Question

如何将以下因子转换为数据框？

dput(d)
structure(1L, .Label = "c(\"CONTEXTLESS\", \"CONTEXTLESS\", \"CONTEXTLESS\", \"CONTEXTLESS\", \"CONTEXTLESS\", \"CONTEXTLESS\", \"CONTEXTLESS\", \"CONTEXTLESS\"), c(\"app1\", \"client\", \"org\", \"app1\", \"DATA_CENTER\", \"PURPOSE\", \"REGION\", \"Test\"), c(NA, \"NONE\", \"Host:Environment:test123\", \"111\", \"222\", \"GENERAL\", \"444\", \"555\")", class = "factor")


> d
[1] c("CONTEXTLESS", "CONTEXTLESS", "CONTEXTLESS", "CONTEXTLESS", "CONTEXTLESS", "CONTEXTLESS", "CONTEXTLESS", "CONTEXTLESS"), c("app1", "client", "org", "app1", "DATA_CENTER", "PURPOSE", "REGION", "Test"), c(NA, "NONE", "Host:Environment:test123", "111", "222", "GENERAL", "444", "555")
Levels: c("CONTEXTLESS", "CONTEXTLESS", "CONTEXTLESS", "CONTEXTLESS", "CONTEXTLESS", "CONTEXTLESS", "CONTEXTLESS", "CONTEXTLESS"), c("app1", "client", "org", "app1", "DATA_CENTER", "PURPOSE", "REGION", "Test"), c(NA, "NONE", "Host:Environment:test123", "111", "222", "GENERAL", "444", "555")

Answer 1

也许尝试一下，让我知道是否适合您。您可以将字符连贯，然后为字符串放置一个公共分隔符。这里的代码：

#Code
d1 <- as.character(d)
d1 <- gsub(', c','*c',d1)
d1 <- strsplit(d1,split='\\*')
df <- as.data.frame(do.call(cbind,d1))

df的输出如下所示：

在获得一些新的详细信息之后，请在清理它们后使用tidyverse来分隔行：

library(tidyverse)
#Code
d1 <- as.character(d)
d1 <- gsub(', c','*c',d1)
d1 <- strsplit(d1,split='\\*')
df <- as.data.frame(do.call(cbind,d1),stringsAsFactors = F)
#Clean
df$V1 <- gsub('\"','',df$V1)
df$V1 <- gsub("c(","",df$V1,fixed=T)
df$V1 <- gsub(")","",df$V1)
#Format
newdf <- df %>% mutate(id=row_number()) %>%
  separate_rows(V1,sep = ',') %>%
  mutate(V1=trimws(V1)) %>%
  group_by(id) %>% mutate(Var=paste0('Var.',row_number())) %>%
  pivot_wider(names_from = Var,values_from=V1)

输出：

# A tibble: 3 x 9
# Groups:   id [3]
     id Var.1      Var.2      Var.3               Var.4     Var.5     Var.6     Var.7     Var.8    
  <int> <chr>      <chr>      <chr>               <chr>     <chr>     <chr>     <chr>     <chr>    
1     1 CONTEXTLE~ CONTEXTLE~ CONTEXTLESS         CONTEXTL~ CONTEXTL~ CONTEXTL~ CONTEXTL~ CONTEXTL~
2     2 app1       client     org                 app1      DATA_CEN~ PURPOSE   REGION    Test     
3     3 NA         NONE       Host:Environment:t~ 111       222       GENERAL   444       555

Answer 2

如果从Duck的前三行代码开始，则可以从以下位置解析（显然）deparse()-d的R代码：

#Code
d1 <- as.character(d)
d1 <- gsub(', c','*c',d1)
d1 <- strsplit(d1,split='\\*')
d1[[1]][2]
#[1] "c(\"app1\", \"client\", \"org\", \"app1\", \"DATA_CENTER\", \"PURPOSE\", \"REGION\", \"Test\")"

如果您使用lapply来串行应用eval(parse(text=.)函数，那么您将获得一个对象，该对象可以被强制转换为长格式，并在注释中提示您想要：

setNames(lapply(d1[[1]], function(x) eval(parse(text=x))),paste0("V", 1:3))

$V1
[1] "CONTEXTLESS" "CONTEXTLESS" "CONTEXTLESS" "CONTEXTLESS" "CONTEXTLESS" "CONTEXTLESS"
[7] "CONTEXTLESS" "CONTEXTLESS"

$V2
[1] "app1"        "client"      "org"         "app1"        "DATA_CENTER" "PURPOSE"    
[7] "REGION"      "Test"       

$V3
[1] NA                         "NONE"                     "Host:Environment:test123"
[4] "111"                      "222"                      "GENERAL"                 
[7] "444"                      "555"

因此：

> d2 <- setNames(lapply(d1[[1]], function(x) eval(parse(text=x))),paste0("V", 1:3))
> data.frame( d2[2:3])
           V2                       V3
1        app1                     <NA>
2      client                     NONE
3         org Host:Environment:test123
4        app1                      111
5 DATA_CENTER                      222
6     PURPOSE                  GENERAL
7      REGION                      444
8        Test                      555

Answer 3

parse中的文本仅为eval和data.frame(...)。无需重新发明轮子。

df <- eval(parse(text = paste0("data.frame(", as.character(d), ")")))
names(df) <- paste0("v", seq_along(df))

输出

           v1          v2                       v3
1 CONTEXTLESS        app1                     <NA>
2 CONTEXTLESS      client                     NONE
3 CONTEXTLESS         org Host:Environment:test123
4 CONTEXTLESS        app1                      111
5 CONTEXTLESS DATA_CENTER                      222
6 CONTEXTLESS     PURPOSE                  GENERAL
7 CONTEXTLESS      REGION                      444
8 CONTEXTLESS        Test                      555

Answer 4

我们可以使用str_match_all提取c(....)之间的所有内容，在逗号上分割字符串并绑定数据集。

do.call(rbind, lapply(stringr::str_match_all(d, 'c\\((.*?)\\)'), function(x) {
  do.call(rbind, strsplit(gsub('"', '',x[, 2]), ',\\s*'))
}))

如何将因子转换为数据框

4 个答案: