如何根据优先词列表（使用 for 循环和条件）查找和替换 df 中的值？

Question

我在数据框中有一列，每个单元格中有多个单词，用“;”分隔（第二列）。

my_dataframe <- data.frame( first_column = c("x", "y", "x", "x", "y"),
                            second_column = c("important; very important; not important",
                                              "not important; important; very important",
                                              "very important; important",
                                              "important; not important",
                                              "not important"))
> my_dataframe
  first_column                            second_column
1            x important; very important; not important
2            y not important; important; very important
3            x                very important; important
4            x                 important; not important
5            y                            not important

我想在每个单元格中保留一个词：最重要的一个。

所以我按优先顺序列出了单词：

reference_importance <- list("very important", "important", "not important")

我希望在第二列获得的内容：

 second_column
1 very important
2 very important
3 very important
4 important
5 not important

我试过了

for (i in 1:dim(my_dataframe)[1]) {
  for (j in 1:length(reference_importance)) {
    if (j %in% my_dataframe$second_column){
      my_dataframe$second_column[i] <- paste(j)
      break}
  }
}

然后我认为问题是它没有考虑用“;”分隔的不同单词。所以我尝试了这个：

for (i in 1:dim(my_dataframe)[1]) {
  value_as_list <- strsplit(my_dataframe$second_column[i], ";")
  print(value_as_list)
  for (j in reference_importance) {
    if (j %in% value_as_list){
      my_dataframe$second_column[i] == j
      break}
  }
}

但是这些并没有改变我专栏中的任何内容...

（我做这个例子是为了简化，但实际上我有一个巨大的 table 有更多的词和可能性。这就是为什么我尝试用循环来做，我不只是分配手动可能的答案。）

Answer 1

基本上使用strsplit和match。

my_dataframe <- transform(my_dataframe, z=strsplit(second_column, '; ') |>
                            lapply(match, reference_importance) |>
                            sapply(min) |>
                            {\(x) unlist(reference_importance)[x]}())
my_dataframe
#   first_column                            second_column              z
# 1            x important; very important; not important very important
# 2            y not important; important; very important very important
# 3            x                very important; important very important
# 4            x                 important; not important      important
# 5            y                            not important  not important

注意： R >= 4.1 使用。

如果你需要一个循环，你可以这样做

spl <- strsplit(my_dataframe$second_column, '; ')
my_dataframe$z <- NA_character_

for (i in seq_along(spl)) {
  my_dataframe$z[i] <- reference_importance[[min(match(spl[[i]], reference_importance))]]
}
my_dataframe
#   first_column                            second_column              z
# 1            x important; very important; not important very important
# 2            y not important; important; very important very important
# 3            x                very important; important very important
# 4            x                 important; not important      important
# 5            y                            not important  not important

当然我使用 z 作为演示目的，实际上您会使用 second_column 而不是 z。

Answer 2

如果你想使用循环，以下方法对我有用：

my_dataframe <- data.frame( first_column = c("x", "y", "x", "x", "y"),
                            second_column = c("important; very important; not important",
                                              "not important; important; very important",
                                              "very important; important",
                                              "important; not important",
                                              "not important"))

reference_importance <- list("very important", "important", "not important")


# add new column for priority word 
my_dataframe <- my_dataframe %>%
  mutate(Priority_importance = NA)

# use a loop to identify highest priority substring
for (i in 1:nrow(my_dataframe)) {
  value_as_list <- strsplit(my_dataframe$second_column[i], ";")
  
  for (j in  1:length(reference_importance)) {
    if (value_as_list == as.character((reference_importance[j]))) { 
      my_dataframe$Priority_importance[i] <- reference_importance[j] # paste importance level 
      break # move to next iteration 
    }
  }
}

my_dataframe

  first_column                            second_column Priority_importance
1            x important; very important; not important      very important
2            y not important; important; very important      very important
3            x                very important; important      very important
4            x                 important; not important           important
5            y                            not important       not important

Answer 3

dplyr 和 tidyr 的一个选项：

my_dataframe %>%
    rowid_to_column() %>%
    separate_rows(second_column, sep = "; ") %>%
    group_by(rowid) %>%
    slice_min(match(second_column, reference_importance))

 rowid first_column second_column 
  <int> <chr>        <chr>         
1     1 x            very important
2     2 y            very important
3     3 x            very important
4     4 x            important     
5     5 y            not important

我使用 reference_importance 作为字符向量而不是列表：

reference_importance <- c("very important", "important", "not important")

Answer 4

另一种可能的解决方案，基于tidyverse：

library(tidyverse)

my_dataframe %>% 
  mutate(id = row_number()) %>% 
  separate_rows(second_column, sep = "\s*;\s*") %>% 
  group_by(id) %>% 
  slice(match(reference_importance, second_column) %>% na.omit() %>% .[1]) %>% 
  ungroup %>% 
  select(-id)

#> # A tibble: 5 × 2
#>   first_column second_column 
#>   <chr>        <chr>         
#> 1 x            very important
#> 2 y            very important
#> 3 x            very important
#> 4 x            important     
#> 5 y            not important

如何根据优先词列表（使用 for 循环和条件）查找和替换 df 中的值？

How to find and replace values in a df according to a list of priority words (with for loop and condition)?

loops

r

list

conditional-statements

dataframe