R 计算多列平均值和一列西格玛的概率密度函数值
鉴于下面的样本数据R 计算多列平均值和一列西格玛的概率密度函数值,r,function,data.table,tidyverse,probability-density,R,Function,Data.table,Tidyverse,Probability Density,鉴于下面的样本数据sampleDT,其中包含均值mean1至mean10和标准偏差sd2,我希望能为创建以下函数提供帮助: 对于平均值mean1到mean10估计值的每一列,提取并添加 在数据框中输入条件密度函数的值 在观察到的变量水平上评估的正态分布 美元工资1使用标准差sd2 使用下面的代码,我可以成功地计算一列平均数,但我看不到如何指定一个函数来同时计算每列平均数 #样本数据 sampleDT<-structure(list(id = 1:10, N = c(10L, 10L, 10
sampleDT
,其中包含均值mean1
至mean10
和标准偏差sd2
,我希望能为创建以下函数提供帮助:
对于平均值mean1
到mean10
估计值的每一列,提取并添加
在数据框中输入条件密度函数的值
在观察到的变量水平上评估的正态分布
美元工资1
使用标准差sd2
使用下面的代码,我可以成功地计算一列平均数,但我看不到如何指定一个函数来同时计算每列平均数
#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"))
sampleDT我们可以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)
表示这是我使用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)] }
sampleDT或放入sapply()
这是dplyr::mutate_at()的典型例子
输出将是一个完整的数据集,您不需要绑定任何内容
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)] }
sapply(grep("^mean", names(sampleDT)), function(x)
dnorm(sampleDT$dollar.wage_1, sampleDT[[x]], sampleDT$sd2))
df %>% mutate_at(vars(matches("mean")),
funs(dens = dnorm(dollar.wage_1, mean = ., sd = sd2)))