R 在FUN中访问lappy索引名
有没有办法在我的lappy()函数中获取列表索引名R 在FUN中访问lappy索引名,r,lapply,names,indices,R,Lapply,Names,Indices,有没有办法在我的lappy()函数中获取列表索引名 n = names(mylist) lapply(mylist, function(list.elem) { cat("What is the name of this list element?\n" }) 我问是否可以在lappy()返回的列表中保留索引名,但我仍然不知道是否有一种简单的方法可以获取自定义函数中的每个元素名。我希望避免对名称本身调用lappy,我更希望在函数参数中获得名称。不幸的是,lappy只提供传递向量的元素。 通常的
n = names(mylist)
lapply(mylist, function(list.elem) { cat("What is the name of this list element?\n" })
我问是否可以在lappy()返回的列表中保留索引名,但我仍然不知道是否有一种简单的方法可以获取自定义函数中的每个元素名。我希望避免对名称本身调用lappy,我更希望在函数参数中获得名称。不幸的是,
lappy
只提供传递向量的元素。
通常的解决方法是传递向量的名称或索引,而不是向量本身
但请注意,您始终可以将额外的参数传递给函数,因此以下方法有效:
x <- list(a=11,b=12,c=13) # Changed to list to address concerns in commments
lapply(seq_along(x), function(y, n, i) { paste(n[[i]], y[[i]]) }, y=x, n=names(x))
更新更简单的示例,相同的结果:
lapply(seq_along(x), function(i) paste(names(x)[[i]], x[[i]]))
在这里,函数使用“全局”变量
x
,并提取每个调用中的名称。Tommy的答案适用于命名向量,但我知道您对列表感兴趣。他似乎在做一个结束,因为他引用了调用环境中的“x”。此函数仅使用传递给函数的参数,因此不对传递的对象的名称进行假设:
x <- list(a=11,b=12,c=13)
lapply(x, function(z) { attributes(deparse(substitute(z)))$names } )
#--------
$a
NULL
$b
NULL
$c
NULL
#--------
names( lapply(x, function(z) { attributes(deparse(substitute(z)))$names } ))
#[1] "a" "b" "c"
what_is_my_name <- function(ZZZ) return(deparse(substitute(ZZZ)))
what_is_my_name(X)
#[1] "X"
what_is_my_name(ZZZ=this)
#[1] "this"
exists("this")
#[1] FALSE
xR版本3.2的更新
免责声明:这是一个黑客把戏,在下一个版本中可能会停止工作
您可以使用以下方法获取索引:
> lapply(list(a=10,b=20), function(x){parent.frame()$i[]})
$a
[1] 1
$b
[1] 2
注意:[]
是这项工作所必需的,因为它诱使R认为符号i
(位于lappy
的计算框架中)可能有更多的引用,从而激活它的延迟复制。没有它,R将不会保留i
的单独副本:
> lapply(list(a=10,b=20), function(x){parent.frame()$i})
$a
[1] 2
$b
[1] 2
可以使用其他奇特的技巧,如函数(x){parent.frame()$i+0}
或函数(x){--parent.frame()$i}
性能影响
强制复制是否会导致性能损失?对以下是基准:
> x <- as.list(seq_len(1e6))
> system.time( y <- lapply(x, function(x){parent.frame()$i[]}) )
user system elapsed
2.38 0.00 2.37
> system.time( y <- lapply(x, function(x){parent.frame()$i[]}) )
user system elapsed
2.45 0.00 2.45
> system.time( y <- lapply(x, function(x){parent.frame()$i[]}) )
user system elapsed
2.41 0.00 2.41
> y[[2]]
[1] 2
> system.time( y <- lapply(x, function(x){parent.frame()$i}) )
user system elapsed
1.92 0.00 1.93
> system.time( y <- lapply(x, function(x){parent.frame()$i}) )
user system elapsed
2.07 0.00 2.09
> system.time( y <- lapply(x, function(x){parent.frame()$i}) )
user system elapsed
1.89 0.00 1.89
> y[[2]]
[1] 1000000
结果:
$a
[1] 1
$b
[1] 2
$c
[1] 3
$a
[1] "a"
$b
[1] "b"
$c
[1] "c"
说明:lappy
创建形式为FUN(X[[1L]],…)
,FUN(X[[2L]],…)
等的调用。因此它传递的参数是X[[i]]
,其中i
是循环中的当前索引。如果我们在求值之前得到它(即,如果我们使用替换
),我们得到未求值的表达式X[[i]]
。这是对[[
函数的调用,带有参数X
(一个符号)和i
(一个整数)。因此substitute(X)[[3]]
精确返回这个整数
有了索引,您可以轻松地访问名称,如果您先这样保存它:
L <- list(a=10,b=10,c=10)
n <- names(L)
lapply(L, function(x)n[substitute(x)[[3]]])
lapply2(letters, function(x, i) paste(x, i))
或者使用第二个技巧::-)
(结果是一样的)
解释2:sys.call(1)
返回lappy(…)
,因此sys.call(1)[[2]]
是用作lappy
的列表参数的表达式。将该表达式传递给eval
将创建一个合法对象,该对象的名称
可以访问。这很棘手,但它可以工作
奖励:获得姓名的第二种方法:
lapply(list(a=10,b=10,c=10), function(x)eval.parent(quote(names(X)))[substitute(x)[[3]]])
请注意,X
在FUN
的父框架中是一个有效的对象,并且引用了lappy
的列表参数,因此我们可以使用eval.parent
来获得它。这基本上使用了与Tommy相同的解决方法,但是使用了Map()
,无需访问存储列表组件名称的全局变量
> x <- list(a=11, b=12, c=13)
> Map(function(x, i) paste(i, x), x, names(x))
$a
[1] "a 11"
$b
[1] "b 12"
$c
[1] "c 13
我的答案与Tommy和caracals的答案方向相同,但避免了将列表另存为其他对象
lapply(seq(3), function(i, y=list(a=14,b=15,c=16)) { paste(names(y)[[i]], y[[i]]) })
结果:
[[1]]
[1] "a 14"
[[2]]
[1] "b 15"
[[3]]
[1] "c 16"
这将列表作为命名参数提供给FUN(而不是lappy)。lappy只需迭代列表的元素(在更改列表长度时,请小心将第一个参数更改为lappy)
注意:将列表直接作为附加参数提供给Lappy也可以:
lapply(seq(3), function(i, y) { paste(names(y)[[i]], y[[i]]) }, y=list(a=14,b=15,c=16))
我有过很多次同样的问题。。。
我开始使用另一种方法…我开始使用mapply
> mapply(function(x, i) paste(i, x), x, names(x))
a b c
"a 11" "b 12" "c 13"
n = names(mylist)
mapply(function(list.elem, names) { }, list.elem = mylist, names = n)
只需循环输入名称
sapply(names(mylist), function(n) {
doSomething(mylist[[n]])
cat(n, '\n')
}
只需编写自己的自定义lappy
函数
lapply2 <- function(X, FUN){
if( length(formals(FUN)) == 1 ){
# No index passed - use normal lapply
R = lapply(X, FUN)
}else{
# Index passed
R = lapply(seq_along(X), FUN=function(i){
FUN(X[[i]], i)
})
}
# Set names
names(R) = names(X)
return(R)
}
您可以尝试使用purr
包中的imap()
来自文档:
> lapply(list(a=10,b=10,c=10), function(x)substitute(x)[[3]])
imap(x,…)是map2(x,名称(x),…)的简写,如果x有名称,或者map2(x,沿着(x),…)没有名称
因此,您可以这样使用它:
library(purrr)
myList <- list(a=11,b=12,c=13)
imap(myList, function(x, y) paste(x, y))
@caracals和@Tommy都是很好的解决方案,这是一个示例,包括list
's和data.frame
's.
r
是列表和数据帧的列表(dput(r[[1]]]
)
目的是取消列出
所有列表,将列表
的名称序列作为一列来识别案例
r=unlist(unlist(r,F),F)
names(r)
[1] "todos.F0.rst1" "todos.F0.rst5" "todos.T0.rst1" "todos.T0.rst5" "random.F0.rst1" "random.F0.rst5"
[7] "random.T0.rst1" "random.T0.rst5"
取消列出列表,但不列出数据框
ra=Reduce(rbind,Map(function(x,y) cbind(case=x,y),names(r),r))
Map
将名称序列作为一列。Reduce
join alldata.frame
's
head(ra)
case algo rst prec rorac prPo pos
1 todos.F0.rst1 Mean 56.4 0.450 25.872 91.2 239
6 todos.F0.rst1 gbm1 41.8 0.438 22.595 77.4 239
4 todos.F0.rst1 GAM2 37.2 0.512 43.256 50.0 172
7 todos.F0.rst1 gbm2 36.8 0.422 18.039 85.4 239
11 todos.F0.rst1 ran2 35.0 0.442 23.810 61.5 239
2 todos.F0.rst1 nai1 29.8 0.544 52.281 33.1 172
附言r[[1]]
:
structure(list(F0 = structure(list(rst1 = structure(list(algo = c("Mean",
"gbm1", "GAM2", "gbm2", "ran2", "nai1", "GAM3", "GAM1", "ran1",
"svm2", "svm1"), rst = c(56.4, 41.8, 37.2, 36.8, 35, 29.8, 28.8,
21.8, 19.4, 14, 0.8), prec = c(0.45, 0.438, 0.512, 0.422, 0.442,
0.544, 0.403, 0.405, 0.406, 0.385, 0.359), rorac = c(25.872,
22.595, 43.256, 18.039, 23.81, 52.281, 12.743, 13.374, 13.566,
7.692, 0.471), prPo = c(91.2, 77.4, 50, 85.4, 61.5, 33.1, 94.6,
68.2, 59.8, 76.2, 71.1), pos = c(239L, 239L, 172L, 239L, 239L,
172L, 239L, 239L, 239L, 239L, 239L)), .Names = c("algo", "rst",
"prec", "rorac", "prPo", "pos"), row.names = c(1L, 6L, 4L, 7L,
11L, 2L, 5L, 3L, 10L, 9L, 8L), class = "data.frame"), rst5 = structure(list(
algo = c("Mean", "gbm2", "gbm1", "GAM3", "GAM2", "ran2",
"nai1", "GAM1", "svm2", "ran1", "svm1"), rst = c(52.4, 46.4,
31.2, 28.8, 28.2, 26.6, 23.6, 20.6, 14.4, 14, 6.2), prec = c(0.441,
0.44, 0.416, 0.403, 0.481, 0.422, 0.519, 0.398, 0.386, 0.39,
0.37), rorac = c(23.604, 23.2, 16.421, 12.743, 34.815, 18.095,
45.385, 11.381, 8.182, 9.091, 3.584), prPo = c(92.9, 83.7,
79.5, 94.6, 47.1, 61.5, 30.2, 75.7, 73.6, 64.4, 72.4), pos = c(239L,
239L, 239L, 239L, 172L, 239L, 172L, 239L, 239L, 239L, 239L
)), .Names = c("algo", "rst", "prec", "rorac", "prPo", "pos"
), row.names = c(1L, 7L, 6L, 5L, 4L, 11L, 2L, 3L, 9L, 10L, 8L
), class = "data.frame")), .Names = c("rst1", "rst5")), T0 = structure(list(
rst1 = structure(list(algo = c("Mean", "ran1", "GAM1", "GAM2",
"gbm1", "svm1", "nai1", "gbm2", "svm2", "ran2"), rst = c(22.6,
19.4, 13.6, 10.2, 9.6, 8, 5.6, 3.4, -0.4, -0.6), prec = c(0.478,
0.452, 0.5, 0.421, 0.423, 0.833, 0.429, 0.373, 0.355, 0.356
), rorac = c(33.731, 26.575, 40, 17.895, 18.462, 133.333,
20, 4.533, -0.526, -0.368), prPo = c(34.4, 52.1, 24.3, 40.7,
37.1, 3.1, 14.4, 53.6, 54.3, 116.4), pos = c(195L, 140L,
140L, 140L, 140L, 195L, 195L, 140L, 140L, 140L)), .Names = c("algo",
"rst", "prec", "rorac", "prPo", "pos"), row.names = c(1L,
9L, 3L, 4L, 5L, 7L, 2L, 6L, 8L, 10L), class = "data.frame"),
rst5 = structure(list(algo = c("gbm1", "ran1", "Mean", "GAM1",
"GAM2", "svm1", "nai1", "svm2", "gbm2", "ran2"), rst = c(17.6,
16.4, 15, 12.8, 9, 6.2, 5.8, -2.6, -3, -9.2), prec = c(0.466,
0.434, 0.435, 0.5, 0.41, 0.8, 0.44, 0.346, 0.345, 0.337),
rorac = c(30.345, 21.579, 21.739, 40, 14.754, 124, 23.2,
-3.21, -3.448, -5.542), prPo = c(41.4, 54.3, 35.4, 22.9,
43.6, 2.6, 12.8, 57.9, 62.1, 118.6), pos = c(140L, 140L,
195L, 140L, 140L, 195L, 195L, 140L, 140L, 140L)), .Names = c("algo",
"rst", "prec", "rorac", "prPo", "pos"), row.names = c(5L,
9L, 1L, 3L, 4L, 7L, 2L, 8L, 6L, 10L), class = "data.frame")), .Names = c("rst1",
"rst5"))), .Names = c("F0", "T0"))
假设我们要计算每个元素的长度
mylist <- list(a=1:4,b=2:9,c=10:20)
mylist
$a
[1] 1 2 3 4
$b
[1] 2 3 4 5 6 7 8 9
$c
[1] 10 11 12 13 14 15 16 17 18 19 20
如果目标是在函数内部使用标签,则通过在两个对象上循环使用mapply()
非常有用;这两个对象是列表元素和列表名称
fun <- function(x,y) paste0(length(x),"_",y)
mapply(fun,mylist,names(mylist))
a b c
"4_a" "8_b" "11_c"
fun@ferdinand kraft给了我们一个很好的把戏,然后告诉我们不应该使用它
因为它没有文档,而且性能开销很大
我不能对第一点有太多的争论,但我想指出,开销
应该很少引起关注
让我们定义活动函数,这样就不必调用复杂表达式
parent.frame()$i[]
但仅.i()
,我们还将创建.n()
以访问
名称,它应该适用于基本函数和purrr函数(可能也适用于大多数其他函数)
.i自定义函数中的'i'参数是如何初始化的?明白了,所以lappy()确实适用于t
head(ra)
case algo rst prec rorac prPo pos
1 todos.F0.rst1 Mean 56.4 0.450 25.872 91.2 239
6 todos.F0.rst1 gbm1 41.8 0.438 22.595 77.4 239
4 todos.F0.rst1 GAM2 37.2 0.512 43.256 50.0 172
7 todos.F0.rst1 gbm2 36.8 0.422 18.039 85.4 239
11 todos.F0.rst1 ran2 35.0 0.442 23.810 61.5 239
2 todos.F0.rst1 nai1 29.8 0.544 52.281 33.1 172
structure(list(F0 = structure(list(rst1 = structure(list(algo = c("Mean",
"gbm1", "GAM2", "gbm2", "ran2", "nai1", "GAM3", "GAM1", "ran1",
"svm2", "svm1"), rst = c(56.4, 41.8, 37.2, 36.8, 35, 29.8, 28.8,
21.8, 19.4, 14, 0.8), prec = c(0.45, 0.438, 0.512, 0.422, 0.442,
0.544, 0.403, 0.405, 0.406, 0.385, 0.359), rorac = c(25.872,
22.595, 43.256, 18.039, 23.81, 52.281, 12.743, 13.374, 13.566,
7.692, 0.471), prPo = c(91.2, 77.4, 50, 85.4, 61.5, 33.1, 94.6,
68.2, 59.8, 76.2, 71.1), pos = c(239L, 239L, 172L, 239L, 239L,
172L, 239L, 239L, 239L, 239L, 239L)), .Names = c("algo", "rst",
"prec", "rorac", "prPo", "pos"), row.names = c(1L, 6L, 4L, 7L,
11L, 2L, 5L, 3L, 10L, 9L, 8L), class = "data.frame"), rst5 = structure(list(
algo = c("Mean", "gbm2", "gbm1", "GAM3", "GAM2", "ran2",
"nai1", "GAM1", "svm2", "ran1", "svm1"), rst = c(52.4, 46.4,
31.2, 28.8, 28.2, 26.6, 23.6, 20.6, 14.4, 14, 6.2), prec = c(0.441,
0.44, 0.416, 0.403, 0.481, 0.422, 0.519, 0.398, 0.386, 0.39,
0.37), rorac = c(23.604, 23.2, 16.421, 12.743, 34.815, 18.095,
45.385, 11.381, 8.182, 9.091, 3.584), prPo = c(92.9, 83.7,
79.5, 94.6, 47.1, 61.5, 30.2, 75.7, 73.6, 64.4, 72.4), pos = c(239L,
239L, 239L, 239L, 172L, 239L, 172L, 239L, 239L, 239L, 239L
)), .Names = c("algo", "rst", "prec", "rorac", "prPo", "pos"
), row.names = c(1L, 7L, 6L, 5L, 4L, 11L, 2L, 3L, 9L, 10L, 8L
), class = "data.frame")), .Names = c("rst1", "rst5")), T0 = structure(list(
rst1 = structure(list(algo = c("Mean", "ran1", "GAM1", "GAM2",
"gbm1", "svm1", "nai1", "gbm2", "svm2", "ran2"), rst = c(22.6,
19.4, 13.6, 10.2, 9.6, 8, 5.6, 3.4, -0.4, -0.6), prec = c(0.478,
0.452, 0.5, 0.421, 0.423, 0.833, 0.429, 0.373, 0.355, 0.356
), rorac = c(33.731, 26.575, 40, 17.895, 18.462, 133.333,
20, 4.533, -0.526, -0.368), prPo = c(34.4, 52.1, 24.3, 40.7,
37.1, 3.1, 14.4, 53.6, 54.3, 116.4), pos = c(195L, 140L,
140L, 140L, 140L, 195L, 195L, 140L, 140L, 140L)), .Names = c("algo",
"rst", "prec", "rorac", "prPo", "pos"), row.names = c(1L,
9L, 3L, 4L, 5L, 7L, 2L, 6L, 8L, 10L), class = "data.frame"),
rst5 = structure(list(algo = c("gbm1", "ran1", "Mean", "GAM1",
"GAM2", "svm1", "nai1", "svm2", "gbm2", "ran2"), rst = c(17.6,
16.4, 15, 12.8, 9, 6.2, 5.8, -2.6, -3, -9.2), prec = c(0.466,
0.434, 0.435, 0.5, 0.41, 0.8, 0.44, 0.346, 0.345, 0.337),
rorac = c(30.345, 21.579, 21.739, 40, 14.754, 124, 23.2,
-3.21, -3.448, -5.542), prPo = c(41.4, 54.3, 35.4, 22.9,
43.6, 2.6, 12.8, 57.9, 62.1, 118.6), pos = c(140L, 140L,
195L, 140L, 140L, 195L, 195L, 140L, 140L, 140L)), .Names = c("algo",
"rst", "prec", "rorac", "prPo", "pos"), row.names = c(5L,
9L, 1L, 3L, 4L, 7L, 2L, 8L, 6L, 10L), class = "data.frame")), .Names = c("rst1",
"rst5"))), .Names = c("F0", "T0"))
mylist <- list(a=1:4,b=2:9,c=10:20)
mylist
$a
[1] 1 2 3 4
$b
[1] 2 3 4 5 6 7 8 9
$c
[1] 10 11 12 13 14 15 16 17 18 19 20
sapply(mylist,length,USE.NAMES=T)
a b c
4 8 11
fun <- function(x,y) paste0(length(x),"_",y)
mapply(fun,mylist,names(mylist))
a b c
"4_a" "8_b" "11_c"