Question

我想根据每个元素中“ a”的数量对列表进行排序。

library("purrr")
data1 <- c("apple","appreciate","available","account","adapt")
data2 <- c("tab","banana","cable","tatabox","aaaaaaa")
list1 <- list(data1,data2)
ca <- lapply(list1, function(x) str_count(x, "a"))
t2 <- Map(split, list1, ca)
t3 <- transpose(t2)
> t3
$`1`
$`1`[[1]]
[1] "apple"   "account"

$`1`[[2]]
[1] "tab"   "cable"


$`2`
$`2`[[1]]
[1] "appreciate" "adapt"     

$`2`[[2]]
[1] "tatabox"


$`3`
$`3`[[1]]
[1] "available"

$`3`[[2]]
[1] "banana"

它丢失了data2中的“ aaaaaaa”。我该如何解决这个问题？

我找到了解决方法：

data1 <- c("apple","appreciate","available","account","adapt")
data2 <- c("tab","banana","cable","tatabox","aaaaaaa","aaaaaaaaaaa")
list1 <- list(data1,data2)
ca <- lapply(list1, function(x) str_count(x, "a"))
k11<- flatten(Map(split, list1, ca))
k1<-split(k11, as.integer(names(k11)))

Answer 1

引用lionel的单词：“ transpose()将列表的列表视为隐式矩形表”。它不是为像当前这样的许多边缘情况设计的。但是，您可以将较长的查询放在开头：transpose(t2[2:1])。

但是，此解决方法无法推广。我更喜欢以下方式---将子列表合并为一个列表，然后再次拆分：

> t3 <- do.call(c, t2)
> split(t3, names(t3))
$`1`
$`1`$`1`
[1] "apple"   "account"

$`1`$`1`
[1] "tab"   "cable"


$`2`
$`2`$`2`
[1] "appreciate" "adapt"     

$`2`$`2`
[1] "tatabox"


$`3`
$`3`$`3`
[1] "available"

$`3`$`3`
[1] "banana"


$`7`
$`7`$`7`
[1] "aaaaaaa"

编辑

用于命名和未命名输入的函数：

data1 <- c("apple","appreciate","available","account","adapt")
data2 <- c("tab","banana","cable","tatabox","aaaaaaa","aaaaaaaaaaa")
list1 <- list(data1,data2)
names(list1) <- c("atf","bdfs")

f <- function(x){
    if(is.null(names(x))){
        names(x) <- make.names(seq_along(x))
    }
    dtf <- stack(x)
    res <- split(dtf, str_count(dtf$values, 'a'))
    lapply(res, function(y) split(y$values, y$ind, drop = TRUE) )
}

f(list1)

Answer 2

我们也可以使用带有purrr函数的管道

map(list1, str_count, "a") %>% 
        map2(list1, ., split) %>%
        flatten %>% 
        split(names(.))
#$`1`
#$`1`$`1`
#[1] "apple"   "account"

#$`1`$`1`
#[1] "tab"   "cable"


#$`2`
#$`2`$`2`
#[1] "appreciate" "adapt"     

#$`2`$`2`
#[1] "tatabox"


#$`3`
#$`3`$`3`
#[1] "available"

#$`3`$`3`
#[1] "banana"


#$`7`
#$`7`$`7`
#[1] "aaaaaaa"

按purrr :: transpose排序列表，但缺少一些值

2 个答案:

编辑