Question

鉴于下面的示例数据sampleDT，其中包含均值mean1至mean10和标准差sd2，对于创建以下功能的帮助，我将不胜感激：< / p>

针对均值mean1至mean10的每一列进行估计，提取和添加到数据帧的条件函数的密度函数的值在变量的观察水平上评估正态分布 dollar.wage_1使用标准偏差sd2。

使用下面的代码，我可以成功地为一列均值进行计算，但是我看不到如何指定一个函数来为均值的每一列同时进行计算。

#sample数据

sampleDT<-structure(list(id = 1:10, N = c(10L, 10L, 10L, 10L, 10L, 10L, 
10L, 10L, 10L, 10L), A = c(62L, 96L, 17L, 41L, 212L, 143L, 143L, 
143L, 73L, 73L), B = c(3L, 1L, 0L, 2L, 170L, 21L, 0L, 33L, 62L, 
17L), C = c(0.05, 0.01, 0, 0.05, 0.8, 0.15, 0, 0.23, 0.85, 0.23
), employer = c(1L, 1L, 0L, 1L, 0L, 1L, 1L, 0L, 0L, 0L), F = c(0L, 
0L, 0L, 0L, 0L, 1L, 1L, 1L, 1L, 1L), G = c(1.94, 1.19, 1.16, 
1.16, 1.13, 1.13, 1.13, 1.13, 1.12, 1.12), H = c(0.14, 0.24, 
0.28, 0.28, 0.21, 0.12, 0.17, 0.07, 0.14, 0.12), dollar.wage_1 = c(1.94, 
1.19, 3.16, 3.16, 1.13, 1.13, 2.13, 1.13, 1.12, 1.12), mean1 = c(1.936652081, 
3.688171386, 3.160993574, 3.768485048, 1.311370546, 0.313760016, 
-1.621000294, 1.13182676, 1.114458025, 1.119315775), mean2 = c(1.946806222, 
3.688885811, 3.15903495, 3.767778705, 1.309663497, 0.316394741, 
-1.618552806, 1.134088181, 1.117600968, 1.120688482), mean3 = c(1.893627954, 
3.689341572, 3.157622975, 3.771231512, 1.324985578, 0.318026311, 
-1.620565712, 1.13301769, 1.120760085, 1.119426932), mean4 = c(1.887509366, 
3.660243949, 3.160911994, 3.738992465, 1.331637143, 0.284716279, 
-1.655368774, 1.137338962, 1.122096234, 1.120837428), mean5 = c(7.071170501, 
3.458558276, 3.156676637, 3.160692822, 1.131841192, 1.126997224, 
1.028924299, 1.219378155, 0.118097115, 1.118108075), mean6 = c(7.010141264, 
3.434098438, 3.160978044, 3.161388054, 1.131706507, 1.131073576, 
1.044957033, 1.202376831, 0.088502176, 1.120101488), mean7 = c(6.918631396, 
3.455412441, 3.064840549, 3.158657611, 1.134281965, 1.131677907, 
1.035688483, 1.181551066, 0.542276222, 1.121549931), mean8 = c(6.980214117, 
3.513440689, 3.175191087, 3.158919334, 1.130088008, 1.131692248, 
1.12222788, 1.235102249, 0.281700405, 1.118473791), mean9 = c(6.708505027, 
3.504542699, 3.173629275, 3.158457814, 1.134560107, 1.129357587, 
1.151489857, 1.219991269, 0.364343124, 1.120228667), mean10 = c(6.883206883, 
3.467216323, 3.174805298, 3.160917024, 1.128835398, 1.128265912, 
1.084046983, 1.214981489, 0.160046133, 1.118496504), sd1 = c(2.6334129999306, 
2.6334129999306, 2.6334129999306, 2.6334129999306, 2.6334129999306, 
2.6334129999306, 2.6334129999306, 2.6334129999306, 2.6334129999306, 
2.6334129999306), sd2 = c(514.02608349227, 101.976862386691, 
8.70627514696715, 4.79710442214283, 2.45930925299156e+49, 2.01406038865916e+30, 
1.8980055884822e+34, 1.65244344266379e+28, 26.9398910547703, 
1.74978644797635)), row.names = c(NA, -10L), spec = structure(list(
    cols = list(id = structure(list(), class = c("collector_integer", 
    "collector")), N = structure(list(), class = c("collector_integer", 
    "collector")), A = structure(list(), class = c("collector_integer", 
    "collector")), B = structure(list(), class = c("collector_integer", 
    "collector")), C = structure(list(), class = c("collector_double", 
    "collector")), employer = structure(list(), class = c("collector_integer", 
    "collector")), F = structure(list(), class = c("collector_integer", 
    "collector")), G = structure(list(), class = c("collector_double", 
    "collector")), H = structure(list(), class = c("collector_double", 
    "collector")), dollar.wage_1 = structure(list(), class = c("collector_double", 
    "collector")), mean1 = structure(list(), class = c("collector_double", 
    "collector")), mean2 = structure(list(), class = c("collector_double", 
    "collector")), mean3 = structure(list(), class = c("collector_double", 
    "collector")), mean4 = structure(list(), class = c("collector_double", 
    "collector")), mean5 = structure(list(), class = c("collector_double", 
    "collector")), mean6 = structure(list(), class = c("collector_double", 
    "collector")), mean7 = structure(list(), class = c("collector_double", 
    "collector")), mean8 = structure(list(), class = c("collector_double", 
    "collector")), mean9 = structure(list(), class = c("collector_double", 
    "collector")), mean10 = structure(list(), class = c("collector_double", 
    "collector"))), default = structure(list(), class = c("collector_guess", 
    "collector"))), class = "col_spec"), class = c("tbl_df", 
"tbl", "data.frame"))

#my方法

sampleDT$dens_test <- dnorm(sampleDT$dollar.wage_1, 
                      mean = sampleDT$mean1,sd = sampleDT$sd2)

在此先感谢您的帮助。

Answer 1

我们可能会更进一步：

means <- as.matrix(sampleDT[, grep("mean", names(sampleDT))])
dnorm(sampleDT$dollar.wage_1, mean = means, sd = sampleDT$sd2)

通过这种方式，我们传递了均值矩阵，而dollar.wage_1和sd2也由于回收而被正确使用。

然后您可以简单地使用sampleDT将此结果添加到cbind。

Answer 2

或放入sapply（）。

sapply(grep("^mean", names(sampleDT)), function(x) 
  dnorm(sampleDT$dollar.wage_1, sampleDT[[x]], sampleDT$sd2))

Answer 3

这是dplyr::mutate_at()的典型情况：

df %>% mutate_at(vars(matches("mean")),
                 funs(dens = dnorm(dollar.wage_1, mean = ., sd = sd2)))

输出将是一个完整的数据集，您不需要绑定任何内容。

Answer 4

这是我使用data.table的方法：

sampleDT <- as.data.table(sampleDT) 

for(i in c(1:10)){   
 sampleDT[, eval(paste0("dnorm",i)):=mapply(dnorm,dollar.wage_1,get(paste0("mean",i)),sd2)] }

计算多列均值和一列sigma的概率密度函数的值

4 个答案: