[英]How to apply a function to more than one list of lists in r?
我有一個包含列表列表的數據。 我想找到最大值。 如圖所示,每個雙精度類型數據的值;
這是我的數據結構;
list(`Cluster 1` = list(Day_1 = list(structure(c(`1` = 0, `2` = 0,
`3` = 0, `4` = 0, `5` = 0, `6` = 0, `7` = 0, `8` = 0, `9` = 0.041,
`10` = 0.673, `11` = 0, `12` = 0.766), .Dim = 12L, .Dimnames = list(
c("1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11",
"12"))), structure(c(`1` = 0, `2` = 0, `3` = 0, `4` = 0,
`5` = 0, `6` = 0.041, `7` = 0.673, `8` = 0.766), .Dim = 8L, .Dimnames = list(
c("1", "2", "3", "4", "5", "6", "7", "8")))), Day_2 = list(
structure(c(`1` = 1.07, `2` = 0, `3` = 1.27, `4` = 0.19,
`5` = 0, `6` = 0, `7` = 0, `8` = 0, `9` = 0, `10` = 0, `11` = 0,
`12` = 0), .Dim = 12L, .Dimnames = list(c("1", "2", "3",
"4", "5", "6", "7", "8", "9", "10", "11", "12"))), structure(c(`1` = 1.07,
`2` = 1.27, `3` = 0.19, `4` = 0, `5` = 0, `6` = 0, `7` = 0,
`8` = 0), .Dim = 8L, .Dimnames = list(c("1", "2", "3", "4",
"5", "6", "7", "8"))))), `Cluster 2` = list(Day_3 = list(
structure(c(`1` = 0, `2` = 0, `3` = 0, `4` = 0, `5` = 0,
`6` = 0, `7` = 0, `8` = 0, `9` = 0.19, `10` = 0, `11` = 0,
`12` = 0), .Dim = 12L, .Dimnames = list(c("1", "2", "3",
"4", "5", "6", "7", "8", "9", "10", "11", "12"))), structure(c(`1` = 0,
`2` = 0, `3` = 0, `4` = 0, `5` = 0, `6` = 0.19, `7` = 0,
`8` = 0), .Dim = 8L, .Dimnames = list(c("1", "2", "3", "4",
"5", "6", "7", "8")))), Day_4 = list(structure(c(`1` = 0.521,
`2` = 0.229, `3` = 0, `4` = 0, `5` = 0, `6` = 0, `7` = 0, `8` = 0,
`9` = 0, `10` = 0, `11` = 0, `12` = 0), .Dim = 12L, .Dimnames = list(
c("1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11",
"12"))), structure(c(`1` = 0.75, `2` = 0, `3` = 0, `4` = 0,
`5` = 0, `6` = 0, `7` = 0, `8` = 0), .Dim = 8L, .Dimnames = list(
c("1", "2", "3", "4", "5", "6", "7", "8"))))))
我試過這段代碼;
nested_lapply <- function(data, fun) {
lapply(data, function(sublist) { lapply(sublist, fun) })
}
但我收到了這個錯誤
res<-nested_lapply(out, max) Error in FUN(X[[i]], ...): invalid 'type' (list) of argument Called from: lapply(sublist, fun)
除了問題:
我需要找到最大值。 所有 [[ 1 ]] 以及 [[2]] 的值。
max1<-max(out$ Cluster 1
$Day_1[ 1 ], out$ Cluster 1
$Day_2[ 1 ], out$ Cluster 2
$Day_3[ 1 ], out$ Cluster 2
$Day_4[ 1 ])
max2<-max(out$ Cluster 1
$Day_1[[2]], out$ Cluster 1
$Day_2[[2]], out$ Cluster 2
$Day_3[[2]], out$ Cluster 2
$Day_4[[2 ]])
我們可以只使用來自base R
rapply
rapply ,它將遞歸地循環三個嵌套list
並從內部vector
中獲取max
rapply(out, max)
如果我們想跨越max
library(dplyr)
library(data.table)
reshape2::melt(out) %>%
group_by(L3) %>%
summarise(value = max(value))
# A tibble: 2 x 2
# L3 value
# <int> <dbl>
#1 1 1.27
#2 2 1.27
或者它可能是
flatten(out) %>%
transpose %>%
map(reduce, pmax)
#[[1]]
# 1 2 3 4 5 6 7 8 9 10 11 12
#1.070 0.229 1.270 0.190 0.000 0.000 0.000 0.000 0.190 0.673 0.000 0.766
#[[2]]
# 1 2 3 4 5 6 7 8
#1.070 1.270 0.190 0.000 0.000 0.190 0.673 0.766
或單個值
flatten(out) %>%
transpose %>%
map_dbl(reduce, max)
#[1] 1.27 1.27
你可以使用這個解決方案
library(tidyverse)
library(purrr)
df %>%
enframe() %>%
unnest_longer(value) %>%
unnest_longer(value) %>%
transmute(name, value_id, out = map_dbl(value, max))
你可以做這樣的事情
lapply(1:length(data), function(X) lapply(data[[X]], FUN))
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.