Split vector at multiplicity of a number

I want to split a vector at a multiplicity of a number so giving vector c(1:100) I would like to receive vectors like:

c(1,11,21,31,41,51,61,71,81,91) 
c(2,12,22,32,42,52,62,72,82,92) 
..... 
c(10,20,30,40,50,60,70,80,90,100)

Note that i want the vector c(1,11,21,31,41,51,61,71,81,91) to be the FIRST in the list of new vectors, because I know that split(1:100,1:100 %% 10)[1] is c(10,20,30,40,50,60,70,80,90,100) which is not what I want

EDIT
The 1:100 example is just an example. In my app I want to split a vector of 256 numbers (random numbers not from 1 to 256) and divide it to 16 new vectors...

edited Nov 14 '18 at 21:15

asked Nov 14 '18 at 20:53

heisenberg7584

505

4

split(1:100,1:100 %% 10), perhaps?

– joran
Nov 14 '18 at 21:01

@joran, post as answer or find a duplicate? (I know, posting an answer is easier ...)

– Ben Bolker
Nov 14 '18 at 21:03

I tried it but then the first new vector is the one with 10,20,30 etc. So split(1:100,1:100 %% 10)[1] = c(10,20............). I want the first out of 10 to be the one with 1,11,21..... and the last with 10,20.30....

– heisenberg7584
Nov 14 '18 at 21:04

There may be a way to do better, but split(1:100,((1:100 %% 10) -1) %% 10) would give the order you want

– Chris
Nov 14 '18 at 21:09

1

better: split(1:100,0:99 %% 10)

– Chris
Nov 14 '18 at 21:12

|
show 1 more comment

I want to split a vector at a multiplicity of a number so giving vector c(1:100) I would like to receive vectors like:

c(1,11,21,31,41,51,61,71,81,91) 
c(2,12,22,32,42,52,62,72,82,92) 
..... 
c(10,20,30,40,50,60,70,80,90,100)

EDIT
The 1:100 example is just an example. In my app I want to split a vector of 256 numbers (random numbers not from 1 to 256) and divide it to 16 new vectors...

edited Nov 14 '18 at 21:15

asked Nov 14 '18 at 20:53

heisenberg7584

505

4

split(1:100,1:100 %% 10), perhaps?

– joran
Nov 14 '18 at 21:01

@joran, post as answer or find a duplicate? (I know, posting an answer is easier ...)

– Ben Bolker
Nov 14 '18 at 21:03

I tried it but then the first new vector is the one with 10,20,30 etc. So split(1:100,1:100 %% 10)[1] = c(10,20............). I want the first out of 10 to be the one with 1,11,21..... and the last with 10,20.30....

– heisenberg7584
Nov 14 '18 at 21:04

There may be a way to do better, but split(1:100,((1:100 %% 10) -1) %% 10) would give the order you want

– Chris
Nov 14 '18 at 21:09

1

better: split(1:100,0:99 %% 10)

– Chris
Nov 14 '18 at 21:12

|
show 1 more comment

I want to split a vector at a multiplicity of a number so giving vector c(1:100) I would like to receive vectors like:

c(1,11,21,31,41,51,61,71,81,91) 
c(2,12,22,32,42,52,62,72,82,92) 
..... 
c(10,20,30,40,50,60,70,80,90,100)

EDIT
The 1:100 example is just an example. In my app I want to split a vector of 256 numbers (random numbers not from 1 to 256) and divide it to 16 new vectors...

edited Nov 14 '18 at 21:15

asked Nov 14 '18 at 20:53

heisenberg7584

505

I want to split a vector at a multiplicity of a number so giving vector c(1:100) I would like to receive vectors like:

c(1,11,21,31,41,51,61,71,81,91) 
c(2,12,22,32,42,52,62,72,82,92) 
..... 
c(10,20,30,40,50,60,70,80,90,100)

EDIT
The 1:100 example is just an example. In my app I want to split a vector of 256 numbers (random numbers not from 1 to 256) and divide it to 16 new vectors...

edited Nov 14 '18 at 21:15

asked Nov 14 '18 at 20:53

heisenberg7584

505

edited Nov 14 '18 at 21:15

asked Nov 14 '18 at 20:53

heisenberg7584

505

edited Nov 14 '18 at 21:15

asked Nov 14 '18 at 20:53

heisenberg7584

505

asked Nov 14 '18 at 20:53

heisenberg7584

505

asked Nov 14 '18 at 20:53

heisenberg7584

505

4

split(1:100,1:100 %% 10), perhaps?

– joran
Nov 14 '18 at 21:01

@joran, post as answer or find a duplicate? (I know, posting an answer is easier ...)

– Ben Bolker
Nov 14 '18 at 21:03

I tried it but then the first new vector is the one with 10,20,30 etc. So split(1:100,1:100 %% 10)[1] = c(10,20............). I want the first out of 10 to be the one with 1,11,21..... and the last with 10,20.30....

– heisenberg7584
Nov 14 '18 at 21:04

There may be a way to do better, but split(1:100,((1:100 %% 10) -1) %% 10) would give the order you want

– Chris
Nov 14 '18 at 21:09

1

better: split(1:100,0:99 %% 10)

– Chris
Nov 14 '18 at 21:12

|
show 1 more comment

4

split(1:100,1:100 %% 10), perhaps?

– joran
Nov 14 '18 at 21:01

@joran, post as answer or find a duplicate? (I know, posting an answer is easier ...)

– Ben Bolker
Nov 14 '18 at 21:03

I tried it but then the first new vector is the one with 10,20,30 etc. So split(1:100,1:100 %% 10)[1] = c(10,20............). I want the first out of 10 to be the one with 1,11,21..... and the last with 10,20.30....

– heisenberg7584
Nov 14 '18 at 21:04

There may be a way to do better, but split(1:100,((1:100 %% 10) -1) %% 10) would give the order you want

– Chris
Nov 14 '18 at 21:09

1

better: split(1:100,0:99 %% 10)

– Chris
Nov 14 '18 at 21:12

split(1:100,1:100 %% 10), perhaps?

– joran
Nov 14 '18 at 21:01

@joran, post as answer or find a duplicate? (I know, posting an answer is easier ...)

– Ben Bolker
Nov 14 '18 at 21:03

I tried it but then the first new vector is the one with 10,20,30 etc. So split(1:100,1:100 %% 10)[1] = c(10,20............). I want the first out of 10 to be the one with 1,11,21..... and the last with 10,20.30....

– heisenberg7584
Nov 14 '18 at 21:04

There may be a way to do better, but split(1:100,((1:100 %% 10) -1) %% 10) would give the order you want

– Chris
Nov 14 '18 at 21:09

better: split(1:100,0:99 %% 10)

– Chris
Nov 14 '18 at 21:12

|
show 1 more comment

3 Answers
3

active

oldest

votes

As in the comment, perfect choice would be to make it like this:

split(1:100,0:99 %% 10)

In such case we can make it really generic so as:

split(vec, 0:(length(vec)-1) %% X)

where vec is the vector we want to divide and X is your multiplicity

answered Nov 14 '18 at 21:43

jake-ferguson

10211

add a comment |

Sounds like you want split(1:100,1:100 %% 10). Some options for ordering include:

x <- split(1:100,1:100 %% 10)
c(tail(x,-1),head(x,1))

or from the comments above,

split(1:100,((1:100 %% 10) -1) %% 10)

answered Nov 14 '18 at 21:12

joran

136k19328388

add a comment |

This is uglier than joran's answer, but you could also do that in a loop, and use grep() to extract those numbers ending in a particular value:

yourlist <- vector("list", 10)

for (i in 1:10) 
 yourlist[[i]] <- grep(paste0(i %% 10, "$"), 1:100)

answered Nov 14 '18 at 21:19

Phil

1,863629

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

As in the comment, perfect choice would be to make it like this:

split(1:100,0:99 %% 10)

In such case we can make it really generic so as:

split(vec, 0:(length(vec)-1) %% X)

where vec is the vector we want to divide and X is your multiplicity

answered Nov 14 '18 at 21:43

jake-ferguson

10211

add a comment |

As in the comment, perfect choice would be to make it like this:

split(1:100,0:99 %% 10)

In such case we can make it really generic so as:

split(vec, 0:(length(vec)-1) %% X)

where vec is the vector we want to divide and X is your multiplicity

answered Nov 14 '18 at 21:43

jake-ferguson

10211

add a comment |

As in the comment, perfect choice would be to make it like this:

split(1:100,0:99 %% 10)

In such case we can make it really generic so as:

split(vec, 0:(length(vec)-1) %% X)

where vec is the vector we want to divide and X is your multiplicity

answered Nov 14 '18 at 21:43

jake-ferguson

10211

As in the comment, perfect choice would be to make it like this:

split(1:100,0:99 %% 10)

In such case we can make it really generic so as:

split(vec, 0:(length(vec)-1) %% X)

where vec is the vector we want to divide and X is your multiplicity

answered Nov 14 '18 at 21:43

jake-ferguson

10211

answered Nov 14 '18 at 21:43

jake-ferguson

10211

answered Nov 14 '18 at 21:43

jake-ferguson

10211

answered Nov 14 '18 at 21:43

jake-ferguson

10211

add a comment |

Sounds like you want split(1:100,1:100 %% 10). Some options for ordering include:

x <- split(1:100,1:100 %% 10)
c(tail(x,-1),head(x,1))

or from the comments above,

split(1:100,((1:100 %% 10) -1) %% 10)

answered Nov 14 '18 at 21:12

joran

136k19328388

add a comment |

Sounds like you want split(1:100,1:100 %% 10). Some options for ordering include:

x <- split(1:100,1:100 %% 10)
c(tail(x,-1),head(x,1))

or from the comments above,

split(1:100,((1:100 %% 10) -1) %% 10)

answered Nov 14 '18 at 21:12

joran

136k19328388

add a comment |

Sounds like you want split(1:100,1:100 %% 10). Some options for ordering include:

x <- split(1:100,1:100 %% 10)
c(tail(x,-1),head(x,1))

or from the comments above,

split(1:100,((1:100 %% 10) -1) %% 10)

answered Nov 14 '18 at 21:12

joran

136k19328388

Sounds like you want split(1:100,1:100 %% 10). Some options for ordering include:

x <- split(1:100,1:100 %% 10)
c(tail(x,-1),head(x,1))

or from the comments above,

split(1:100,((1:100 %% 10) -1) %% 10)

answered Nov 14 '18 at 21:12

joran

136k19328388

answered Nov 14 '18 at 21:12

joran

136k19328388

answered Nov 14 '18 at 21:12

joran

136k19328388

answered Nov 14 '18 at 21:12

joran

136k19328388

add a comment |

This is uglier than joran's answer, but you could also do that in a loop, and use grep() to extract those numbers ending in a particular value:

yourlist <- vector("list", 10)

for (i in 1:10) 
 yourlist[[i]] <- grep(paste0(i %% 10, "$"), 1:100)

answered Nov 14 '18 at 21:19

Phil

1,863629

add a comment |

This is uglier than joran's answer, but you could also do that in a loop, and use grep() to extract those numbers ending in a particular value:

yourlist <- vector("list", 10)

for (i in 1:10) 
 yourlist[[i]] <- grep(paste0(i %% 10, "$"), 1:100)

answered Nov 14 '18 at 21:19

Phil

1,863629

add a comment |

This is uglier than joran's answer, but you could also do that in a loop, and use grep() to extract those numbers ending in a particular value:

yourlist <- vector("list", 10)

for (i in 1:10) 
 yourlist[[i]] <- grep(paste0(i %% 10, "$"), 1:100)

answered Nov 14 '18 at 21:19

Phil

1,863629

This is uglier than joran's answer, but you could also do that in a loop, and use grep() to extract those numbers ending in a particular value:

yourlist <- vector("list", 10)

for (i in 1:10) 
 yourlist[[i]] <- grep(paste0(i %% 10, "$"), 1:100)

answered Nov 14 '18 at 21:19

Phil

1,863629

answered Nov 14 '18 at 21:19

Phil

1,863629

answered Nov 14 '18 at 21:19

Phil

1,863629

answered Nov 14 '18 at 21:19

Phil

1,863629

add a comment |

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