Question

我想对照4个列表（a，b，c，d）（在数据框的一列中）检查一个单词：

if df$word is in a then df$code <- 1
if df$word is in b then df$code <- 2
if df$word is in c then df$code <- 3
if df$word is in d then df$code <- 4

if df$word is in a & b then df$code <- 1 2
if df$word is in a & c then df$code <- 1 3
if df$word is in a & d then df$code <- 1 4
if df$word is in b & c then df$code <- 2 3
if df$word is in b & d then df$code <- 2 4
if df$word is in c & d then df$code <- 3 4

等

最有效的方法是什么？

示例

df <- data.frame(word = c("book", "worm", "digital", "context"))

a <- c("book", "context")
b <- c("book", "worm", "context")
c <- c("digital", "worm", "context")
d <- c("context")

预期输出：

book    1 2
worm    2 3
digital 3
context 1 2 3 4

Answer 1

我们可以使用双sapply循环，其中对于数据帧中的每个元素，我们都检出which列表元素，并获得相应的列表编号。

lst <- list(a, b, c, d)
df$output <- sapply(df$V1, function(x) paste0(which(sapply(lst, 
                           function(y) any(grepl(x,y)))), collapse = ","))

df
#       V1  output
#1    book     1,2
#2    worm     2,3
#3 digital       3
#4 context 1,2,3,4

数据

df <- read.table(text = "book
      worm
      digital
      context")

Answer 2

尝试一下：

df <- data.frame(x =c("book", "worm","digital", "context"))

a <- c("book", "context")
b<- c("book", "worm", "context")
c <- c("digital", "worm", "context")
d <- c("context")


anno <- function(x){
  rslt = ""
  if (x %in% a) rslt =paste0(rslt," 1")
  if (x %in% b) rslt =paste0(rslt," 2")
  if (x %in% c) rslt =paste0(rslt," 3")
  if (x %in% d) rslt =paste0(rslt," 4")
  return(stringr::str_trim(rslt))
}

df$code <- sapply(df$x, anno)
df
#>         x    code
#> 1    book     1 2
#> 2    worm     2 3
#> 3 digital       3
#> 4 context 1 2 3 4

由reprex package（v0.2.0.9000）创建于2018-08-17。

Answer 3

这也可以分两个步骤完成：

合并四个列表，并将其整形为长格式
与df

一起聚集

使用data.table：

library(data.table)
long <-setDT(melt(list(a, b, c, d), value.name = "word"))
long[setDT(df), on = "word", by = .EACHI, .(code = toString(L1))][]

      word       code
1:    book       1, 2
2:    worm       2, 3
3: digital          3
4: context 1, 2, 3, 4

根据4个列表检查data.frame列（字符）

3 个答案: