將一列中的多個文本值轉換為不同的列

我有一個來自調查平臺的大 df 檔案。列包含一個問題（作為標題）以及所有可能的答案。我試圖將每個答案列分成一組新的列（答案名稱作為新列的名稱）。

之后，我希望每個新列都指示該值是否出現在原始列（帶有“1”）中，以便更輕松地處理資料

     df<-data.frame("name"= c("John","mark","bell","elsa"),"what do you like to 
     eat"=c("apple","fries apple","peach","bread"))

原始檔案

我正在使用這個有效的代碼，但我確信必須有更有效/更簡單的方法來做到這一點，因為我有超過 50 個這樣的列。

    df<-df %>%
  separate(what.do.you.like.to.eat, c("apple","fries","peach","bread",NA ), remove = F)
df[,3:6]<-""
{
  df[,3] =  with(df, ifelse(grepl("apple", df$what.do.you.like.to.eat,ignore.case = T), 
                                             paste('1', df[,3]), 
                                             paste("", df[,3])))
  df[,4] =  with(df, ifelse(grepl("fries", df$what.do.you.like.to.eat,ignore.case = T), 
                            paste('1', df[,4]), 
                            paste("", df[,4])))
  df[,5] =  with(df, ifelse(grepl("peach", df$what.do.you.like.to.eat,ignore.case = T), 
                            paste('1', df[,5]), 
                            paste("", df[,5])))
  df[,6] =  with(df, ifelse(grepl("bread", df$what.do.you.like.to.eat,ignore.case = T), 
                            paste('1', df[,6]), 
                            paste("", df[,6])))
}

欲望輸出

uj5u.com熱心網友回復：

您可以使用purrr::map來應用您的答案向量，并針對每個答案檢查它們在字串中的存在。

library(tidyverse)

df <- data.frame(
  name = c("John", "mark", "bell", "elsa"),
  "what do you like to eat" = c("apple", "fries apple", "peach", "bread")
)

ans <- c("apple", "fries", "peach", "bread")

map_dfc(ans,~ transmute(df, !!sym(.x) := str_detect(what.do.you.like.to.eat, .x))) %>%
  bind_cols(df, .)
#>   name what.do.you.like.to.eat apple fries peach bread
#> 1 John                   apple  TRUE FALSE FALSE FALSE
#> 2 mark             fries apple  TRUE  TRUE FALSE FALSE
#> 3 bell                   peach FALSE FALSE  TRUE FALSE
#> 4 elsa                   bread FALSE FALSE FALSE  TRUE

uj5u.com熱心網友回復：

好的，我已經完成了，告訴我這是否適合您：

my_df <- data.frame("name" = c("John","mark","bell","elsa"),
                "what do you like to eat" = c("apple","fries apple","peach","bread"),
                stringsAsFactors = FALSE)
my_var <- unique(sort(str_split(string = my_df$what.do.you.like.to.eat, pattern = " ", simplify = TRUE)))
my_pos <- which(my_var == "")
if (length(my_pos)) {
  my_var <- my_var[-my_pos]
}
my_col <- c(colnames(my_df), my_var)
my_miss <- setdiff(my_col, colnames(my_df))
my_df[my_miss] <- NA
my_f <- function(x, y) {
  my_var <- grep(pattern = colnames(my_df)[x], x = my_df[, y])
  if (length(my_var)) {
    my_df[my_var, x] <<- 1
  }
}
lapply(3:ncol(my_df), function(x) my_f(x, 2))

你可以把這部分改成這樣：

my_df <- data.frame("name" = c("John","mark","bell","elsa"),
                "what do you like to eat" = c("i like apple","i love fries apple","i'm kind of peach","bread all the way"),
                stringsAsFactors = FALSE)
my_var <- unique(sort(str_split(string = 
my_df$what.do.you.like.to.eat, pattern = " ", simplify = TRUE)))
my_food <- c("apple", "fries", "bread", "peach")
my_var <- my_var[which(my_var %in% my_food)]
my_pos <- which(my_var == "")
if (length(my_pos)) {
  my_var <- my_var[-my_pos]
}

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