计算多列均值和单列 sigma 的概率密度函数值答案

【问题标题】：Calculate the values of a probability density function for multiple columns of means and a single column of sigma计算多列均值和单列 sigma 的概率密度函数值
【发布时间】：2019-07-04 05:56:18
【问题描述】：

鉴于下面的示例数据sampleDT，其中包含mean1 到mean10 的平均值和标准差sd2，我将不胜感激创建一个函数：

对于每列均值mean1 到mean10estimate，提取并添加数据框的密度函数的值的条件在观察到的变量水平上评估的正态分布 dollar.wage_1 使用标准差 sd2。

使用下面的代码，我可以成功计算单列均值，但我看不到如何指定一个函数来同时计算每一列均值。

#样本数据

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"))

#我的方法

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

提前感谢您的帮助。

【问题讨论】：

标签： r function data.table tidyverse probability-density

【解决方案1】：

我们可能会进一步采用您的方法：

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

通过这种方式，我们传递了一个均值矩阵，而dollar.wage_1 和sd2 也由于回收而被正确使用。

然后你可以简单地将这个结果添加到sampleDTcbind。

【讨论】：

【解决方案2】：

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)] }

【讨论】：

【解决方案3】：

或者放入一个 sapply()。

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

【讨论】：

【解决方案4】：

这是dplyr::mutate_at()的典型案例：

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，你不需要绑定任何东西。

【讨论】：