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[英]How to create a new list that contain each individual matrix column from an old list in R?
[英]Creating a new list from an old list (R)
作為 python 用戶,到目前為止,R 還是比較直觀的。 但是,我遇到了從舊列表創建新列表的問題。
具體來說,我有一個清單:
> x
[[1]]
[1] "Heads" "Tails" "Tails" "Heads" "Tails" "Heads"
[7] "Tails" "Tails" "Tails" "Heads" "Heads" "Tails"
... .... ... etc.
這是長度 100。我想遍歷這個列表並創建一個列表 y,其值為 'Heads' = 1,'Tails' = -1。 到目前為止我的嘗試:
> for (val in x) {
ifelse(val == 'Heads', c(y,1), c(y,-1))}
其中附加了一個空列表 y。 但是,結果是長度為 2 的列表。有人可以幫我嗎?
您可以使用命名向量:
c("Heads" = 1, "Tails" = -1)[x[[1]]]
> Heads Tails Tails Heads Tails Heads Tails Tails Tails Heads Heads Tails
1 -1 -1 1 -1 1 -1 -1 -1 1 1 -1
或者使用 boolean 轉換為數字來獲取向量(優雅但可能不是最有效的):
(x[[1]] == "Heads") * 2 - 1
> [1] 1 -1 -1 1 -1 1 -1 -1 -1 1 1 -1
據我了解,您有一個包含兩個唯一值的嵌套列表。 您可以為此使用lapply
:
x <- list(list("Heads","Tails","Heads"),list("Heads","Tails","Tails"))
lapply(x, function(x){y <- ifelse(x == "Heads", 1, -1); names(y) <- x; y})
# [[1]]
# Heads Tails Heads
# 1 -1 1
#
# [[2]]
# Heads Tails Tails
# 1 -1 -1
或者,如果您不需要名稱,請使用rapply
:
x <- list(list("Heads","Tails","Heads"),list("Heads","Tails","Tails"))
rapply(x, function(x){ifelse(x == "Heads", 1, -1)})
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