R 中的 apply
函数是一种简化 for 循环以获得输出的好方法。是否有等效的函数可以帮助人们在替换向量值时避免 for 循环?这可以通过示例更好地理解......
# Take this list for example
x = list( list(a=1,b=2), list(a=3,b=4), list(a=5,b=6) )
# To get all of the "a" elements from each list, I can do
vapply(x,"[[",1,"a")
[1] 1 3 5
# If I want to change all of the "a" elements, I cannot do
vapply(x,"[[",1,"a") = 10:12
Error in vapply(x, "[[", 1, "a") = 10:12 :
could not find function "vapply<-"
# (this error was expected)
# Instead I must do something like this...
new.a = 10:12
for(i in seq_along(x)) x[[i]]$a = new.a[i]
有没有比使用循环更简单或更快的替代方法?
最佳答案
一种选择是首先 unlist
列表 x
,然后替换名为 "a"
的值,然后 relist
基于 u
的列表结构的新列表 x
。
u <- unlist(x)
u[names(u) == "a"] <- 10:12
relist(u, x)