How to count the number of underscores and split the string on the middle one only?

I would like to count the number of underscores and split the string into two different strings at the middle underscore.

strings <- c('aa_bb_cc_dd_ee_ff', 'cc_hh_ff_zz", "bb_dd")

Desired Output:

First        Last

"aa_bb_cc"   "dd_ee_ff"

"cc_hh"      "ff_zz"

"bb"         "dd"

edited Jan 18 at 20:41

IceCreamToucan

9,3161816

asked Jan 18 at 18:58

ldan

192

5

Possible duplicate of Split on first/nth occurrence of delimiter

– markus
Jan 18 at 19:07

1

What happens when there are an even number of underscores (e.g., aa_bb_cc)?

– Lyngbakr
Jan 18 at 19:09

add a comment |

I would like to count the number of underscores and split the string into two different strings at the middle underscore.

strings <- c('aa_bb_cc_dd_ee_ff', 'cc_hh_ff_zz", "bb_dd")

Desired Output:

First        Last

"aa_bb_cc"   "dd_ee_ff"

"cc_hh"      "ff_zz"

"bb"         "dd"

edited Jan 18 at 20:41

IceCreamToucan

9,3161816

asked Jan 18 at 18:58

ldan

192

5

Possible duplicate of Split on first/nth occurrence of delimiter

– markus
Jan 18 at 19:07

1

What happens when there are an even number of underscores (e.g., aa_bb_cc)?

– Lyngbakr
Jan 18 at 19:09

add a comment |

I would like to count the number of underscores and split the string into two different strings at the middle underscore.

strings <- c('aa_bb_cc_dd_ee_ff', 'cc_hh_ff_zz", "bb_dd")

Desired Output:

First        Last

"aa_bb_cc"   "dd_ee_ff"

"cc_hh"      "ff_zz"

"bb"         "dd"

edited Jan 18 at 20:41

IceCreamToucan

9,3161816

asked Jan 18 at 18:58

ldan

192

I would like to count the number of underscores and split the string into two different strings at the middle underscore.

strings <- c('aa_bb_cc_dd_ee_ff', 'cc_hh_ff_zz", "bb_dd")

Desired Output:

First        Last

"aa_bb_cc"   "dd_ee_ff"

"cc_hh"      "ff_zz"

"bb"         "dd"

r string

edited Jan 18 at 20:41

IceCreamToucan

9,3161816

asked Jan 18 at 18:58

ldan

192

edited Jan 18 at 20:41

IceCreamToucan

9,3161816

asked Jan 18 at 18:58

ldan

192

edited Jan 18 at 20:41

IceCreamToucan

9,3161816

edited Jan 18 at 20:41

IceCreamToucan

9,3161816

edited Jan 18 at 20:41

IceCreamToucan

9,3161816

asked Jan 18 at 18:58

ldan

192

asked Jan 18 at 18:58

ldan

192

asked Jan 18 at 18:58

ldan

192

5

Possible duplicate of Split on first/nth occurrence of delimiter

– markus
Jan 18 at 19:07

1

What happens when there are an even number of underscores (e.g., aa_bb_cc)?

– Lyngbakr
Jan 18 at 19:09

add a comment |

5

Possible duplicate of Split on first/nth occurrence of delimiter

– markus
Jan 18 at 19:07

1

What happens when there are an even number of underscores (e.g., aa_bb_cc)?

– Lyngbakr
Jan 18 at 19:09

Possible duplicate of Split on first/nth occurrence of delimiter

– markus
Jan 18 at 19:07

What happens when there are an even number of underscores (e.g., aa_bb_cc)?

– Lyngbakr
Jan 18 at 19:09

add a comment |

3 Answers
3

active

oldest

votes

Here's a cludgy solution that assumes that there are always an odd number of underscores.

# Load libraries

library(stringr)



# Define function

even_split <- function(s){

  # Split string

  tmp <- str_split(s, "_")



  lapply(tmp, function(x){

    # Patch string back together in two pieces

    c(paste(x[1:(length(x)/2)], collapse = "_"),

      paste(x[(1+length(x)/2):length(x)], collapse = "_"))

  })

}



# Example

strings <- c('aa_bb_cc_dd_ee_ff', 'cc_hh_ff_zz', 'bb_dd')



# Test function

even_split(strings)

#> [[1]]

#> [1] "aa_bb_cc" "dd_ee_ff"

#> 

#> [[2]]

#> [1] "cc_hh" "ff_zz"

#> 

#> [[3]]

#> [1] "bb" "dd"

^{Created on 2019-01-18 by the reprex package (v0.2.1)}

answered Jan 18 at 19:19

Lyngbakr

4,63311325

add a comment |

Adapting nhahtdh's answer here, all you need to do is add a step to count the underscores (done here with str_count) and return the median number of underscores.

library(stringr)



strsplit(

  strings, 

  paste0("^[^_]*(?:_[^_]*){", str_count(strings, '_') %/% 2, "}\K_"), 

  perl = TRUE)



# [[1]]

# [1] "aa_bb_cc" "dd_ee_ff"

# 

# [[2]]

# [1] "cc_hh" "ff_zz"

# 

# [[3]]

# [1] "bb" "dd"

edited Jan 18 at 20:28

answered Jan 18 at 19:44

IceCreamToucan

9,3161816

add a comment |

This assumes an odd number of underscores, and 99 or fewer.

library(stringr)

library(strex)

strings <- c('aa_bb_cc_dd_ee_ff', 'cc_hh_ff_zz', 'bb_dd')



splitMiddleUnderscore <- function(x){

    nUnderscore <- str_count(x, '_')

    middleUnderscore <- match(nUnderscore, seq(1, 99, 2))

    str1 <- str_before_nth(x, '_',  middleUnderscore)

    str2 <- str_after_nth(x, '_', middleUnderscore)

    c(str1, str2)

}



lapply(strings, splitMiddleUnderscore)



#[[1]]

#[1] "aa_bb_cc" "dd_ee_ff"



#[[2]]

#[1] "cc_hh" "ff_zz"



#[[3]]

#[1] "bb" "dd"

answered Jan 18 at 19:36

Bill O'Brien

576

1

you can use middleUnderscore <- str_count(x, '_') %/% 2 + 1 to avoid the "99 or fewer" requirement.

– IceCreamToucan
Jan 18 at 20:11

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

Here's a cludgy solution that assumes that there are always an odd number of underscores.

# Load libraries

library(stringr)



# Define function

even_split <- function(s){

  # Split string

  tmp <- str_split(s, "_")



  lapply(tmp, function(x){

    # Patch string back together in two pieces

    c(paste(x[1:(length(x)/2)], collapse = "_"),

      paste(x[(1+length(x)/2):length(x)], collapse = "_"))

  })

}



# Example

strings <- c('aa_bb_cc_dd_ee_ff', 'cc_hh_ff_zz', 'bb_dd')



# Test function

even_split(strings)

#> [[1]]

#> [1] "aa_bb_cc" "dd_ee_ff"

#> 

#> [[2]]

#> [1] "cc_hh" "ff_zz"

#> 

#> [[3]]

#> [1] "bb" "dd"

^{Created on 2019-01-18 by the reprex package (v0.2.1)}

answered Jan 18 at 19:19

Lyngbakr

4,63311325

add a comment |

Here's a cludgy solution that assumes that there are always an odd number of underscores.

# Load libraries

library(stringr)



# Define function

even_split <- function(s){

  # Split string

  tmp <- str_split(s, "_")



  lapply(tmp, function(x){

    # Patch string back together in two pieces

    c(paste(x[1:(length(x)/2)], collapse = "_"),

      paste(x[(1+length(x)/2):length(x)], collapse = "_"))

  })

}



# Example

strings <- c('aa_bb_cc_dd_ee_ff', 'cc_hh_ff_zz', 'bb_dd')



# Test function

even_split(strings)

#> [[1]]

#> [1] "aa_bb_cc" "dd_ee_ff"

#> 

#> [[2]]

#> [1] "cc_hh" "ff_zz"

#> 

#> [[3]]

#> [1] "bb" "dd"

^{Created on 2019-01-18 by the reprex package (v0.2.1)}

answered Jan 18 at 19:19

Lyngbakr

4,63311325

add a comment |

Here's a cludgy solution that assumes that there are always an odd number of underscores.

# Load libraries

library(stringr)



# Define function

even_split <- function(s){

  # Split string

  tmp <- str_split(s, "_")



  lapply(tmp, function(x){

    # Patch string back together in two pieces

    c(paste(x[1:(length(x)/2)], collapse = "_"),

      paste(x[(1+length(x)/2):length(x)], collapse = "_"))

  })

}



# Example

strings <- c('aa_bb_cc_dd_ee_ff', 'cc_hh_ff_zz', 'bb_dd')



# Test function

even_split(strings)

#> [[1]]

#> [1] "aa_bb_cc" "dd_ee_ff"

#> 

#> [[2]]

#> [1] "cc_hh" "ff_zz"

#> 

#> [[3]]

#> [1] "bb" "dd"

^{Created on 2019-01-18 by the reprex package (v0.2.1)}

answered Jan 18 at 19:19

Lyngbakr

4,63311325

Here's a cludgy solution that assumes that there are always an odd number of underscores.

# Load libraries

library(stringr)



# Define function

even_split <- function(s){

  # Split string

  tmp <- str_split(s, "_")



  lapply(tmp, function(x){

    # Patch string back together in two pieces

    c(paste(x[1:(length(x)/2)], collapse = "_"),

      paste(x[(1+length(x)/2):length(x)], collapse = "_"))

  })

}



# Example

strings <- c('aa_bb_cc_dd_ee_ff', 'cc_hh_ff_zz', 'bb_dd')



# Test function

even_split(strings)

#> [[1]]

#> [1] "aa_bb_cc" "dd_ee_ff"

#> 

#> [[2]]

#> [1] "cc_hh" "ff_zz"

#> 

#> [[3]]

#> [1] "bb" "dd"

^{Created on 2019-01-18 by the reprex package (v0.2.1)}

answered Jan 18 at 19:19

Lyngbakr

4,63311325

answered Jan 18 at 19:19

Lyngbakr

4,63311325

answered Jan 18 at 19:19

Lyngbakr

4,63311325

answered Jan 18 at 19:19

Lyngbakr

4,63311325

add a comment |

Adapting nhahtdh's answer here, all you need to do is add a step to count the underscores (done here with str_count) and return the median number of underscores.

library(stringr)



strsplit(

  strings, 

  paste0("^[^_]*(?:_[^_]*){", str_count(strings, '_') %/% 2, "}\K_"), 

  perl = TRUE)



# [[1]]

# [1] "aa_bb_cc" "dd_ee_ff"

# 

# [[2]]

# [1] "cc_hh" "ff_zz"

# 

# [[3]]

# [1] "bb" "dd"

edited Jan 18 at 20:28

answered Jan 18 at 19:44

IceCreamToucan

9,3161816

add a comment |

Adapting nhahtdh's answer here, all you need to do is add a step to count the underscores (done here with str_count) and return the median number of underscores.

library(stringr)



strsplit(

  strings, 

  paste0("^[^_]*(?:_[^_]*){", str_count(strings, '_') %/% 2, "}\K_"), 

  perl = TRUE)



# [[1]]

# [1] "aa_bb_cc" "dd_ee_ff"

# 

# [[2]]

# [1] "cc_hh" "ff_zz"

# 

# [[3]]

# [1] "bb" "dd"

edited Jan 18 at 20:28

answered Jan 18 at 19:44

IceCreamToucan

9,3161816

add a comment |

Adapting nhahtdh's answer here, all you need to do is add a step to count the underscores (done here with str_count) and return the median number of underscores.

library(stringr)



strsplit(

  strings, 

  paste0("^[^_]*(?:_[^_]*){", str_count(strings, '_') %/% 2, "}\K_"), 

  perl = TRUE)



# [[1]]

# [1] "aa_bb_cc" "dd_ee_ff"

# 

# [[2]]

# [1] "cc_hh" "ff_zz"

# 

# [[3]]

# [1] "bb" "dd"

edited Jan 18 at 20:28

answered Jan 18 at 19:44

IceCreamToucan

9,3161816

Adapting nhahtdh's answer here, all you need to do is add a step to count the underscores (done here with str_count) and return the median number of underscores.

library(stringr)



strsplit(

  strings, 

  paste0("^[^_]*(?:_[^_]*){", str_count(strings, '_') %/% 2, "}\K_"), 

  perl = TRUE)



# [[1]]

# [1] "aa_bb_cc" "dd_ee_ff"

# 

# [[2]]

# [1] "cc_hh" "ff_zz"

# 

# [[3]]

# [1] "bb" "dd"

edited Jan 18 at 20:28

answered Jan 18 at 19:44

IceCreamToucan

9,3161816

edited Jan 18 at 20:28

answered Jan 18 at 19:44

IceCreamToucan

9,3161816

answered Jan 18 at 19:44

IceCreamToucan

9,3161816

answered Jan 18 at 19:44

IceCreamToucan

9,3161816

add a comment |

This assumes an odd number of underscores, and 99 or fewer.

library(stringr)

library(strex)

strings <- c('aa_bb_cc_dd_ee_ff', 'cc_hh_ff_zz', 'bb_dd')



splitMiddleUnderscore <- function(x){

    nUnderscore <- str_count(x, '_')

    middleUnderscore <- match(nUnderscore, seq(1, 99, 2))

    str1 <- str_before_nth(x, '_',  middleUnderscore)

    str2 <- str_after_nth(x, '_', middleUnderscore)

    c(str1, str2)

}



lapply(strings, splitMiddleUnderscore)



#[[1]]

#[1] "aa_bb_cc" "dd_ee_ff"



#[[2]]

#[1] "cc_hh" "ff_zz"



#[[3]]

#[1] "bb" "dd"

answered Jan 18 at 19:36

Bill O'Brien

576

1

you can use middleUnderscore <- str_count(x, '_') %/% 2 + 1 to avoid the "99 or fewer" requirement.

– IceCreamToucan
Jan 18 at 20:11

add a comment |

This assumes an odd number of underscores, and 99 or fewer.

library(stringr)

library(strex)

strings <- c('aa_bb_cc_dd_ee_ff', 'cc_hh_ff_zz', 'bb_dd')



splitMiddleUnderscore <- function(x){

    nUnderscore <- str_count(x, '_')

    middleUnderscore <- match(nUnderscore, seq(1, 99, 2))

    str1 <- str_before_nth(x, '_',  middleUnderscore)

    str2 <- str_after_nth(x, '_', middleUnderscore)

    c(str1, str2)

}



lapply(strings, splitMiddleUnderscore)



#[[1]]

#[1] "aa_bb_cc" "dd_ee_ff"



#[[2]]

#[1] "cc_hh" "ff_zz"



#[[3]]

#[1] "bb" "dd"

answered Jan 18 at 19:36

Bill O'Brien

576

1

you can use middleUnderscore <- str_count(x, '_') %/% 2 + 1 to avoid the "99 or fewer" requirement.

– IceCreamToucan
Jan 18 at 20:11

add a comment |

This assumes an odd number of underscores, and 99 or fewer.

library(stringr)

library(strex)

strings <- c('aa_bb_cc_dd_ee_ff', 'cc_hh_ff_zz', 'bb_dd')



splitMiddleUnderscore <- function(x){

    nUnderscore <- str_count(x, '_')

    middleUnderscore <- match(nUnderscore, seq(1, 99, 2))

    str1 <- str_before_nth(x, '_',  middleUnderscore)

    str2 <- str_after_nth(x, '_', middleUnderscore)

    c(str1, str2)

}



lapply(strings, splitMiddleUnderscore)



#[[1]]

#[1] "aa_bb_cc" "dd_ee_ff"



#[[2]]

#[1] "cc_hh" "ff_zz"



#[[3]]

#[1] "bb" "dd"

answered Jan 18 at 19:36

Bill O'Brien

576

This assumes an odd number of underscores, and 99 or fewer.

library(stringr)

library(strex)

strings <- c('aa_bb_cc_dd_ee_ff', 'cc_hh_ff_zz', 'bb_dd')



splitMiddleUnderscore <- function(x){

    nUnderscore <- str_count(x, '_')

    middleUnderscore <- match(nUnderscore, seq(1, 99, 2))

    str1 <- str_before_nth(x, '_',  middleUnderscore)

    str2 <- str_after_nth(x, '_', middleUnderscore)

    c(str1, str2)

}



lapply(strings, splitMiddleUnderscore)



#[[1]]

#[1] "aa_bb_cc" "dd_ee_ff"



#[[2]]

#[1] "cc_hh" "ff_zz"



#[[3]]

#[1] "bb" "dd"

answered Jan 18 at 19:36

Bill O'Brien

576

answered Jan 18 at 19:36

Bill O'Brien

576

answered Jan 18 at 19:36

Bill O'Brien

576

answered Jan 18 at 19:36

Bill O'Brien

576

1

you can use middleUnderscore <- str_count(x, '_') %/% 2 + 1 to avoid the "99 or fewer" requirement.

– IceCreamToucan
Jan 18 at 20:11

add a comment |

1

you can use middleUnderscore <- str_count(x, '_') %/% 2 + 1 to avoid the "99 or fewer" requirement.

– IceCreamToucan
Jan 18 at 20:11

you can use middleUnderscore <- str_count(x, '_') %/% 2 + 1 to avoid the "99 or fewer" requirement.

– IceCreamToucan
Jan 18 at 20:11

add a comment |

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