使用dataframe按键重命名列表元素

我有一个如下列表：

$fec9
[1] "yes"

$`39c1`
[1] "no"

$d387
[1] "yes"

$`0065`
[1] "yes"

以及具有与列表元素匹配的键的数据帧，如下所示：

dataframe <- data.frame(key = c('39c1', 'fec9', 'p731' '0065', 'd387'),
                        label = c('trash', 'wash car', 'cook dinner', 'mow lawn', 'vacuum'))

---

dataframe假设data.frame的列是
字符
，然后我们得到'list'和'key'的
名称
的
元素的
相交
并使用该元素为
列表
值分配相应的'label'

nm1 <- intersect(names(list), dataframe$key)
list[nm1] <- dataframe$label[dataframe$key %in% nm1]

nm1我希望您正在寻找：

#method 1
#get common key from dataframe
df <- df[df$key %in% names(list), ]
list <- list[df$key] #getting the same order as of dataframe
names(list) <- df$label

#method 2
#if you want to preserve the order of labels: 
df <- df[df$key %in% names(list), ]
row.names(df) <- df$key
names(list) <- df[names(list), ]$label

下面是一个使用
merge
+
stack
+
setNames

with(
  merge(stack(lst), df, by.x = "ind", by.y = "key"),
  setNames(as.list(values), label)
)

给

$`mow lawn`
[1] "yes"

$trash
[1] "no"

$vacuum
[1] "yes"

$`wash car`
[1] "yes"

数据

list <- list(fec9 = 'yes', `39c1` = 'no', 'd387' = 'yes', `0065` = 'yes')
df <- data.frame(key = c('39c1', 'fec9', 'p731', '0065', 'd387'),  
                        label = c('trash', 'wash car', 'cook dinner', 'mow lawn', 'vacuum'), stringsAsFactors = FALSE)

> dput(lst)
list(fec9 = "yes", `39c1` = "no", d387 = "yes", `0065` = "yes")

> dput(df)
structure(list(key = c("39c1", "fec9", "p731", "0065", "d387"
), label = c("trash", "wash car", "cook dinner", "mow lawn",
"vacuum")), class = "data.frame", row.names = c(NA, -5L))

with(
  merge(stack(lst), df, by.x = "ind", by.y = "key"),
  setNames(as.list(values), label)
)

$`mow lawn`
[1] "yes"

$trash
[1] "no"

$vacuum
[1] "yes"

$`wash car`
[1] "yes"

> dput(lst)
list(fec9 = "yes", `39c1` = "no", d387 = "yes", `0065` = "yes")

> dput(df)
structure(list(key = c("39c1", "fec9", "p731", "0065", "d387"
), label = c("trash", "wash car", "cook dinner", "mow lawn",
"vacuum")), class = "data.frame", row.names = c(NA, -5L))