R - 提取匹配的字符串,拆分成多个列,这些列由字典向量匹配

时间:2022-03-02 21:43:46

I want to extract specific string of favor column in target data which is matched by a dictionary. Here is my data:

我想在目标数据中提取特定字符串的favyl列,该字符串与字典匹配。这是我的数据:

dictionary <- c("apple", "banana", "orange", "grape")

target <- data.frame("user" = c("A", "B", "C"),
                     "favor" = c("I like apple and banana", "grape and kiwi", "orange, banana and grape are the best"))
target
  user                                 favor
1    A               I like apple and banana
2    B                        grape and kiwi
3    C orange, banana and grape are the best

And below is my expected outcome result. I want to automatically create the column based on the most favors I matched in dictionary(in my case, 3), and extract the string I match in dictionary.

以下是我预期的结果结果。我想基于我在字典中匹配的最喜欢(在我的例子中,3)自动创建列,并提取我在字典中匹配的字符串。

result <- data.frame("user" = c("A", "B", "C"), 
                     "favor_1" = c("apple", "grape", "orange"), 
                     "favor_2" = c("banana", "", "banana"), 
                     "favor_3" = c("", "", "grape"))
result

  user favor_1 favor_2 favor_3
1    A   apple  banana        
2    B   grape                
3    C  orange  banana   grape

Any help will be thankful.

任何帮助都会感激不尽。

2 个解决方案

#1


1  

Your best bet is probably to apply str_extract_all to each row.

您最好的选择可能是将str_extract_all应用于每一行。

library(stringr)
result <- t(apply(target, 1,
                  function(x) str_extract_all(x[['favor']], dictionary, simplify = TRUE)))

     [,1]    [,2]     [,3]     [,4]   
[1,] "apple" "banana" ""       ""     
[2,] ""      ""       ""       "grape"
[3,] ""      "banana" "orange" "grape"

#2


3  

# Remove all words from `target$favor` that are not in the dictionary
result <- lapply(strsplit(target$favor, ',| '), function(x) { x[x %in% dictionary] })
result
# [[1]]
# [1] "apple"  "banana"
# 
# [[2]]
# [1] "grape" 
# 
# [[3]]
# [1] "orange" "banana" "grape" 

# Fill in NAs when the rows have different numbers of items
result <- lapply(result, `length<-`, max(lengths(result)))

# Rebuild the data.frame using the list of words in each row
cbind(target[ , 'user', drop = F], do.call(rbind, result))
#   user      1      2     3
# 1    A  apple banana  <NA>
# 2    B  grape   <NA>  <NA>
# 3    C orange banana grape

Note that I read in target with stringsAsFactors = FALSE so that strsplit can work.

请注意,我使用stringsAsFactors = FALSE读取目标,以便strsplit可以工作。

#1


1  

Your best bet is probably to apply str_extract_all to each row.

您最好的选择可能是将str_extract_all应用于每一行。

library(stringr)
result <- t(apply(target, 1,
                  function(x) str_extract_all(x[['favor']], dictionary, simplify = TRUE)))

     [,1]    [,2]     [,3]     [,4]   
[1,] "apple" "banana" ""       ""     
[2,] ""      ""       ""       "grape"
[3,] ""      "banana" "orange" "grape"

#2


3  

# Remove all words from `target$favor` that are not in the dictionary
result <- lapply(strsplit(target$favor, ',| '), function(x) { x[x %in% dictionary] })
result
# [[1]]
# [1] "apple"  "banana"
# 
# [[2]]
# [1] "grape" 
# 
# [[3]]
# [1] "orange" "banana" "grape" 

# Fill in NAs when the rows have different numbers of items
result <- lapply(result, `length<-`, max(lengths(result)))

# Rebuild the data.frame using the list of words in each row
cbind(target[ , 'user', drop = F], do.call(rbind, result))
#   user      1      2     3
# 1    A  apple banana  <NA>
# 2    B  grape   <NA>  <NA>
# 3    C orange banana grape

Note that I read in target with stringsAsFactors = FALSE so that strsplit can work.

请注意,我使用stringsAsFactors = FALSE读取目标,以便strsplit可以工作。