list 基于逻辑条件的列表中的子集元素
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Subset elements in a list based on a logical condition
提问by jrara
How can I subset a list based on a condition (TRUE, FALSE) in another list? Please, see my example below:
如何根据另一个列表中的条件（TRUE、FALSE）对列表进行子集化？请看我下面的例子：
l <- list(a=c(1,2,3), b=c(4,5,6,5), c=c(3,4,5,6))
l
$a
[1] 1 2 3
$b
[1] 4 5 6 5
$c
[1] 3 4 5 6
cond <- lapply(l, function(x) length(x) > 3)
cond
$a
[1] FALSE
$b
[1] TRUE
$c
[1] TRUE
> l[cond]
Error in l[cond] : invalid subscript type 'list'
l[cond] 中的错误：无效的下标类型“列表”
采纳答案by James
[is expecting a vector, so use unliston cond:
[期待一个向量，所以使用unliston cond：
l[unlist(cond)]
$b
[1] 4 5 6 5
$c
[1] 3 4 5 6
回答by Jason Morgan
This is what the Filterfunction was made for:
这是该Filter功能的用途：
Filter(function(x) length(x) > 3, l)
$b
[1] 4 5 6 5
$c
[1] 3 4 5 6
回答by PatrickR
Another way is to use sapplyinstead of lapply.
另一种方法是使用sapply代替lapply。
cond <- sapply(l, function(x) length(x) > 3)
l[cond]
回答by NPE
> l[as.logical(cond)]
$b
[1] 4 5 6 5
$c
[1] 3 4 5 6
回答by jazzurro
I recently learned lengths(), which gets the length of each element of a list. This allows us to avoid making another list including logical values as the OP tried.
我最近学习了lengths()，它获取列表中每个元素的长度。这使我们能够避免在 OP 尝试时制作另一个包含逻辑值的列表。
lengths(l)
#a b c
#3 4 4
Using this in a logical condition, we can subset list elements in l.
在逻辑条件中使用它，我们可以在 中对列表元素进行子集化l。
l[lengths(l) > 3]
$b
[1] 4 5 6 5
$c
[1] 3 4 5 6
回答by Andre Elrico
cond <- lapply(l, length) > 3
l[cond]
