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Add number in the string after each letter

What is the difference between String and string in C#?How to validate an email address using a regular expression?How do I read / convert an InputStream into a String in Java?Case insensitive 'Contains(string)'Convert bytes to a string?How do I make the first letter of a string uppercase in JavaScript?How to replace all occurrences of a string in JavaScriptHow to check whether a string contains a substring in JavaScript?Does Python have a string 'contains' substring method?Why is char[] preferred over String for passwords?

7

1

I have several strings with a fixed format.

The format is one letter followed by a number, e.g., A3B1C7D1.

However, if the number behind a letter is 1, the string is written as A3BC7D.

What I want to do is to insert number 1, and convert the string from A3BC7D to A3B1C7D1.

My example data is

strings <- c("A", "A3BC3", "A2B1C")

What I want to get is:

strings_new <- c("A1", "A3B1C3", "A2B1C1")

Thanks a lot!

r regex string

share|improve this question

edited 2 hours ago

markus

13.1k1234

asked 2 hours ago

DongDong

1466

add a comment | 

7

1

I have several strings with a fixed format.

The format is one letter followed by a number, e.g., A3B1C7D1.

However, if the number behind a letter is 1, the string is written as A3BC7D.

What I want to do is to insert number 1, and convert the string from A3BC7D to A3B1C7D1.

My example data is

strings <- c("A", "A3BC3", "A2B1C")

What I want to get is:

strings_new <- c("A1", "A3B1C3", "A2B1C1")

Thanks a lot!

r regex string

share|improve this question

edited 2 hours ago

markus

13.1k1234

asked 2 hours ago

DongDong

1466

add a comment | 

I have several strings with a fixed format.

The format is one letter followed by a number, e.g., A3B1C7D1.

However, if the number behind a letter is 1, the string is written as A3BC7D.

What I want to do is to insert number 1, and convert the string from A3BC7D to A3B1C7D1.

My example data is

strings <- c("A", "A3BC3", "A2B1C")

What I want to get is:

strings_new <- c("A1", "A3B1C3", "A2B1C1")

Thanks a lot!

r regex string

share|improve this question

edited 2 hours ago

markus

13.1k1234

asked 2 hours ago

DongDong

1466

strings <- c("A", "A3BC3", "A2B1C")

strings_new <- c("A1", "A3B1C3", "A2B1C1")

share|improve this question

edited 2 hours ago

markus

13.1k1234

asked 2 hours ago

DongDong

1466

share|improve this question

edited 2 hours ago

markus

13.1k1234

asked 2 hours ago

DongDong

1466

edited 2 hours ago

markus

13.1k1234

edited 2 hours ago

markus

13.1k1234

asked 2 hours ago

DongDong

1466

asked 2 hours ago

DongDong

1466

4 Answers
4

active

oldest

votes

3

Another option:

gsub("([A-Za-z])(?=[A-Za-z])|([A-Za-z])$", "\1\21", strings, perl = T)

Output:

[1] "A1"     "A3B1C3" "A2B1C1"

Or if you only have capitals, just:

gsub("([A-Z])(?=[A-Z])|([A-Z])$", "\1\21", strings, perl = T)

Basically this finds letters that are either followed by another letter or are at the end of string, and replaces them with themselves while at the same time adds the desired number, 1 in this case.

share|improve this answer

edited 2 hours ago

answered 2 hours ago

arg0nautarg0naut

4,1191315

add a comment | 

7

IIUC you could do:

gsub("(\D)(\D|$)", "\11\2", strings)

#[1] "A1"     "A3B1C3" "A2B1C1"

---

We create two capture groups, the first is a signle non-digit (\D), the second is either a single non-digit or the end of the string (\D|$). Whenever there's a match in a string, this means two letters (non-digits) follow each other without a number in between. Hence, we use the replacement \11\2 to replace it with group1, then a 1, then group2 (which can also be the end of the string).

share|improve this answer

edited 2 hours ago

answered 2 hours ago

docendo discimusdocendo discimus

51.9k1179117

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  If I run gsub("(\D)(\D|$)", "\11\2", "AB"), I got A1B
  
  – Dong
  2 mins ago
  

  
  
  
  

add a comment | 

1

Find all (uppercase) letters ([A-Z]) that is not followed by a number and replace it with that string + 1:

gsub("([A-Z])(?![0-9])", "\11", strings, perl = TRUE)

# [1] "A1"     "A3B1C3" "A2B1C1"

share|improve this answer

edited 4 mins ago

answered 29 mins ago

sindri_baldursindri_baldur

7,7941032

add a comment | 

0

strings[!grepl("[0-9]$",strings)]=paste0(strings[!grepl("[0-9]$",strings)],"1")

[1] "A1"     "A3BC3"  "A2B1C1"

We first grep all positions that do not end with a number, and paste a 1 to them.

share|improve this answer

answered 2 hours ago

boskiboski

490210

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  
  Almost. Check the second element it should be "A3B1C3"
  
  – markus
  2 hours ago
  
  
  

  
  
  
  

add a comment | 

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3

Another option:

gsub("([A-Za-z])(?=[A-Za-z])|([A-Za-z])$", "\1\21", strings, perl = T)

Output:

[1] "A1"     "A3B1C3" "A2B1C1"

Or if you only have capitals, just:

gsub("([A-Z])(?=[A-Z])|([A-Z])$", "\1\21", strings, perl = T)

Basically this finds letters that are either followed by another letter or are at the end of string, and replaces them with themselves while at the same time adds the desired number, 1 in this case.

share|improve this answer

edited 2 hours ago

answered 2 hours ago

arg0nautarg0naut

4,1191315

add a comment | 

3

Another option:

gsub("([A-Za-z])(?=[A-Za-z])|([A-Za-z])$", "\1\21", strings, perl = T)

Output:

[1] "A1"     "A3B1C3" "A2B1C1"

Or if you only have capitals, just:

gsub("([A-Z])(?=[A-Z])|([A-Z])$", "\1\21", strings, perl = T)

Basically this finds letters that are either followed by another letter or are at the end of string, and replaces them with themselves while at the same time adds the desired number, 1 in this case.

share|improve this answer

edited 2 hours ago

answered 2 hours ago

arg0nautarg0naut

4,1191315

add a comment | 

Another option:

gsub("([A-Za-z])(?=[A-Za-z])|([A-Za-z])$", "\1\21", strings, perl = T)

Output:

[1] "A1"     "A3B1C3" "A2B1C1"

Or if you only have capitals, just:

gsub("([A-Z])(?=[A-Z])|([A-Z])$", "\1\21", strings, perl = T)

Basically this finds letters that are either followed by another letter or are at the end of string, and replaces them with themselves while at the same time adds the desired number, 1 in this case.

share|improve this answer

edited 2 hours ago

answered 2 hours ago

arg0nautarg0naut

4,1191315

gsub("([A-Za-z])(?=[A-Za-z])|([A-Za-z])$", "\1\21", strings, perl = T)

[1] "A1"     "A3B1C3" "A2B1C1"

gsub("([A-Z])(?=[A-Z])|([A-Z])$", "\1\21", strings, perl = T)

share|improve this answer

edited 2 hours ago

answered 2 hours ago

arg0nautarg0naut

4,1191315

answered 2 hours ago

arg0nautarg0naut

4,1191315

answered 2 hours ago

arg0nautarg0naut

4,1191315

7

IIUC you could do:

gsub("(\D)(\D|$)", "\11\2", strings)

#[1] "A1"     "A3B1C3" "A2B1C1"

---

We create two capture groups, the first is a signle non-digit (\D), the second is either a single non-digit or the end of the string (\D|$). Whenever there's a match in a string, this means two letters (non-digits) follow each other without a number in between. Hence, we use the replacement \11\2 to replace it with group1, then a 1, then group2 (which can also be the end of the string).

share|improve this answer

edited 2 hours ago

answered 2 hours ago

docendo discimusdocendo discimus

51.9k1179117

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  If I run gsub("(\D)(\D|$)", "\11\2", "AB"), I got A1B
  
  – Dong
  2 mins ago
  

  
  
  
  

add a comment | 

7

IIUC you could do:

gsub("(\D)(\D|$)", "\11\2", strings)

#[1] "A1"     "A3B1C3" "A2B1C1"

---

We create two capture groups, the first is a signle non-digit (\D), the second is either a single non-digit or the end of the string (\D|$). Whenever there's a match in a string, this means two letters (non-digits) follow each other without a number in between. Hence, we use the replacement \11\2 to replace it with group1, then a 1, then group2 (which can also be the end of the string).

share|improve this answer

edited 2 hours ago

answered 2 hours ago

docendo discimusdocendo discimus

51.9k1179117

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  If I run gsub("(\D)(\D|$)", "\11\2", "AB"), I got A1B
  
  – Dong
  2 mins ago
  

  
  
  
  

add a comment | 

IIUC you could do:

gsub("(\D)(\D|$)", "\11\2", strings)

#[1] "A1"     "A3B1C3" "A2B1C1"

---

We create two capture groups, the first is a signle non-digit (\D), the second is either a single non-digit or the end of the string (\D|$). Whenever there's a match in a string, this means two letters (non-digits) follow each other without a number in between. Hence, we use the replacement \11\2 to replace it with group1, then a 1, then group2 (which can also be the end of the string).

share|improve this answer

edited 2 hours ago

answered 2 hours ago

docendo discimusdocendo discimus

51.9k1179117

gsub("(\D)(\D|$)", "\11\2", strings)

#[1] "A1"     "A3B1C3" "A2B1C1"

share|improve this answer

edited 2 hours ago

answered 2 hours ago

docendo discimusdocendo discimus

51.9k1179117

answered 2 hours ago

docendo discimusdocendo discimus

51.9k1179117

answered 2 hours ago

docendo discimusdocendo discimus

51.9k1179117

1

Find all (uppercase) letters ([A-Z]) that is not followed by a number and replace it with that string + 1:

gsub("([A-Z])(?![0-9])", "\11", strings, perl = TRUE)

# [1] "A1"     "A3B1C3" "A2B1C1"

share|improve this answer

edited 4 mins ago

answered 29 mins ago

sindri_baldursindri_baldur

7,7941032

add a comment | 

1

Find all (uppercase) letters ([A-Z]) that is not followed by a number and replace it with that string + 1:

gsub("([A-Z])(?![0-9])", "\11", strings, perl = TRUE)

# [1] "A1"     "A3B1C3" "A2B1C1"

share|improve this answer

edited 4 mins ago

answered 29 mins ago

sindri_baldursindri_baldur

7,7941032

add a comment | 

Find all (uppercase) letters ([A-Z]) that is not followed by a number and replace it with that string + 1:

gsub("([A-Z])(?![0-9])", "\11", strings, perl = TRUE)

# [1] "A1"     "A3B1C3" "A2B1C1"

share|improve this answer

edited 4 mins ago

answered 29 mins ago

sindri_baldursindri_baldur

7,7941032

gsub("([A-Z])(?![0-9])", "\11", strings, perl = TRUE)

# [1] "A1"     "A3B1C3" "A2B1C1"

share|improve this answer

edited 4 mins ago

answered 29 mins ago

sindri_baldursindri_baldur

7,7941032

answered 29 mins ago

sindri_baldursindri_baldur

7,7941032

answered 29 mins ago

sindri_baldursindri_baldur

7,7941032

0

strings[!grepl("[0-9]$",strings)]=paste0(strings[!grepl("[0-9]$",strings)],"1")

[1] "A1"     "A3BC3"  "A2B1C1"

We first grep all positions that do not end with a number, and paste a 1 to them.

share|improve this answer

answered 2 hours ago

boskiboski

490210

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  
  Almost. Check the second element it should be "A3B1C3"
  
  – markus
  2 hours ago
  
  
  

  
  
  
  

add a comment | 

0

strings[!grepl("[0-9]$",strings)]=paste0(strings[!grepl("[0-9]$",strings)],"1")

[1] "A1"     "A3BC3"  "A2B1C1"

We first grep all positions that do not end with a number, and paste a 1 to them.

share|improve this answer

answered 2 hours ago

boskiboski

490210

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  
  Almost. Check the second element it should be "A3B1C3"
  
  – markus
  2 hours ago
  
  
  

  
  
  
  

add a comment | 

strings[!grepl("[0-9]$",strings)]=paste0(strings[!grepl("[0-9]$",strings)],"1")

[1] "A1"     "A3BC3"  "A2B1C1"

We first grep all positions that do not end with a number, and paste a 1 to them.

share|improve this answer

answered 2 hours ago

boskiboski

490210

strings[!grepl("[0-9]$",strings)]=paste0(strings[!grepl("[0-9]$",strings)],"1")

[1] "A1"     "A3BC3"  "A2B1C1"

share|improve this answer

answered 2 hours ago

boskiboski

490210

answered 2 hours ago

boskiboski

490210

answered 2 hours ago

boskiboski

490210