替换列表列表中的元素

Question

applyR中的函数是简化循环以获得输出的好方法.是否有一个等效函数可以帮助人们在替换向量值时避免循环？通过实例更好地理解这一点......

# 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]

使用循环是否有更简单或更快的替代方案？

Answer 1

一种选择是首先unlist列表x,然后替换名为的值"a",然后根据列表结构替换relist新列表.ux

u <- unlist(x)
u[names(u) == "a"] <- 10:12
relist(u, x)