我正在尝试将数据帧解析为一种列表格式，其中A列是链接到B列中一组bin的一组“ bin”，其中C列是从bin A移到bin B的比例。手动创建列表格式很容易，例如。
df.list <- list()
df.list$litter <- data.frame('lost'=50,'pot'=15,'pool'=35)
df.list$lost <- data.frame('pool'=30, 'pot'=10, 'lost'=60)
df.list$bin <- data.frame(uncollected=4, collected=96)
df.list$dump <- data.frame(litter=50, burn=20)
df.list$uncollected <- data.frame(litter=25, burn=55, dump=20)
df.list$collected <- data.frame(litter=3.7, dump=54.3, recycle=30, burn=12 )
但是我有一个大型数据库，并且以csv格式存在多个年份和多个位置，在这里我需要将每年和位置分开，然后列出一个类似于以下每年输出的列表。
> df.list
$litter
lost pot pool
1 50 15 35
$lost
pool pot lost
1 30 10 60
$bin
uncollected collected
1 4 96
$dump
litter burn
1 50 20
$uncollected
litter burn dump
1 25 55 20
$collected
litter dump recycle burn
1 3.7 54.3 30 12
我尝试了其他已发布的示例，但是我不知道如何在 item 列中隔离每个“ bin”，然后嵌套 node 和以列表格式在项目中关联值 ... 这是我正在使用的数据框的模拟：
df <- data.frame(year=sort(c(rep(c(2010, 2011, 2012), 15))),
item=c(rep(rep(c("litter", "lost", "bin", "uncollected", "collected"), c(3,3,2,3,4)),3)),
node=rep(c("lost", "pot" , "pool", "pool" ,"pot" , "lost", "collected", "uncollected",
"litter", "burn" , "dump" , "litter" , "dump", "recycle", "burn"),3),
value=rep(c(50.0, 15.0 ,35.0, 30.0, 10.0, 60.0, 96.0, 4.0, 25.0, 55.0 , 20.0, 3.7, 54.3,30.0, 12.0),3))
非常感谢！
答案 0 :(得分：1)
我已经将其敲定到位，但这应该可以满足您的要求。我将在这里笨拙地进行循环，但是我敢肯定可以通过lapply进行修改以提高效率。
outlist <- list()
for(i in unique(df$year)){
for(j in unique(df$item)){
list.item.name <- paste(j,i,sep='.')
tdf <- df[which(df$year==i & df$item==j),c('node','value')]
newdf <- data.frame(t(tdf))
names(newdf) <- newdf$node
rownames(newdf) <- NULL
outlist[[list.item.name]] <- newdf
}
}
这将产生：
> outlist
$`litter.2010`
1 lost pot pool
2 50 15 35
$lost.2010
1 pool pot lost
2 30 10 60
$bin.2010
1 collected uncollected
2 96 4
$uncollected.2010
1 litter burn dump
2 25 55 20
$collected.2010
1 litter dump recycle burn
2 3.7 54.3 30.0 12.0
$litter.2011
1 lost pot pool
2 50 15 35
$lost.2011
1 pool pot lost
2 30 10 60
$bin.2011
1 collected uncollected
2 96 4
$uncollected.2011
1 litter burn dump
2 25 55 20
$collected.2011
1 litter dump recycle burn
2 3.7 54.3 30.0 12.0
$litter.2012
1 lost pot pool
2 50 15 35
$lost.2012
1 pool pot lost
2 30 10 60
$bin.2012
1 collected uncollected
2 96 4
$uncollected.2012
1 litter burn dump
2 25 55 20
$collected.2012
1 litter dump recycle burn
2 3.7 54.3 30.0 12.0
答案 1 :(得分：0)
两次调用dlply()是否可以提供所需的输出？
df1 <- dlply(df,.(year),function(x){
x %>% select(-year) %>%
dlply(.,.(item))
})
df1
$`2010`
$bin
item node value
1 bin collected 96
2 bin uncollected 4
$collected
item node value
1 collected litter 3.7
2 collected dump 54.3
3 collected recycle 30.0
4 collected burn 12.0
$litter
item node value
1 litter lost 50
2 litter pot 15
3 litter pool 35
$lost
item node value
1 lost pool 30
2 lost pot 10
3 lost lost 60
...
str(df1)
List of 3
$ 2010:List of 5
..$ bin :'data.frame': 2 obs. of 3 variables:
.. ..$ item : Factor w/ 5 levels "bin","collected",..: 1 1
.. ..$ node : Factor w/ 9 levels "burn","collected",..: 2 9
.. ..$ value: num [1:2] 96 4
.. ..- attr(*, "vars")= chr "item"
..$ collected :'data.frame': 4 obs. of 3 variables:
.. ..$ item : Factor w/ 5 levels "bin","collected",..: 2 2 2 2
.. ..$ node : Factor w/ 9 levels "burn","collected",..: 4 3 8 1
.. ..$ value: num [1:4] 3.7 54.3 30 12
.. ..- attr(*, "vars")= chr "item"
..$ litter :'data.frame': 3 obs. of 3 variables:
.. ..$ item : Factor w/ 5 levels "bin","collected",..: 3 3 3
.. ..$ node : Factor w/ 9 levels "burn","collected",..: 5 7 6
.. ..$ value: num [1:3] 50 15 35
.. ..- attr(*, "vars")= chr "item"
..$ lost :'data.frame': 3 obs. of 3 variables:
.. ..$ item : Factor w/ 5 levels "bin","collected",..: 4 4 4
.. ..$ node : Factor w/ 9 levels "burn","collected",..: 6 7 5
.. ..$ value: num [1:3] 30 10 60
.. ..- attr(*, "vars")= chr "item"
..$ uncollected:'data.frame': 3 obs. of 3 variables:
.. ..$ item : Factor w/ 5 levels "bin","collected",..: 5 5 5
.. ..$ node : Factor w/ 9 levels "burn","collected",..: 4 1 3
.. ..$ value: num [1:3] 25 55 20
.. ..- attr(*, "vars")= chr "item"
..- attr(*, "class")= chr [1:2] "split" "list"
...
编辑
首先尝试使用spread()，然后将其传递给两个dlply()调用。我认为这与您所需的输出有点接近。您可能可以修改dlply()中的调用以删除具有NA值的变量。
df %>% spread(node,value) %>% dlply(.,.(year),function(x){
x %>% select(-year) %>%
dlply(.,.(item))
})
$`2010`
$bin
item burn collected dump litter lost pool pot recycle uncollected
1 bin NA 96 NA NA NA NA NA NA 4
$collected
item burn collected dump litter lost pool pot recycle uncollected
1 collected 12 NA 54.3 3.7 NA NA NA 30 NA
$litter
item burn collected dump litter lost pool pot recycle uncollected
1 litter NA NA NA NA 50 35 15 NA NA
$lost
item burn collected dump litter lost pool pot recycle uncollected
1 lost NA NA NA NA 60 30 10 NA NA
$uncollected
item burn collected dump litter lost pool pot recycle uncollected
1 uncollected 55 NA 20 25 NA NA NA NA NA
$`2011`
$bin
item burn collected dump litter lost pool pot recycle uncollected
1 bin NA 96 NA NA NA NA NA NA 4
$collected
item burn collected dump litter lost pool pot recycle uncollected
1 collected 12 NA 54.3 3.7 NA NA NA 30 NA
$litter
item burn collected dump litter lost pool pot recycle uncollected
1 litter NA NA NA NA 50 35 15 NA NA
$lost
item burn collected dump litter lost pool pot recycle uncollected
1 lost NA NA NA NA 60 30 10 NA NA
$uncollected
item burn collected dump litter lost pool pot recycle uncollected
1 uncollected 55 NA 20 25 NA NA NA NA NA
$`2012`
$bin
item burn collected dump litter lost pool pot recycle uncollected
1 bin NA 96 NA NA NA NA NA NA 4
$collected
item burn collected dump litter lost pool pot recycle uncollected
1 collected 12 NA 54.3 3.7 NA NA NA 30 NA
$litter
item burn collected dump litter lost pool pot recycle uncollected
1 litter NA NA NA NA 50 35 15 NA NA
$lost
item burn collected dump litter lost pool pot recycle uncollected
1 lost NA NA NA NA 60 30 10 NA NA
$uncollected
item burn collected dump litter lost pool pot recycle uncollected
1 uncollected 55 NA 20 25 NA NA NA NA NA
答案 2 :(得分：-1)
修改：不确定投反对票。无论如何，这是一个类似的输出，它提供矢量而不是data.frames：
by(df[, -(1:2)]
, INDICES = list(df$item, df$year)
, FUN = function(DF) {x <- DF$value; names(x) = as.character(DF$node); x}
)
这提供了相似的输出，但附加了一年，输出是一堆data.frames。
split(df, f = list(df$item, df$year))
$bin.2010
year item node value
7 2010 bin collected 96
8 2010 bin uncollected 4
$collected.2010
year item node value
12 2010 collected litter 3.7
13 2010 collected dump 54.3
14 2010 collected recycle 30.0
15 2010 collected burn 12.0
$litter.2010
year item node value
1 2010 litter lost 50
2 2010 litter pot 15
3 2010 litter pool 35
$lost.2010
year item node value
4 2010 lost pool 30
5 2010 lost pot 10
6 2010 lost lost 60
$uncollected.2010
year item node value
9 2010 uncollected litter 25
10 2010 uncollected burn 55
11 2010 uncollected dump 20
$bin.2011
year item node value
22 2011 bin collected 96
23 2011 bin uncollected 4
#rest truncated