问题描述
我希望仅将函数应用于嵌套列表的某些元素
"peerDependencies": {
"react": "*","react-native": "*"
}
所需的输出如下所示: 将sum函数应用于列表的$ forecast元素
l <- list()
l$a$forecast <- rnorm(3)
l$a$model <- "arima"
l$b$forecast <- rnorm(3)
l$b$model <- "prophet"
我尝试过这样的事情:
fcst <- c(sum(l$a$forecast),sum(l$b$forecast))
mdl <- c(l$a$model,l$b$model)
df <- data.frame(fcst,mdl)
解决方法
do.call(
rbind,lapply(
l,function(x) list(fcst = sum(x$forecast),model = x$model)
)
)
由于您知道返回对象的确切尺寸,因此可以在这种情况下使用vapply
来改善性能:
vapply(
l,FUN = function(x) list(fcst = sum(x$forecast),model = x$model),FUN.VALUE = list(fcst = numeric(1),model = character(1))
)
但是,生成的对象可能很难使用。
,在rrapply()
软件包中结合使用rrapply
和dplyr的bind_rows()
的另一种方法。这也扩展到包含更深层嵌套级别的列表。
rrapply::rrapply(l,condition = function(x,.xname) .xname == "forecast",f = sum) %>%
dplyr::bind_rows()
#> # A tibble: 2 x 2
#> forecast model
#> <dbl> <chr>
#> 1 -1.28 arima
#> 2 1.10 prophet
数据
set.seed(1)
l <- list(
a = list(forecast = rnorm(3),model = "arima"),b = list(forecast = rnorm(3),model = "prophet")
)
,
您可以获得带有对象letters
的字母,然后在循环中使用其输出:
n = 2 #number of lists you have
sumfore = model = vector()
for(i in letters[seq(1,n,1)]){
sumfore[i] = sum(l[[i]]$forecast)
model[i] =l[[i]]$model}
newdf = data.frame(sumfore,model)