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0

I am looking for a function that allows me to add a new column to add the values called ID to a string, that is:

I have a list of words with your ID:

car = 9112
red = 9512
employee = 6117
sky = 2324

words<- c("car", "sky", "red", "employee", "domestic")
match<- c("car", "red", "domestic", "employee", "sky")

the comparison is made by reading in an excel file, if it finds the value equal to my vector words, it replaces the word with its ID, but leaves the original word

 x10<- c(words)# string

words.corpus <- c(L4$`match`) # pattern
idwords.corpus <- c(L4$`ID`) # replace
words.corpus <- paste0("\A",idwords.corpus, "\z|\A", words.corpus,"\z")

vect.corpus <- idwords.corpus
names(vect.corpus) <- words.corpus

data15 <- str_replace_all(x10, vect.corpus)

result:

data15:

" 9112", "2324", "9512", "6117", "employee"

What I'm looking for is to add a new column with the ID, instead of replacing the word with the ID

words ID
car 9112
red 9512
employee 6117
sky 2324
domestic domestic

r string tm

share|improve this question

asked Mar 22 at 16:36

Max TCMax TC

325

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  difficult to help you fully without more info. But you could use a data.frame or data.table. data15 <- cbind(x10,ID=str_replace_all(x10, vect.corpus))
  
  – DJJ
  Mar 22 at 16:55
  

  
  
  
  

add a comment | 

0

I am looking for a function that allows me to add a new column to add the values called ID to a string, that is:

I have a list of words with your ID:

car = 9112
red = 9512
employee = 6117
sky = 2324

words<- c("car", "sky", "red", "employee", "domestic")
match<- c("car", "red", "domestic", "employee", "sky")

the comparison is made by reading in an excel file, if it finds the value equal to my vector words, it replaces the word with its ID, but leaves the original word

 x10<- c(words)# string

words.corpus <- c(L4$`match`) # pattern
idwords.corpus <- c(L4$`ID`) # replace
words.corpus <- paste0("\A",idwords.corpus, "\z|\A", words.corpus,"\z")

vect.corpus <- idwords.corpus
names(vect.corpus) <- words.corpus

data15 <- str_replace_all(x10, vect.corpus)

result:

data15:

" 9112", "2324", "9512", "6117", "employee"

What I'm looking for is to add a new column with the ID, instead of replacing the word with the ID

words ID
car 9112
red 9512
employee 6117
sky 2324
domestic domestic

r string tm

share|improve this question

asked Mar 22 at 16:36

Max TCMax TC

325

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  difficult to help you fully without more info. But you could use a data.frame or data.table. data15 <- cbind(x10,ID=str_replace_all(x10, vect.corpus))
  
  – DJJ
  Mar 22 at 16:55
  

  
  
  
  

add a comment | 

I am looking for a function that allows me to add a new column to add the values called ID to a string, that is:

I have a list of words with your ID:

car = 9112
red = 9512
employee = 6117
sky = 2324

words<- c("car", "sky", "red", "employee", "domestic")
match<- c("car", "red", "domestic", "employee", "sky")

the comparison is made by reading in an excel file, if it finds the value equal to my vector words, it replaces the word with its ID, but leaves the original word

 x10<- c(words)# string

words.corpus <- c(L4$`match`) # pattern
idwords.corpus <- c(L4$`ID`) # replace
words.corpus <- paste0("\A",idwords.corpus, "\z|\A", words.corpus,"\z")

vect.corpus <- idwords.corpus
names(vect.corpus) <- words.corpus

data15 <- str_replace_all(x10, vect.corpus)

result:

data15:

" 9112", "2324", "9512", "6117", "employee"

What I'm looking for is to add a new column with the ID, instead of replacing the word with the ID

words ID
car 9112
red 9512
employee 6117
sky 2324
domestic domestic

r string tm

share|improve this question

asked Mar 22 at 16:36

Max TCMax TC

325

car = 9112
red = 9512
employee = 6117
sky = 2324

words<- c("car", "sky", "red", "employee", "domestic")
match<- c("car", "red", "domestic", "employee", "sky")

 x10<- c(words)# string

words.corpus <- c(L4$`match`) # pattern
idwords.corpus <- c(L4$`ID`) # replace
words.corpus <- paste0("\A",idwords.corpus, "\z|\A", words.corpus,"\z")

vect.corpus <- idwords.corpus
names(vect.corpus) <- words.corpus

data15 <- str_replace_all(x10, vect.corpus)

" 9112", "2324", "9512", "6117", "employee"

words ID
car 9112
red 9512
employee 6117
sky 2324
domestic domestic

share|improve this question

asked Mar 22 at 16:36

Max TCMax TC

325

share|improve this question

asked Mar 22 at 16:36

Max TCMax TC

325

asked Mar 22 at 16:36

Max TCMax TC

325

asked Mar 22 at 16:36

Max TCMax TC

325

1 Answer
1

active

oldest

votes

0

I'd use data.table for fast lookup based on the fixed words value. While it's not 100% clear what you are asking for, it sounds like you want to replace words with an index value if there is a match, or leave the word as a word if not. This code will do that:

library("data.table")

# associate your ids with fixed word matches in a named numeric vector
ids <- data.table(
 word = c("car", "red", "employee", "sky"),
 ID = c(9112, 9512, 6117, 2324)
)
setkey(ids, word)

# this is what you would read in
data <- data.table(
 word = c("car", "sky", "red", "employee", "domestic", "sky")
)
setkey(data, word)

data <- ids[data]
# replace NAs from no match with word
data[, ID := ifelse(is.na(ID), word, ID)]

data
## word ID
## 1: car 9112
## 2: domestic domestic
## 3: employee 6117
## 4: red 9512
## 5: sky 2324
## 6: sky 2324

Here the "domestic" is not matched so it remains as the word in the ID column. I also repeated "sky" to show how this will work for every instance of a word.

If you want to preserve the original sort order, you could create an index variable before the merge, and then reorder the output by that index variable.

share|improve this answer

answered Mar 26 at 5:17

Ken BenoitKen Benoit

7,7081635

add a comment | 

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0

I'd use data.table for fast lookup based on the fixed words value. While it's not 100% clear what you are asking for, it sounds like you want to replace words with an index value if there is a match, or leave the word as a word if not. This code will do that:

library("data.table")

# associate your ids with fixed word matches in a named numeric vector
ids <- data.table(
 word = c("car", "red", "employee", "sky"),
 ID = c(9112, 9512, 6117, 2324)
)
setkey(ids, word)

# this is what you would read in
data <- data.table(
 word = c("car", "sky", "red", "employee", "domestic", "sky")
)
setkey(data, word)

data <- ids[data]
# replace NAs from no match with word
data[, ID := ifelse(is.na(ID), word, ID)]

data
## word ID
## 1: car 9112
## 2: domestic domestic
## 3: employee 6117
## 4: red 9512
## 5: sky 2324
## 6: sky 2324

Here the "domestic" is not matched so it remains as the word in the ID column. I also repeated "sky" to show how this will work for every instance of a word.

If you want to preserve the original sort order, you could create an index variable before the merge, and then reorder the output by that index variable.

share|improve this answer

answered Mar 26 at 5:17

Ken BenoitKen Benoit

7,7081635

add a comment | 

0

I'd use data.table for fast lookup based on the fixed words value. While it's not 100% clear what you are asking for, it sounds like you want to replace words with an index value if there is a match, or leave the word as a word if not. This code will do that:

library("data.table")

# associate your ids with fixed word matches in a named numeric vector
ids <- data.table(
 word = c("car", "red", "employee", "sky"),
 ID = c(9112, 9512, 6117, 2324)
)
setkey(ids, word)

# this is what you would read in
data <- data.table(
 word = c("car", "sky", "red", "employee", "domestic", "sky")
)
setkey(data, word)

data <- ids[data]
# replace NAs from no match with word
data[, ID := ifelse(is.na(ID), word, ID)]

data
## word ID
## 1: car 9112
## 2: domestic domestic
## 3: employee 6117
## 4: red 9512
## 5: sky 2324
## 6: sky 2324

Here the "domestic" is not matched so it remains as the word in the ID column. I also repeated "sky" to show how this will work for every instance of a word.

If you want to preserve the original sort order, you could create an index variable before the merge, and then reorder the output by that index variable.

share|improve this answer

answered Mar 26 at 5:17

Ken BenoitKen Benoit

7,7081635

add a comment | 

I'd use data.table for fast lookup based on the fixed words value. While it's not 100% clear what you are asking for, it sounds like you want to replace words with an index value if there is a match, or leave the word as a word if not. This code will do that:

library("data.table")

# associate your ids with fixed word matches in a named numeric vector
ids <- data.table(
 word = c("car", "red", "employee", "sky"),
 ID = c(9112, 9512, 6117, 2324)
)
setkey(ids, word)

# this is what you would read in
data <- data.table(
 word = c("car", "sky", "red", "employee", "domestic", "sky")
)
setkey(data, word)

data <- ids[data]
# replace NAs from no match with word
data[, ID := ifelse(is.na(ID), word, ID)]

data
## word ID
## 1: car 9112
## 2: domestic domestic
## 3: employee 6117
## 4: red 9512
## 5: sky 2324
## 6: sky 2324

Here the "domestic" is not matched so it remains as the word in the ID column. I also repeated "sky" to show how this will work for every instance of a word.

If you want to preserve the original sort order, you could create an index variable before the merge, and then reorder the output by that index variable.

share|improve this answer

answered Mar 26 at 5:17

Ken BenoitKen Benoit

7,7081635

library("data.table")

# associate your ids with fixed word matches in a named numeric vector
ids <- data.table(
 word = c("car", "red", "employee", "sky"),
 ID = c(9112, 9512, 6117, 2324)
)
setkey(ids, word)

# this is what you would read in
data <- data.table(
 word = c("car", "sky", "red", "employee", "domestic", "sky")
)
setkey(data, word)

data <- ids[data]
# replace NAs from no match with word
data[, ID := ifelse(is.na(ID), word, ID)]

data
## word ID
## 1: car 9112
## 2: domestic domestic
## 3: employee 6117
## 4: red 9512
## 5: sky 2324
## 6: sky 2324

share|improve this answer

answered Mar 26 at 5:17

Ken BenoitKen Benoit

7,7081635

answered Mar 26 at 5:17

Ken BenoitKen Benoit

7,7081635

answered Mar 26 at 5:17

Ken BenoitKen Benoit

7,7081635