13450863211111, 1.13010411911111,
2.13010411911111, 1.13010411911111, 1.12303833911111, 1.12303833911111
), group = c(1L, 1L, 0L, 1L, 0L, 1L, 1L, 0L,
0L, 0L), treat = c(0L, 0L, 0L, 0L, 0L, 1L, 1L, 1L, 1L, 1L), c1 = c(1.941115288,
1.186583128, 1.159882668, 1.159882668, 1.133397521, 1.128993008,
1.128993008, 1.128993008, 1.121927228, 1.121927228), var.w_6 = c(1.939115288, 1.184583128,
3.157882668, 3.157882668, 1.131397521, 1.126993008, 2.126993008,
1.126993008, 1.119927228, 1.119927228), var.w_7 = c(1.94278195466667,
1.18824979466667, 3.16154933466667, 3.16154933466667, 1.13506418766667,
1.13065967466667, 2.13065967466667, 1.13065967466667, 1.12359389466667,
1.12359389466667), c2 = c(0.1438,
0.237, 0.2774, 0.2774, 0.2093, 0.1206, 0.1707, 0.0699, 0.1351,
0.1206), var.w_1 = c(1.941115288, 1.186583128, 3.159882668, 3.159882668,
1.133397521, 1.128993008, 2.128993008, 1.128993008, 1.121927228,
1.121927228), var.w_2 = c(1.931115288, 1.176583128, 3.149882668,
3.149882668, 1.123397521, 1.118993008, 2.118993008, 1.118993008,
1.111927228, 1.111927228), var.w_3 = c(1.946115288, 1.191583128,
3.164882668, 3.164882668, 1.138397521, 1.133993008, 2.133993008,
1.133993008, 1.126927228, 1.126927228), var.w_4 = c(1.93778195466667,
1.18324979466667, 3.15654933466667, 3.15654933466667, 1.13006418766667,
1.12565967466667, 2.12565967466667, 1.12565967466667, 1.11859389466667,
1.11859389466667), var.w_5 = c(1.943615288, 1.189083128, 3.162382668,
3.162382668, 1.135897521, 1.131493008, 2.131493008, 1.131493008,
1.124427228, 1.124427228)), class = "data.frame", row.names = c(NA, -10L))
#my尝试
//based on the comment by @akrun - this does not keep the other variables as specified above
myvars <- sample(grep("var\\.w_", names(sampleDT), value = TRUE), 5)
sampleDT_test <- sampleDT[myvars]
在此先感谢您的帮助
道歉,不得不进入一次会议。因此,我认为您可以调整akrun的解决方案并保留示例数据帧的第一列。让我知道这是如何在整个数据帧上扩展的。另外,谢谢你进一步澄清。
> # Subsetting the variable names not matching your pattern using grepl
> names(sampleDT)[!grepl("var\\.w_", names(sampleDT))]
[1] "n" "r" "p" "group" "treat" "c1" "c2"
>
> # Combine that with akrun's solution
> myvars <- c(names(sampleDT)[!grepl("var\\.w_", names(sampleDT))],
+ sample(grep("var\\.w_", names(sampleDT), value = TRUE), 5))
> head(sampleDT[myvars])
n r p group treat c1 c2 var.w_6 var.w_1 var.w_4 var.w_3 var.w_8
1 62 3 0.04838710 1 0 1.941115 0.1438 1.939115 1.941115 1.937782 1.946115 1.942544
2 96 1 0.01041667 1 0 1.186583 0.2370 1.184583 1.186583 1.183250 1.191583 1.188012
3 17 0 0.00000000 0 0 1.159883 0.2774 3.157883 3.159883 3.156549 3.164883 3.161311
4 41 2 0.04878049 1 0 1.159883 0.2774 3.157883 3.159883 3.156549 3.164883 3.161311
5 212 170 0.80188679 0 0 1.133398 0.2093 1.131398 1.133398 1.130064 1.138398 1.134826
6 143 21 0.14685315 1 1 1.128993 0.1206 1.126993 1.128993 1.125660 1.133993 1.130422