将函数重命名为r上的列表时出错

我的数据：

library(tidyverse)

1:6%>%
  str_c('var',.)%>%
  map(~assign(.,runif(30,20,100),envir=globalenv()))
tibble<-as_tibble(
  bind_cols(mget(ls(pattern='*v')))
)
cluster<-kmeans(tibble,centers=3)
cluster
tibble$kmeans<-as.factor(cluster[['cluster']])
head(tibble)

第一个功能：

mylist%>%
  lapply(function(x){
    lapply(x,names(x)[str_which(names(x),regex('KMEa',ignore_case=T))])<- 'variable'
})

mylist%>%
  map(.,~names(.)[str_which(names(.),regex('KM',ignore_case=T))])<-'variable'

mylist%>%
lappy（函数（x）{
lapply（x，names（x）[str_which（names（x），regex（'KMEa'，ignore_case=T）））元素在
mylist
中是数字向量，而对于
rename_at
来说，我们需要一个tible对象。因此，首先我们需要将该数字向量转换为tible对象，然后我们可以一起使用
map
和
rename_at

library(tidyverse)

mylist[] <- lapply(mylist, function(x) as.tibble(t(x)))
map(mylist, ~ rename_at(., vars(contains('kmeans')), ~ 'variable'))


#$dataset1
# A tibble: 1 x 3
#      a variable     c
#  <dbl>    <dbl> <dbl>
#1     1        2     3

#$dataset2
# A tibble: 1 x 3
#      a variable     c
#  <dbl>    <dbl> <dbl>
#1     4        5     6

#$dataset2
# A tibble: 1 x 3
#      r variable     p
#  <dbl>    <dbl> <dbl>
#1     7        8     9

库（tidyverse）
mylist[]在函数中，重命名后应返回x以获得所需结果-

mylist<-list(dataset1=c(a=1,kmeans=2,c=3),
             dataset2=c(a=4,kmeans=5,c=6),
             dataset2=c(r=7,kmeans=8,p=9))

fn <- function(x) {names(x)[str_which(names(x),regex('KMEa',ignore_case=T))] <- 'variable'
                  return(x)}
lapply(mylist,fn) 

mylist
library(tidyverse)

mylist[] <- lapply(mylist, function(x) as.tibble(t(x)))
map(mylist, ~ rename_at(., vars(contains('kmeans')), ~ 'variable'))


#$dataset1
# A tibble: 1 x 3
#      a variable     c
#  <dbl>    <dbl> <dbl>
#1     1        2     3

#$dataset2
# A tibble: 1 x 3
#      a variable     c
#  <dbl>    <dbl> <dbl>
#1     4        5     6

#$dataset2
# A tibble: 1 x 3
#      r variable     p
#  <dbl>    <dbl> <dbl>
#1     7        8     9

mylist<-list(dataset1=c(a=1,kmeans=2,c=3),
             dataset2=c(a=4,kmeans=5,c=6),
             dataset2=c(r=7,kmeans=8,p=9))

fn <- function(x) {names(x)[str_which(names(x),regex('KMEa',ignore_case=T))] <- 'variable'
                  return(x)}
lapply(mylist,fn) 

mylist %>% lapply( function(x) {names(x)[str_which(names(x),regex('KMEa',ignore_case=T))] <- 'variable'
return(x)})