在 R 中循环一组数字生成命令
Looping over a set of number generating commands in R
下面，我展示了一段代码，用于在 R 中生成一些项目分数。但是，似乎有相当多的不必要的重复才能达到最终的 data.
我想知道在 R 中是否有更紧凑的方法来实现相同的 data？
set.seed(8649)
N = 10
latent = rnorm(N)
##### generate latent responses to items
item1 = latent + rnorm(N, mean=0, sd=0.2)
item2 = latent + rnorm(N, mean=0, sd=0.3)
item3 = latent + rnorm(N, mean=0, sd=0.5)
item4 = latent + rnorm(N, mean=0, sd=1.0)
item5 = latent + rnorm(N, mean=0, sd=1.2)
##### convert latent responses to ordered categories
item1 = findInterval(item1, vec=c(-Inf,-2.5,-1, 1,2.5,Inf))
item2 = findInterval(item2, vec=c(-Inf,-2.5,-1, 1,2.5,Inf))
item3 = findInterval(item3, vec=c(-Inf,-3, -2, 2,3, Inf))
item4 = findInterval(item4, vec=c(-Inf,-3, -2, 2,3, Inf))
item5 = findInterval(item5, vec=c(-Inf,-3.5,-3,-1,0.5,Inf))
data = cbind(item1, item2, item3, item4, item5)
我们可以在 list 中创建第一组 'item'，可变部分为 'sd'
# // loop over the sd vector and create the list of random numbers in a list
lst1 <- lapply(c(0.2, 0.3, 0.5, 1, 1.2), function(x)
latent + rnorm(N, mean = 0, sd = x))
# // set the names of the list if needed
names(lst1) <- paste0("item", seq_along(lst1))
使用Map循环遍历'lst1'和vec的相应元素作为list（'veclst'）并应用findInterval
data.frame(Map(findInterval, lst1, vec = veclst))
# item1 item2 item3 item4 item5
#1 4 4 3 3 5
#2 3 3 3 3 5
#3 3 3 3 3 5
#4 4 4 4 3 5
#5 2 2 3 2 4
#6 3 3 3 3 3
#7 3 3 3 3 4
#8 2 2 2 2 2
#9 4 4 5 4 5
#10 3 3 3 3 4
或者tidyverse
也一样
library(purrr)
library(dplyr)
library(stringr)
map2_dfc(c(0.2, 0.3, 0.5, 1, 1.2), veclst, ~
findInterval(latent + rnorm(N, mean = 0, sd = .x), vec = .y)) %>%
set_names(str_c('item', seq_along(.)))
-输出
# A tibble: 10 x 5
# item1 item2 item3 item4 item5
# <int> <int> <int> <int> <int>
# 1 4 4 3 4 4
# 2 3 3 3 3 5
# 3 3 3 3 3 4
# 4 3 4 3 3 5
# 5 2 2 3 3 4
# 6 3 3 3 3 5
# 7 3 3 3 3 4
# 8 2 2 2 1 2
# 9 4 4 4 5 5
#10 3 3 3 3 4
更新
如果我们正在创建一个函数，请确保 latent 是基于函数内传递的新 'N' 创建的，因为它会导致长度不同。在 OP' post 中显示的原始代码中，length 是 10 并且 latent 是基于
创建的
make_likert <- function(N.judge = 10, item.sds, cut_points, seed = NULL){
set.seed(seed)
latent <- rnorm(N.judge)
lst1 <- lapply(item.sds, function(x) latent + rnorm(n = N.judge, sd = x))
names(lst1) <- paste0("item", seq_along(lst1))
data.frame(Map(findInterval, lst1, cut_points))
}
make_likert(N.judge=13, item.sds = item.sds, cut_points = cut_points)
# item1 item2 item3 item4 item5 item6 item7 item8 item9 item10
#1 4 3 2 3 1 3 3 5 4 4
#2 5 3 3 3 5 3 3 5 5 5
#3 2 2 3 3 3 3 3 4 3 5
#4 3 5 3 3 3 3 4 5 4 3
#5 3 1 4 3 4 3 1 2 4 4
#6 2 1 3 3 1 3 3 5 5 4
#7 3 2 3 3 3 3 5 1 5 3
#8 3 1 2 3 5 3 3 5 3 5
#9 4 3 2 3 3 4 3 1 5 5
#10 3 2 3 3 1 3 4 3 3 2
#11 4 5 1 3 5 3 1 3 5 5
#12 3 5 3 3 3 3 5 3 5 5
#13 3 4 5 3 3 3 1 1 5 3
数据
veclst <- rep(list(c(-Inf,-2.5,-1, 1,2.5,Inf),
c(-Inf,-3, -2, 2,3, Inf),
c(-Inf,-3.5,-3,-1,0.5,Inf)),
c(2, 2, 1))
您可以使用 mapply。例如。像这样：
mapply(findInterval, vec = v_arg,
x = lapply(sig_arg, rnorm, mean = latent, n = N))
#R> [,1] [,2] [,3] [,4] [,5]
#R> [1,] 4 4 3 3 3
#R> [2,] 3 3 3 4 5
#R> [3,] 3 3 3 4 4
#R> [4,] 4 4 3 3 5
#R> [5,] 2 2 3 3 3
#R> [6,] 3 3 3 3 5
#R> [7,] 3 3 3 3 4
#R> [8,] 1 1 2 3 3
#R> [9,] 4 4 3 3 5
#R> [10,] 3 3 3 3 4
如果您想要列名，请使用例如：
mapply(findInterval,
setNames(lapply(sig_arg, rnorm, mean = latent, n = N),
paste0("item", seq_along(sig_arg))),
v_arg)
#R> item1 item2 item3 item4 item5
#R> [1,] 4 4 3 3 3
#R> [2,] 3 3 3 4 5
#R> [3,] 3 3 3 4 4
#R> [4,] 4 4 3 3 5
#R> [5,] 2 2 3 3 3
#R> [6,] 3 3 3 3 5
#R> [7,] 3 3 3 3 4
#R> [8,] 1 1 2 3 3
#R> [9,] 4 4 3 3 5
#R> [10,] 3 3 3 3 4
您可以将其封装到一个函数中，这样您就可以像这样更改 N、标准偏差和断点：
sim_scores <- function(N, sigs, cuts)
mapply(findInterval,
setNames(lapply(sigs, rnorm, mean = rnorm(N), n = N),
paste0("item", seq_along(cuts))),
cuts)
# use the function
sim_scores(10L, sig_arg, v_arg)
#R> item1 item2 item3 item4 item5
#R> [1,] 3 3 3 2 3
#R> [2,] 3 3 3 3 5
#R> [3,] 3 3 3 3 4
#R> [4,] 5 4 4 4 5
#R> [5,] 3 3 3 3 3
#R> [6,] 3 3 3 3 5
#R> [7,] 3 4 3 3 5
#R> [8,] 2 2 3 3 3
#R> [9,] 3 3 3 3 4
#R> [10,] 2 2 3 3 2
sim_scores(4L, sig_arg[1:2], v_arg[1:2])
#R> item1 item2
#R> [1,] 2 2
#R> [2,] 3 3
#R> [3,] 2 3
#R> [4,] 3 3
数据
sig_arg <- c(.2, .3, .5, 1, 1.5)
v_arg <- list(c(-Inf,-2.5,-1, 1,2.5,Inf),
c(-Inf,-2.5,-1, 1,2.5,Inf),
c(-Inf,-3, -2, 2,3, Inf),
c(-Inf,-3, -2, 2,3, Inf),
c(-Inf,-3.5,-3,-1,0.5,Inf))
下面，我展示了一段代码，用于在 R 中生成一些项目分数。但是，似乎有相当多的不必要的重复才能达到最终的 data.
我想知道在 R 中是否有更紧凑的方法来实现相同的 data？
set.seed(8649)
N = 10
latent = rnorm(N)
##### generate latent responses to items
item1 = latent + rnorm(N, mean=0, sd=0.2)
item2 = latent + rnorm(N, mean=0, sd=0.3)
item3 = latent + rnorm(N, mean=0, sd=0.5)
item4 = latent + rnorm(N, mean=0, sd=1.0)
item5 = latent + rnorm(N, mean=0, sd=1.2)
##### convert latent responses to ordered categories
item1 = findInterval(item1, vec=c(-Inf,-2.5,-1, 1,2.5,Inf))
item2 = findInterval(item2, vec=c(-Inf,-2.5,-1, 1,2.5,Inf))
item3 = findInterval(item3, vec=c(-Inf,-3, -2, 2,3, Inf))
item4 = findInterval(item4, vec=c(-Inf,-3, -2, 2,3, Inf))
item5 = findInterval(item5, vec=c(-Inf,-3.5,-3,-1,0.5,Inf))
data = cbind(item1, item2, item3, item4, item5)
我们可以在 list 中创建第一组 'item'，可变部分为 'sd'
# // loop over the sd vector and create the list of random numbers in a list
lst1 <- lapply(c(0.2, 0.3, 0.5, 1, 1.2), function(x)
latent + rnorm(N, mean = 0, sd = x))
# // set the names of the list if needed
names(lst1) <- paste0("item", seq_along(lst1))
使用Map循环遍历'lst1'和vec的相应元素作为list（'veclst'）并应用findInterval
data.frame(Map(findInterval, lst1, vec = veclst))
# item1 item2 item3 item4 item5
#1 4 4 3 3 5
#2 3 3 3 3 5
#3 3 3 3 3 5
#4 4 4 4 3 5
#5 2 2 3 2 4
#6 3 3 3 3 3
#7 3 3 3 3 4
#8 2 2 2 2 2
#9 4 4 5 4 5
#10 3 3 3 3 4
或者tidyverse
library(purrr)
library(dplyr)
library(stringr)
map2_dfc(c(0.2, 0.3, 0.5, 1, 1.2), veclst, ~
findInterval(latent + rnorm(N, mean = 0, sd = .x), vec = .y)) %>%
set_names(str_c('item', seq_along(.)))
-输出
# A tibble: 10 x 5
# item1 item2 item3 item4 item5
# <int> <int> <int> <int> <int>
# 1 4 4 3 4 4
# 2 3 3 3 3 5
# 3 3 3 3 3 4
# 4 3 4 3 3 5
# 5 2 2 3 3 4
# 6 3 3 3 3 5
# 7 3 3 3 3 4
# 8 2 2 2 1 2
# 9 4 4 4 5 5
#10 3 3 3 3 4
更新
如果我们正在创建一个函数，请确保 latent 是基于函数内传递的新 'N' 创建的，因为它会导致长度不同。在 OP' post 中显示的原始代码中，length 是 10 并且 latent 是基于
make_likert <- function(N.judge = 10, item.sds, cut_points, seed = NULL){
set.seed(seed)
latent <- rnorm(N.judge)
lst1 <- lapply(item.sds, function(x) latent + rnorm(n = N.judge, sd = x))
names(lst1) <- paste0("item", seq_along(lst1))
data.frame(Map(findInterval, lst1, cut_points))
}
make_likert(N.judge=13, item.sds = item.sds, cut_points = cut_points)
# item1 item2 item3 item4 item5 item6 item7 item8 item9 item10
#1 4 3 2 3 1 3 3 5 4 4
#2 5 3 3 3 5 3 3 5 5 5
#3 2 2 3 3 3 3 3 4 3 5
#4 3 5 3 3 3 3 4 5 4 3
#5 3 1 4 3 4 3 1 2 4 4
#6 2 1 3 3 1 3 3 5 5 4
#7 3 2 3 3 3 3 5 1 5 3
#8 3 1 2 3 5 3 3 5 3 5
#9 4 3 2 3 3 4 3 1 5 5
#10 3 2 3 3 1 3 4 3 3 2
#11 4 5 1 3 5 3 1 3 5 5
#12 3 5 3 3 3 3 5 3 5 5
#13 3 4 5 3 3 3 1 1 5 3
数据
veclst <- rep(list(c(-Inf,-2.5,-1, 1,2.5,Inf),
c(-Inf,-3, -2, 2,3, Inf),
c(-Inf,-3.5,-3,-1,0.5,Inf)),
c(2, 2, 1))
您可以使用 mapply。例如。像这样：
mapply(findInterval, vec = v_arg,
x = lapply(sig_arg, rnorm, mean = latent, n = N))
#R> [,1] [,2] [,3] [,4] [,5]
#R> [1,] 4 4 3 3 3
#R> [2,] 3 3 3 4 5
#R> [3,] 3 3 3 4 4
#R> [4,] 4 4 3 3 5
#R> [5,] 2 2 3 3 3
#R> [6,] 3 3 3 3 5
#R> [7,] 3 3 3 3 4
#R> [8,] 1 1 2 3 3
#R> [9,] 4 4 3 3 5
#R> [10,] 3 3 3 3 4
如果您想要列名，请使用例如：
mapply(findInterval,
setNames(lapply(sig_arg, rnorm, mean = latent, n = N),
paste0("item", seq_along(sig_arg))),
v_arg)
#R> item1 item2 item3 item4 item5
#R> [1,] 4 4 3 3 3
#R> [2,] 3 3 3 4 5
#R> [3,] 3 3 3 4 4
#R> [4,] 4 4 3 3 5
#R> [5,] 2 2 3 3 3
#R> [6,] 3 3 3 3 5
#R> [7,] 3 3 3 3 4
#R> [8,] 1 1 2 3 3
#R> [9,] 4 4 3 3 5
#R> [10,] 3 3 3 3 4
您可以将其封装到一个函数中，这样您就可以像这样更改 N、标准偏差和断点：
sim_scores <- function(N, sigs, cuts)
mapply(findInterval,
setNames(lapply(sigs, rnorm, mean = rnorm(N), n = N),
paste0("item", seq_along(cuts))),
cuts)
# use the function
sim_scores(10L, sig_arg, v_arg)
#R> item1 item2 item3 item4 item5
#R> [1,] 3 3 3 2 3
#R> [2,] 3 3 3 3 5
#R> [3,] 3 3 3 3 4
#R> [4,] 5 4 4 4 5
#R> [5,] 3 3 3 3 3
#R> [6,] 3 3 3 3 5
#R> [7,] 3 4 3 3 5
#R> [8,] 2 2 3 3 3
#R> [9,] 3 3 3 3 4
#R> [10,] 2 2 3 3 2
sim_scores(4L, sig_arg[1:2], v_arg[1:2])
#R> item1 item2
#R> [1,] 2 2
#R> [2,] 3 3
#R> [3,] 2 3
#R> [4,] 3 3
数据
sig_arg <- c(.2, .3, .5, 1, 1.5)
v_arg <- list(c(-Inf,-2.5,-1, 1,2.5,Inf),
c(-Inf,-2.5,-1, 1,2.5,Inf),
c(-Inf,-3, -2, 2,3, Inf),
c(-Inf,-3, -2, 2,3, Inf),
c(-Inf,-3.5,-3,-1,0.5,Inf))