我有以下数据框,如下所示(列表为3列)。
A tibble: 14 x 4
clinic_name drop_in_hours appointment_hours services
<chr> <list> <list> <list>
1 Birth Control and Sexual Health Centre <list [1]> <list [1]> <list [1]>
2 Black Creek Community Health Centre (Sheridan Mall Site) <list [1]> <list [1]> <list [1]>
3 Black Creek Community Health Centre (Yorkgate mall Site) <list [1]> <list [1]> <list [1]>
4 Crossways Clinic <list [1]> <list [1]> <list [1]>
5 Hassle Free Clinic <list [1]> <list [1]> <list [1]>
6 Immigrant Women's Health Center <list [1]> <list [1]> <list [1]>
7 Rexdale Community Health Center <list [1]> <list [1]> <list [1]>
8 Rexdale Youth Resource Center <list [1]> <list [1]> <list [1]>
9 Scarborough Sexual Health Clinic <list [1]> <list [1]> <list [1]>
10 Special Treatment Clinic <list [1]> <list [1]> <list [1]>
11 Taibu Community Health Center <list [1]> <list [1]> <list [1]>
12 The Gate <list [1]> <list [1]> <list [1]>
13 The Jane Street Clinic <list [1]> <list [1]> <list [1]>
14 The Talk Shop <list [1]> <list [1]> <list [1]>
我想将其输出为csv文件。我注意到数据框的列不应该是R中的列表。所以我做了一些google并找到了这个save data.frames with list-column所以我试了一下:
library(tidyverse)
df %>%
mutate(drop_in_hours = map_chr(drop_in_hours, ~ capture.output(dput(.))),
appointment_hours = map_chr(appointment_hours, ~ capture.output(dput(.))),
services = map_chr(services, ~ capture.output(dput(.))) ) %>%
write_csv("health.csv")
但是我收到了一个错误,我在这里错过了一些东西吗?
Error in mutate_impl(.data, dots) :
Evaluation error: Result 4 is not a length 1 atomic vector
答案 0 :(得分:4)
创建包含列表列的tibble:
cancelAndHoldAtTime
编写一个将任何列表列转换为字符类型的通用函数:
library(tibble)
clinic_name <- c('bobo center', 'yoyo plaza', 'lolo market')
drop_in_hours <- list(c("Monday: 2 pm - 5 pm", "Tuesday: 4 pm - 7 pm"))
appointment_hours <- list(c("Monday: 1 pm - 2 pm", "Tuesday: 2 pm - 3 pm"))
services <- list(c("skin graft", "chicken heart replacement"))
tibb <- data_frame(clinic_name, drop_in_hours, appointment_hours, services)
print(tibb)
将功能应用于与列表列同步:
set_lists_to_chars <- function(x) {
if(class(x) == 'list') {
y <- paste(unlist(x[1]), sep='', collapse=', ')
} else {
y <- x
}
return(y)
}
将新格式化的数据框写为csv文件:
new_frame <- data.frame(lapply(tibb, set_lists_to_chars), stringsAsFactors = F)
new_frame
这是一个csv文件,列表列扩展为常规字符串。
这是一个包罗万象的功能。只需传递你的tibble和文件名:
write.csv(new_frame, file='Desktop/clinics.csv')
<强>用法:强>
tibble_with_lists_to_csv <- function(tibble_object, file_path_name) {
set_lists_to_chars <- function(x) {
if(class(x) == 'list') { y <- paste(unlist(x[1]), sep='', collapse=', ') } else { y <- x }
return(y) }
new_frame <- data.frame(lapply(tibb, set_lists_to_chars), stringsAsFactors = F)
write.csv(new_frame, file=file_path_name)
}
答案 1 :(得分:1)
您是否有任何特定原因要将列保存为列表?或者,您可以使用unnest
并将其保存在csv中。以下示例
library(tidyverse)
df_list<-data_frame(abc = letters[1:3], lst = list(1:3, 1:3, 1:3))
df_list %>% unnest() %>% write.csv("list.csv")
此外,当您阅读文件时,您可以nest
将其恢复
df <- read.csv("list.csv")[ ,2:3]
df %>% nest(lst)
答案 2 :(得分:1)
exploratory::list_to_text()
会将list
列转换为character
列。默认值为sep = ", "
,如果写入.csv,我建议将其更改为其他内容。
devtools::install_github("exploratory-io/exploratory_func")
list_to_text <- function(column, sep = ", "){
loadNamespace("stringr")
ret <- sapply(column, function(x) {
ret <- stringr::str_c(x, collapse = sep)
if(identical(ret, character(0))){
# if it's character(0)
NA
} else {
ret
}
})
as.character(ret)
}
https://github.com/exploratory-io/exploratory_func/blob/master/LICENSE.md
答案 3 :(得分:1)
这是另一个可能更简单的选择。
根据数据,逗号分隔的值可能会变得复杂,因此我使用了条形|
来分隔列表列中的值:
library(tidyverse)
starwars %>%
rowwise() %>%
mutate_if(is.list, ~paste(unlist(.), collapse = '|')) %>%
write.csv('df_starwars.csv', row.names = FALSE)
starwars
是dplyr
示例数据帧之一。