问题描述
基于具有列表xyz值的子列表的名称,我想从子列表中提取xyz值的样本。注意:列表不是从1
开始。
示例数据
set.seed(123)
data <- list('4' = list(x = rnorm(5),y = rnorm(5),z = rnorm(5)),'5' = list(x = rnorm(5),'6' = list(x = rnorm(5),'7' = list(x = rnorm(5),'8' = list(x = rnorm(5),z = rnorm(5)))
get_elements <- function(data,i) {
#select the main list
tmp <- data[[i]]
#Check the length of each sublist,select minimum value
#and sample 1 number from 1 to that number
rand_int <- sample(min(lengths(tmp)),1)
#select that element from each sub-list
sapply(tmp,`[[`,rand_int)
}
功能示例
# Show list number 8
data[['8']]
#> $x
#> [1] 0.3796395 -0.5023235 -0.3332074 -1.0185754 -1.0717912
#> $y
#> [1] 0.30352864 0.44820978 0.05300423 0.92226747 2.05008469
#> $z
#> [1] -0.4910312 -2.3091689 1.0057385 -0.7092008 -0.6880086
# Extract random combination from list 8
get_elements(data,'8')
#> x y z
#> -0.33320738 0.05300423 1.00573852
重写功能
使用与上述相同的功能,我将i
替换为'i'
:
get_elements <- function(data,i) {
tmp <- data[['i']] # <-- changed i to 'i'
rand_int <- sample(min(lengths(tmp)),1)
sapply(tmp,rand_int)
}
错误和问题
get_elements(data,8)
以min(lengths(tmp))为单位的警告:min没有非缺失的参数; 返回Inf list()
解决方法
尝试一下:
get_elements <- function(data,i)
{
tmp <- data[[paste(i)]]
rand_int <- sample(min(lengths(tmp)),1)
sapply(tmp,`[[`,rand_int)
}
您的初始代码失败的原因是因为“ i”被理解为“字符i”,而不是“将变量i转换为字符”。参见:
i = 1
print("i") # i
print(i) # 1