Count vectors in list with the same elements in r
2
I have a list of ~8000 vectors, and I would like to know how many duplicates there are of these 8000 vectors, but the order of the elements in each could be different.
for example:
list <- c()
list[[1]] <- c(1,2,3)
list[[2]] <- c(2,1,3)
list[[3]] <- c(3,2,1)
list[[4]] <- c(4,5)
list[[5]] <- c(5,4)
list[[6]] <- c(1,2,3,5)
should give me a count of 3 for c(1,2,3) and 2 for c(4,5) and 1 for c(1,2,3,5)
I'd like the count of each of the duplicates, not just how many are duplicated.
r list
asked Nov 16 '18 at 19:52
HibaShaban
20110
add a comment |
2
I have a list of ~8000 vectors, and I would like to know how many duplicates there are of these 8000 vectors, but the order of the elements in each could be different.
for example:
list <- c()
list[[1]] <- c(1,2,3)
list[[2]] <- c(2,1,3)
list[[3]] <- c(3,2,1)
list[[4]] <- c(4,5)
list[[5]] <- c(5,4)
list[[6]] <- c(1,2,3,5)
should give me a count of 3 for c(1,2,3) and 2 for c(4,5) and 1 for c(1,2,3,5)
I'd like the count of each of the duplicates, not just how many are duplicated.
r list
asked Nov 16 '18 at 19:52
HibaShaban
20110
1
Do you needlength(unique(list))Also, if you need all the duplicates thensum(duplicated(lst)|duplicated(lst, fromLast = TRUE))
– akrun
Nov 16 '18 at 19:54
in a sense, but I would like some sort of function that tells me which list was duplicated how many times, in the example above it would bec(1,2,3) = 3or something like that
– HibaShaban
Nov 16 '18 at 20:00
1
What you are showing is not a list of lists but a list of vectors
– prosoitos
Nov 16 '18 at 20:05
1
You have a typo:Listshould beliston your first line
– prosoitos
Nov 16 '18 at 20:38
1
yes, thanks. Was trying to make it reproducible but that part slipped through.
– HibaShaban
Nov 16 '18 at 20:41
add a comment |
2
2
2
I have a list of ~8000 vectors, and I would like to know how many duplicates there are of these 8000 vectors, but the order of the elements in each could be different.
for example:
list <- c()
list[[1]] <- c(1,2,3)
list[[2]] <- c(2,1,3)
list[[3]] <- c(3,2,1)
list[[4]] <- c(4,5)
list[[5]] <- c(5,4)
list[[6]] <- c(1,2,3,5)
should give me a count of 3 for c(1,2,3) and 2 for c(4,5) and 1 for c(1,2,3,5)
I'd like the count of each of the duplicates, not just how many are duplicated.
r list
asked Nov 16 '18 at 19:52
HibaShaban
20110
I have a list of ~8000 vectors, and I would like to know how many duplicates there are of these 8000 vectors, but the order of the elements in each could be different.
for example:
list <- c()
list[[1]] <- c(1,2,3)
list[[2]] <- c(2,1,3)
list[[3]] <- c(3,2,1)
list[[4]] <- c(4,5)
list[[5]] <- c(5,4)
list[[6]] <- c(1,2,3,5)
should give me a count of 3 for c(1,2,3) and 2 for c(4,5) and 1 for c(1,2,3,5)
I'd like the count of each of the duplicates, not just how many are duplicated.
r list
r list
asked Nov 16 '18 at 19:52
HibaShaban
20110
asked Nov 16 '18 at 19:52
HibaShaban
20110
edited Nov 18 '18 at 13:15
asked Nov 16 '18 at 19:52
HibaShaban
20110
asked Nov 16 '18 at 19:52
HibaShaban
20110
asked Nov 16 '18 at 19:52
HibaShaban
20110
20110
1
Do you needlength(unique(list))Also, if you need all the duplicates thensum(duplicated(lst)|duplicated(lst, fromLast = TRUE))
– akrun
Nov 16 '18 at 19:54
in a sense, but I would like some sort of function that tells me which list was duplicated how many times, in the example above it would bec(1,2,3) = 3or something like that
– HibaShaban
Nov 16 '18 at 20:00
1
What you are showing is not a list of lists but a list of vectors
– prosoitos
Nov 16 '18 at 20:05
1
You have a typo:Listshould beliston your first line
– prosoitos
Nov 16 '18 at 20:38
1
yes, thanks. Was trying to make it reproducible but that part slipped through.
– HibaShaban
Nov 16 '18 at 20:41
add a comment |
1
Do you needlength(unique(list))Also, if you need all the duplicates thensum(duplicated(lst)|duplicated(lst, fromLast = TRUE))
– akrun
Nov 16 '18 at 19:54
in a sense, but I would like some sort of function that tells me which list was duplicated how many times, in the example above it would bec(1,2,3) = 3or something like that
– HibaShaban
Nov 16 '18 at 20:00
1
What you are showing is not a list of lists but a list of vectors
– prosoitos
Nov 16 '18 at 20:05
1
You have a typo:Listshould beliston your first line
– prosoitos
Nov 16 '18 at 20:38
1
yes, thanks. Was trying to make it reproducible but that part slipped through.
– HibaShaban
Nov 16 '18 at 20:41
1
1
Do you need
length(unique(list)) Also, if you need all the duplicates then sum(duplicated(lst)|duplicated(lst, fromLast = TRUE))– akrun
Nov 16 '18 at 19:54
Do you need
length(unique(list)) Also, if you need all the duplicates then sum(duplicated(lst)|duplicated(lst, fromLast = TRUE))– akrun
Nov 16 '18 at 19:54
in a sense, but I would like some sort of function that tells me which list was duplicated how many times, in the example above it would be
c(1,2,3) = 3 or something like that– HibaShaban
Nov 16 '18 at 20:00
in a sense, but I would like some sort of function that tells me which list was duplicated how many times, in the example above it would be
c(1,2,3) = 3 or something like that– HibaShaban
Nov 16 '18 at 20:00
1
1
What you are showing is not a list of lists but a list of vectors
– prosoitos
Nov 16 '18 at 20:05
What you are showing is not a list of lists but a list of vectors
– prosoitos
Nov 16 '18 at 20:05
1
1
You have a typo:
List should be list on your first line– prosoitos
Nov 16 '18 at 20:38
You have a typo:
List should be list on your first line– prosoitos
Nov 16 '18 at 20:38
1
1
yes, thanks. Was trying to make it reproducible but that part slipped through.
– HibaShaban
Nov 16 '18 at 20:41
yes, thanks. Was trying to make it reproducible but that part slipped through.
– HibaShaban
Nov 16 '18 at 20:41
add a comment |
2 Answers
2
active
oldest
votes
1
library(tidyverse)
library(gtools)
get_perm <- function(v) {
m <- permutations(n = length(v), r = length(v), v = v, set = F)
m[order(c(m))]
}
all <- map(list, get_perm)
unique <- map(list, get_perm) %>% unique()
res_vec <- c()
element <- c()
for(i in seq_along(unique)) {
element[[i]] <- unique[[i]] %>% unique() %>% paste(collapse = ",")
res_vec[[i]] <- all %in% unique[i] %>% sum()
}
tibble(
elements = unlist(element),
numbers = res_vec
)
Result
# A tibble: 3 x 2
elements numbers
<chr> <int>
1 1,2,3 3
2 4,5 2
3 1,2,3,5 1
elements contains all the individual elements of the vectors for each group and numbers are the numbers of vectors you have in each group.
answered Nov 17 '18 at 0:47
prosoitos
935219
add a comment |
1
We create a function to take vector as an argument ('val'), then loop through the list with sapply, check if all the 'valare%in%the 'x', andsumthe logicalvector`
f1 <- function(lst, val) sum(sapply(lst, function(x) all(val %in% x)))
f1(list, c(1, 2, 3))
[#1] 3
f1(list, c(4, 5))
#[1] 2
answered Nov 16 '18 at 20:02
akrun
398k13187260
1
Thanks! This worked
– HibaShaban
Nov 16 '18 at 20:34
So actually this half worked.. I'm not getting extra counts when the vectors contain more elements, examplec<-(1,2,3,5)is being counted withc(1,2,3)how would I fix that?
– HibaShaban
Nov 16 '18 at 22:00
I've resorted to sorting the vectors and then using == instead of %in%. that's the only way I can think of for now. Would appreciate any other solutions.
– HibaShaban
Nov 16 '18 at 22:51
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
library(tidyverse)
library(gtools)
get_perm <- function(v) {
m <- permutations(n = length(v), r = length(v), v = v, set = F)
m[order(c(m))]
}
all <- map(list, get_perm)
unique <- map(list, get_perm) %>% unique()
res_vec <- c()
element <- c()
for(i in seq_along(unique)) {
element[[i]] <- unique[[i]] %>% unique() %>% paste(collapse = ",")
res_vec[[i]] <- all %in% unique[i] %>% sum()
}
tibble(
elements = unlist(element),
numbers = res_vec
)
Result
# A tibble: 3 x 2
elements numbers
<chr> <int>
1 1,2,3 3
2 4,5 2
3 1,2,3,5 1
elements contains all the individual elements of the vectors for each group and numbers are the numbers of vectors you have in each group.
answered Nov 17 '18 at 0:47
prosoitos
935219
add a comment |
1
library(tidyverse)
library(gtools)
get_perm <- function(v) {
m <- permutations(n = length(v), r = length(v), v = v, set = F)
m[order(c(m))]
}
all <- map(list, get_perm)
unique <- map(list, get_perm) %>% unique()
res_vec <- c()
element <- c()
for(i in seq_along(unique)) {
element[[i]] <- unique[[i]] %>% unique() %>% paste(collapse = ",")
res_vec[[i]] <- all %in% unique[i] %>% sum()
}
tibble(
elements = unlist(element),
numbers = res_vec
)
Result
# A tibble: 3 x 2
elements numbers
<chr> <int>
1 1,2,3 3
2 4,5 2
3 1,2,3,5 1
elements contains all the individual elements of the vectors for each group and numbers are the numbers of vectors you have in each group.
answered Nov 17 '18 at 0:47
prosoitos
935219
add a comment |
1
1
1
library(tidyverse)
library(gtools)
get_perm <- function(v) {
m <- permutations(n = length(v), r = length(v), v = v, set = F)
m[order(c(m))]
}
all <- map(list, get_perm)
unique <- map(list, get_perm) %>% unique()
res_vec <- c()
element <- c()
for(i in seq_along(unique)) {
element[[i]] <- unique[[i]] %>% unique() %>% paste(collapse = ",")
res_vec[[i]] <- all %in% unique[i] %>% sum()
}
tibble(
elements = unlist(element),
numbers = res_vec
)
Result
# A tibble: 3 x 2
elements numbers
<chr> <int>
1 1,2,3 3
2 4,5 2
3 1,2,3,5 1
elements contains all the individual elements of the vectors for each group and numbers are the numbers of vectors you have in each group.
answered Nov 17 '18 at 0:47
prosoitos
935219
library(tidyverse)
library(gtools)
get_perm <- function(v) {
m <- permutations(n = length(v), r = length(v), v = v, set = F)
m[order(c(m))]
}
all <- map(list, get_perm)
unique <- map(list, get_perm) %>% unique()
res_vec <- c()
element <- c()
for(i in seq_along(unique)) {
element[[i]] <- unique[[i]] %>% unique() %>% paste(collapse = ",")
res_vec[[i]] <- all %in% unique[i] %>% sum()
}
tibble(
elements = unlist(element),
numbers = res_vec
)
Result
# A tibble: 3 x 2
elements numbers
<chr> <int>
1 1,2,3 3
2 4,5 2
3 1,2,3,5 1
elements contains all the individual elements of the vectors for each group and numbers are the numbers of vectors you have in each group.
answered Nov 17 '18 at 0:47
prosoitos
935219
edited Nov 17 '18 at 5:46
answered Nov 17 '18 at 0:47
prosoitos
935219
answered Nov 17 '18 at 0:47
prosoitos
935219
answered Nov 17 '18 at 0:47
prosoitos
935219
935219
add a comment |
add a comment |
1
We create a function to take vector as an argument ('val'), then loop through the list with sapply, check if all the 'valare%in%the 'x', andsumthe logicalvector`
f1 <- function(lst, val) sum(sapply(lst, function(x) all(val %in% x)))
f1(list, c(1, 2, 3))
[#1] 3
f1(list, c(4, 5))
#[1] 2
answered Nov 16 '18 at 20:02
akrun
398k13187260
1
Thanks! This worked
– HibaShaban
Nov 16 '18 at 20:34
So actually this half worked.. I'm not getting extra counts when the vectors contain more elements, examplec<-(1,2,3,5)is being counted withc(1,2,3)how would I fix that?
– HibaShaban
Nov 16 '18 at 22:00
I've resorted to sorting the vectors and then using == instead of %in%. that's the only way I can think of for now. Would appreciate any other solutions.
– HibaShaban
Nov 16 '18 at 22:51
add a comment |
1
We create a function to take vector as an argument ('val'), then loop through the list with sapply, check if all the 'valare%in%the 'x', andsumthe logicalvector`
f1 <- function(lst, val) sum(sapply(lst, function(x) all(val %in% x)))
f1(list, c(1, 2, 3))
[#1] 3
f1(list, c(4, 5))
#[1] 2
answered Nov 16 '18 at 20:02
akrun
398k13187260
1
Thanks! This worked
– HibaShaban
Nov 16 '18 at 20:34
So actually this half worked.. I'm not getting extra counts when the vectors contain more elements, examplec<-(1,2,3,5)is being counted withc(1,2,3)how would I fix that?
– HibaShaban
Nov 16 '18 at 22:00
I've resorted to sorting the vectors and then using == instead of %in%. that's the only way I can think of for now. Would appreciate any other solutions.
– HibaShaban
Nov 16 '18 at 22:51
add a comment |
1
1
1
We create a function to take vector as an argument ('val'), then loop through the list with sapply, check if all the 'valare%in%the 'x', andsumthe logicalvector`
f1 <- function(lst, val) sum(sapply(lst, function(x) all(val %in% x)))
f1(list, c(1, 2, 3))
[#1] 3
f1(list, c(4, 5))
#[1] 2
answered Nov 16 '18 at 20:02
akrun
398k13187260
We create a function to take vector as an argument ('val'), then loop through the list with sapply, check if all the 'valare%in%the 'x', andsumthe logicalvector`
f1 <- function(lst, val) sum(sapply(lst, function(x) all(val %in% x)))
f1(list, c(1, 2, 3))
[#1] 3
f1(list, c(4, 5))
#[1] 2
answered Nov 16 '18 at 20:02
akrun
398k13187260
edited Nov 16 '18 at 20:30
answered Nov 16 '18 at 20:02
akrun
398k13187260
answered Nov 16 '18 at 20:02
akrun
398k13187260
answered Nov 16 '18 at 20:02
akrun
398k13187260
398k13187260
1
Thanks! This worked
– HibaShaban
Nov 16 '18 at 20:34
So actually this half worked.. I'm not getting extra counts when the vectors contain more elements, examplec<-(1,2,3,5)is being counted withc(1,2,3)how would I fix that?
– HibaShaban
Nov 16 '18 at 22:00
I've resorted to sorting the vectors and then using == instead of %in%. that's the only way I can think of for now. Would appreciate any other solutions.
– HibaShaban
Nov 16 '18 at 22:51
add a comment |
1
Thanks! This worked
– HibaShaban
Nov 16 '18 at 20:34
So actually this half worked.. I'm not getting extra counts when the vectors contain more elements, examplec<-(1,2,3,5)is being counted withc(1,2,3)how would I fix that?
– HibaShaban
Nov 16 '18 at 22:00
I've resorted to sorting the vectors and then using == instead of %in%. that's the only way I can think of for now. Would appreciate any other solutions.
– HibaShaban
Nov 16 '18 at 22:51
1
1
Thanks! This worked
– HibaShaban
Nov 16 '18 at 20:34
Thanks! This worked
– HibaShaban
Nov 16 '18 at 20:34
So actually this half worked.. I'm not getting extra counts when the vectors contain more elements, example
c<-(1,2,3,5) is being counted with c(1,2,3) how would I fix that?– HibaShaban
Nov 16 '18 at 22:00
So actually this half worked.. I'm not getting extra counts when the vectors contain more elements, example
c<-(1,2,3,5) is being counted with c(1,2,3) how would I fix that?– HibaShaban
Nov 16 '18 at 22:00
I've resorted to sorting the vectors and then using == instead of %in%. that's the only way I can think of for now. Would appreciate any other solutions.
– HibaShaban
Nov 16 '18 at 22:51
I've resorted to sorting the vectors and then using == instead of %in%. that's the only way I can think of for now. Would appreciate any other solutions.
– HibaShaban
Nov 16 '18 at 22:51
add a comment |
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1
Do you need
length(unique(list))Also, if you need all the duplicates thensum(duplicated(lst)|duplicated(lst, fromLast = TRUE))– akrun
Nov 16 '18 at 19:54
in a sense, but I would like some sort of function that tells me which list was duplicated how many times, in the example above it would be
c(1,2,3) = 3or something like that– HibaShaban
Nov 16 '18 at 20:00
1
What you are showing is not a list of lists but a list of vectors
– prosoitos
Nov 16 '18 at 20:05
1
You have a typo:
Listshould beliston your first line– prosoitos
Nov 16 '18 at 20:38
1
yes, thanks. Was trying to make it reproducible but that part slipped through.
– HibaShaban
Nov 16 '18 at 20:41