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...
r
asked Nov 14 '18 at 20:53
heisenberg7584heisenberg7584
505
|
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)
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...
r
asked Nov 14 '18 at 20:53
heisenberg7584heisenberg7584
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. Sosplit(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, butsplit(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)
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...
r
asked Nov 14 '18 at 20:53
heisenberg7584heisenberg7584
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)
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...
r
r
asked Nov 14 '18 at 20:53
heisenberg7584heisenberg7584
505
asked Nov 14 '18 at 20:53
heisenberg7584heisenberg7584
505
edited Nov 14 '18 at 21:15
heisenberg7584
asked Nov 14 '18 at 20:53
heisenberg7584heisenberg7584
505
asked Nov 14 '18 at 20:53
heisenberg7584heisenberg7584
505
asked Nov 14 '18 at 20:53
heisenberg7584heisenberg7584
505
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. Sosplit(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, butsplit(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. Sosplit(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, butsplit(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
4
4
split(1:100,1:100 %% 10), perhaps?– joran
Nov 14 '18 at 21:01
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
@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
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
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
1
better:
split(1:100,0:99 %% 10)– Chris
Nov 14 '18 at 21:12
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-fergusonjake-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
joranjoran
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
PhilPhil
1,863629
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
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-fergusonjake-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-fergusonjake-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-fergusonjake-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-fergusonjake-ferguson
10211
answered Nov 14 '18 at 21:43
jake-fergusonjake-ferguson
10211
answered Nov 14 '18 at 21:43
jake-fergusonjake-ferguson
10211
answered Nov 14 '18 at 21:43
jake-fergusonjake-ferguson
10211
10211
add a comment |
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
joranjoran
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
joranjoran
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
joranjoran
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
joranjoran
136k19328388
answered Nov 14 '18 at 21:12
joranjoran
136k19328388
answered Nov 14 '18 at 21:12
joranjoran
136k19328388
answered Nov 14 '18 at 21:12
joranjoran
136k19328388
136k19328388
add a comment |
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
PhilPhil
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, "$"), 1:100)
answered Nov 14 '18 at 21:19
PhilPhil
1,863629
add a comment |
This is uglier than joran's answer, but you could also do that in a loop, and use grep() to extract those numbers ending in a particular value:
yourlist <- vector("list", 10)
for (i in 1:10)
yourlist[[i]] <- grep(paste0(i %% 10, "$"), 1:100)
answered Nov 14 '18 at 21:19
PhilPhil
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
PhilPhil
1,863629
answered Nov 14 '18 at 21:19
PhilPhil
1,863629
answered Nov 14 '18 at 21:19
PhilPhil
1,863629
answered Nov 14 '18 at 21:19
PhilPhil
1,863629
1,863629
add a comment |
add a comment |
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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