R program questions
I am trying to get some unique combinations of two variables.
For each value of x, I would like to have this unique y value, and drop those have several y values. But several x values could share same y value.
For example,a=data.frame(x=c(1,1,2,4,5,5),y=c(2,3,3,3,6,6)),
and I would like to get the output like:
b=data.frame(x=c(2,4,5),y=c(3,3,6))
I have tried unique(), but it does not help this situation.
Thank you!
r
edited Nov 14 '18 at 19:16
joran
136k19328388
asked Nov 14 '18 at 19:10
Lilia FengLilia Feng
374
add a comment |
I am trying to get some unique combinations of two variables.
For each value of x, I would like to have this unique y value, and drop those have several y values. But several x values could share same y value.
For example,a=data.frame(x=c(1,1,2,4,5,5),y=c(2,3,3,3,6,6)),
and I would like to get the output like:
b=data.frame(x=c(2,4,5),y=c(3,3,6))
I have tried unique(), but it does not help this situation.
Thank you!
r
edited Nov 14 '18 at 19:16
joran
136k19328388
asked Nov 14 '18 at 19:10
Lilia FengLilia Feng
374
add a comment |
I am trying to get some unique combinations of two variables.
For each value of x, I would like to have this unique y value, and drop those have several y values. But several x values could share same y value.
For example,a=data.frame(x=c(1,1,2,4,5,5),y=c(2,3,3,3,6,6)),
and I would like to get the output like:
b=data.frame(x=c(2,4,5),y=c(3,3,6))
I have tried unique(), but it does not help this situation.
Thank you!
r
edited Nov 14 '18 at 19:16
joran
136k19328388
asked Nov 14 '18 at 19:10
Lilia FengLilia Feng
374
I am trying to get some unique combinations of two variables.
For each value of x, I would like to have this unique y value, and drop those have several y values. But several x values could share same y value.
For example,a=data.frame(x=c(1,1,2,4,5,5),y=c(2,3,3,3,6,6)),
and I would like to get the output like:
b=data.frame(x=c(2,4,5),y=c(3,3,6))
I have tried unique(), but it does not help this situation.
Thank you!
r
r
edited Nov 14 '18 at 19:16
joran
136k19328388
asked Nov 14 '18 at 19:10
Lilia FengLilia Feng
374
edited Nov 14 '18 at 19:16
joran
136k19328388
asked Nov 14 '18 at 19:10
Lilia FengLilia Feng
374
edited Nov 14 '18 at 19:16
joran
136k19328388
edited Nov 14 '18 at 19:16
joran
136k19328388
edited Nov 14 '18 at 19:16
joran
136k19328388
136k19328388
asked Nov 14 '18 at 19:10
Lilia FengLilia Feng
374
asked Nov 14 '18 at 19:10
Lilia FengLilia Feng
374
asked Nov 14 '18 at 19:10
Lilia FengLilia Feng
374
374
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
First we use unique to omit repeated rows with the same x and y values (keeping only one copy of each). Any repeated x values that are left have different y values, so we want to get rid of them. We use the standard way to remove all copies of any duplicated values as in this R-FAQ.
a=data.frame(x=c(1,1,2,4,5,5),y=c(2,3,3,3,6,6))
b = unique(a)
b = b[!duplicated(b$x) & !duplicated(b$x, fromLast = TRUE), ]
b
# x y
# 3 2 3
# 4 4 3
# 5 5 6
Fans of dplyr would probably do it like this, producing the same result.
library(dplyr)
a %>%
group_by(x) %>%
filter(n_distinct(y) == 1) %>%
distinct
answered Nov 14 '18 at 19:23
GregorGregor
66.8k1092177
add a comment |
Using dplyr:
library(dplyr)
a <- data.frame(x=c(1,1,2,4,5,5),y=c(2,3,3,3,6,6))
a %>%
distinct() %>%
add_count(x) %>% # adds an implicit group_by(x)
filter(n == 1) %>%
select(-n)
#> # A tibble: 3 x 2
#> # Groups: x [3]
#> x y
#> <dbl> <dbl>
#> 1 2 3
#> 2 4 3
#> 3 5 6
Created on 2018-11-14 by the reprex package (v0.2.1)
answered Nov 14 '18 at 19:28
amandaamanda
111411
Good point! Will make the edit.
– amanda
Nov 14 '18 at 20:33
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
First we use unique to omit repeated rows with the same x and y values (keeping only one copy of each). Any repeated x values that are left have different y values, so we want to get rid of them. We use the standard way to remove all copies of any duplicated values as in this R-FAQ.
a=data.frame(x=c(1,1,2,4,5,5),y=c(2,3,3,3,6,6))
b = unique(a)
b = b[!duplicated(b$x) & !duplicated(b$x, fromLast = TRUE), ]
b
# x y
# 3 2 3
# 4 4 3
# 5 5 6
Fans of dplyr would probably do it like this, producing the same result.
library(dplyr)
a %>%
group_by(x) %>%
filter(n_distinct(y) == 1) %>%
distinct
answered Nov 14 '18 at 19:23
GregorGregor
66.8k1092177
add a comment |
First we use unique to omit repeated rows with the same x and y values (keeping only one copy of each). Any repeated x values that are left have different y values, so we want to get rid of them. We use the standard way to remove all copies of any duplicated values as in this R-FAQ.
a=data.frame(x=c(1,1,2,4,5,5),y=c(2,3,3,3,6,6))
b = unique(a)
b = b[!duplicated(b$x) & !duplicated(b$x, fromLast = TRUE), ]
b
# x y
# 3 2 3
# 4 4 3
# 5 5 6
Fans of dplyr would probably do it like this, producing the same result.
library(dplyr)
a %>%
group_by(x) %>%
filter(n_distinct(y) == 1) %>%
distinct
answered Nov 14 '18 at 19:23
GregorGregor
66.8k1092177
add a comment |
First we use unique to omit repeated rows with the same x and y values (keeping only one copy of each). Any repeated x values that are left have different y values, so we want to get rid of them. We use the standard way to remove all copies of any duplicated values as in this R-FAQ.
a=data.frame(x=c(1,1,2,4,5,5),y=c(2,3,3,3,6,6))
b = unique(a)
b = b[!duplicated(b$x) & !duplicated(b$x, fromLast = TRUE), ]
b
# x y
# 3 2 3
# 4 4 3
# 5 5 6
Fans of dplyr would probably do it like this, producing the same result.
library(dplyr)
a %>%
group_by(x) %>%
filter(n_distinct(y) == 1) %>%
distinct
answered Nov 14 '18 at 19:23
GregorGregor
66.8k1092177
First we use unique to omit repeated rows with the same x and y values (keeping only one copy of each). Any repeated x values that are left have different y values, so we want to get rid of them. We use the standard way to remove all copies of any duplicated values as in this R-FAQ.
a=data.frame(x=c(1,1,2,4,5,5),y=c(2,3,3,3,6,6))
b = unique(a)
b = b[!duplicated(b$x) & !duplicated(b$x, fromLast = TRUE), ]
b
# x y
# 3 2 3
# 4 4 3
# 5 5 6
Fans of dplyr would probably do it like this, producing the same result.
library(dplyr)
a %>%
group_by(x) %>%
filter(n_distinct(y) == 1) %>%
distinct
answered Nov 14 '18 at 19:23
GregorGregor
66.8k1092177
edited Nov 14 '18 at 19:29
answered Nov 14 '18 at 19:23
GregorGregor
66.8k1092177
answered Nov 14 '18 at 19:23
GregorGregor
66.8k1092177
answered Nov 14 '18 at 19:23
GregorGregor
66.8k1092177
66.8k1092177
add a comment |
add a comment |
Using dplyr:
library(dplyr)
a <- data.frame(x=c(1,1,2,4,5,5),y=c(2,3,3,3,6,6))
a %>%
distinct() %>%
add_count(x) %>% # adds an implicit group_by(x)
filter(n == 1) %>%
select(-n)
#> # A tibble: 3 x 2
#> # Groups: x [3]
#> x y
#> <dbl> <dbl>
#> 1 2 3
#> 2 4 3
#> 3 5 6
Created on 2018-11-14 by the reprex package (v0.2.1)
answered Nov 14 '18 at 19:28
amandaamanda
111411
Good point! Will make the edit.
– amanda
Nov 14 '18 at 20:33
add a comment |
Using dplyr:
library(dplyr)
a <- data.frame(x=c(1,1,2,4,5,5),y=c(2,3,3,3,6,6))
a %>%
distinct() %>%
add_count(x) %>% # adds an implicit group_by(x)
filter(n == 1) %>%
select(-n)
#> # A tibble: 3 x 2
#> # Groups: x [3]
#> x y
#> <dbl> <dbl>
#> 1 2 3
#> 2 4 3
#> 3 5 6
Created on 2018-11-14 by the reprex package (v0.2.1)
answered Nov 14 '18 at 19:28
amandaamanda
111411
Good point! Will make the edit.
– amanda
Nov 14 '18 at 20:33
add a comment |
Using dplyr:
library(dplyr)
a <- data.frame(x=c(1,1,2,4,5,5),y=c(2,3,3,3,6,6))
a %>%
distinct() %>%
add_count(x) %>% # adds an implicit group_by(x)
filter(n == 1) %>%
select(-n)
#> # A tibble: 3 x 2
#> # Groups: x [3]
#> x y
#> <dbl> <dbl>
#> 1 2 3
#> 2 4 3
#> 3 5 6
Created on 2018-11-14 by the reprex package (v0.2.1)
answered Nov 14 '18 at 19:28
amandaamanda
111411
Using dplyr:
library(dplyr)
a <- data.frame(x=c(1,1,2,4,5,5),y=c(2,3,3,3,6,6))
a %>%
distinct() %>%
add_count(x) %>% # adds an implicit group_by(x)
filter(n == 1) %>%
select(-n)
#> # A tibble: 3 x 2
#> # Groups: x [3]
#> x y
#> <dbl> <dbl>
#> 1 2 3
#> 2 4 3
#> 3 5 6
Created on 2018-11-14 by the reprex package (v0.2.1)
answered Nov 14 '18 at 19:28
amandaamanda
111411
edited Nov 14 '18 at 20:34
answered Nov 14 '18 at 19:28
amandaamanda
111411
answered Nov 14 '18 at 19:28
amandaamanda
111411
answered Nov 14 '18 at 19:28
amandaamanda
111411
111411
Good point! Will make the edit.
– amanda
Nov 14 '18 at 20:33
add a comment |
Good point! Will make the edit.
– amanda
Nov 14 '18 at 20:33
Good point! Will make the edit.
– amanda
Nov 14 '18 at 20:33
Good point! Will make the edit.
– amanda
Nov 14 '18 at 20:33
add a comment |
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