Adding Column and column names dynamically in R









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0
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I have an R data.table like this:



id
1: 1
2: 29
3: 26
4: 6
5: 1
6: 14`


And I would like to add columns dynamically by presence of this Id in a row.
Result:



 id is_present_1 ... is_present_6....
1: 1 1 0
2: 2 0 0
3: 3 0 0
4: 4 0 1
5: 5 0 0
6: 6 0 1


I tried to write a function, or using mutate and paste:



ids <- c(1, 2, 3, 4, 5, 6)
for (i in length(ids))
df %>% mutate(paste("is_present",id[i]) = ifelse(id == ids[i],1,0))


I get an error:
Error: unexpected '=' in:
"for (i in length(ids)) {
df %>% mutate(paste("is_present",id[i]) ="



Can someone help with this?










share|improve this question





















  • You need to use tidy evaluation to create dynamic names within dplyr verbs. See some examples here: stackoverflow.com/q/49700912/786542 & maraaverick.rbind.io/2017/08/tidyeval-resource-roundup
    – Tung
    Nov 12 at 7:47














up vote
0
down vote

favorite












I have an R data.table like this:



id
1: 1
2: 29
3: 26
4: 6
5: 1
6: 14`


And I would like to add columns dynamically by presence of this Id in a row.
Result:



 id is_present_1 ... is_present_6....
1: 1 1 0
2: 2 0 0
3: 3 0 0
4: 4 0 1
5: 5 0 0
6: 6 0 1


I tried to write a function, or using mutate and paste:



ids <- c(1, 2, 3, 4, 5, 6)
for (i in length(ids))
df %>% mutate(paste("is_present",id[i]) = ifelse(id == ids[i],1,0))


I get an error:
Error: unexpected '=' in:
"for (i in length(ids)) {
df %>% mutate(paste("is_present",id[i]) ="



Can someone help with this?










share|improve this question





















  • You need to use tidy evaluation to create dynamic names within dplyr verbs. See some examples here: stackoverflow.com/q/49700912/786542 & maraaverick.rbind.io/2017/08/tidyeval-resource-roundup
    – Tung
    Nov 12 at 7:47












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have an R data.table like this:



id
1: 1
2: 29
3: 26
4: 6
5: 1
6: 14`


And I would like to add columns dynamically by presence of this Id in a row.
Result:



 id is_present_1 ... is_present_6....
1: 1 1 0
2: 2 0 0
3: 3 0 0
4: 4 0 1
5: 5 0 0
6: 6 0 1


I tried to write a function, or using mutate and paste:



ids <- c(1, 2, 3, 4, 5, 6)
for (i in length(ids))
df %>% mutate(paste("is_present",id[i]) = ifelse(id == ids[i],1,0))


I get an error:
Error: unexpected '=' in:
"for (i in length(ids)) {
df %>% mutate(paste("is_present",id[i]) ="



Can someone help with this?










share|improve this question













I have an R data.table like this:



id
1: 1
2: 29
3: 26
4: 6
5: 1
6: 14`


And I would like to add columns dynamically by presence of this Id in a row.
Result:



 id is_present_1 ... is_present_6....
1: 1 1 0
2: 2 0 0
3: 3 0 0
4: 4 0 1
5: 5 0 0
6: 6 0 1


I tried to write a function, or using mutate and paste:



ids <- c(1, 2, 3, 4, 5, 6)
for (i in length(ids))
df %>% mutate(paste("is_present",id[i]) = ifelse(id == ids[i],1,0))


I get an error:
Error: unexpected '=' in:
"for (i in length(ids)) {
df %>% mutate(paste("is_present",id[i]) ="



Can someone help with this?







dynamic dplyr data.table






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 11 at 19:31









Jenny

103




103











  • You need to use tidy evaluation to create dynamic names within dplyr verbs. See some examples here: stackoverflow.com/q/49700912/786542 & maraaverick.rbind.io/2017/08/tidyeval-resource-roundup
    – Tung
    Nov 12 at 7:47
















  • You need to use tidy evaluation to create dynamic names within dplyr verbs. See some examples here: stackoverflow.com/q/49700912/786542 & maraaverick.rbind.io/2017/08/tidyeval-resource-roundup
    – Tung
    Nov 12 at 7:47















You need to use tidy evaluation to create dynamic names within dplyr verbs. See some examples here: stackoverflow.com/q/49700912/786542 & maraaverick.rbind.io/2017/08/tidyeval-resource-roundup
– Tung
Nov 12 at 7:47




You need to use tidy evaluation to create dynamic names within dplyr verbs. See some examples here: stackoverflow.com/q/49700912/786542 & maraaverick.rbind.io/2017/08/tidyeval-resource-roundup
– Tung
Nov 12 at 7:47












1 Answer
1






active

oldest

votes

















up vote
0
down vote



accepted










Not sure if I understand the question. But i think you are looking for a map function from purr with dynamic columns names. If the logic is wrong you can just adjust inside the function.



library(tidyverse)
library(data.table)

map_dfc(df$row, function(x)
nm <- paste("is_present_", x, sep = "")
df %>%
mutate(!!nm := ifelse(id == x, 1, 0))) %>%
select(contains("is_present_"))


results in:



 is_present_1 is_present_2 is_present_3 is_present_4 is_present_5 is_present_6
1 1 0 0 0 0 0
2 0 0 0 0 0 0
3 0 0 0 0 0 0
4 0 0 0 0 0 1
5 1 0 0 0 0 0
6 0 0 0 0 0 0


Sample data:



 df <- fread("
id
1: 1
2: 29
3: 26
4: 6
5: 1
6: 14") %>%
select(2) %>%
rownames_to_column("row")





share|improve this answer




















  • Thank you for your help :)
    – Jenny
    Nov 13 at 10:41










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










Not sure if I understand the question. But i think you are looking for a map function from purr with dynamic columns names. If the logic is wrong you can just adjust inside the function.



library(tidyverse)
library(data.table)

map_dfc(df$row, function(x)
nm <- paste("is_present_", x, sep = "")
df %>%
mutate(!!nm := ifelse(id == x, 1, 0))) %>%
select(contains("is_present_"))


results in:



 is_present_1 is_present_2 is_present_3 is_present_4 is_present_5 is_present_6
1 1 0 0 0 0 0
2 0 0 0 0 0 0
3 0 0 0 0 0 0
4 0 0 0 0 0 1
5 1 0 0 0 0 0
6 0 0 0 0 0 0


Sample data:



 df <- fread("
id
1: 1
2: 29
3: 26
4: 6
5: 1
6: 14") %>%
select(2) %>%
rownames_to_column("row")





share|improve this answer




















  • Thank you for your help :)
    – Jenny
    Nov 13 at 10:41














up vote
0
down vote



accepted










Not sure if I understand the question. But i think you are looking for a map function from purr with dynamic columns names. If the logic is wrong you can just adjust inside the function.



library(tidyverse)
library(data.table)

map_dfc(df$row, function(x)
nm <- paste("is_present_", x, sep = "")
df %>%
mutate(!!nm := ifelse(id == x, 1, 0))) %>%
select(contains("is_present_"))


results in:



 is_present_1 is_present_2 is_present_3 is_present_4 is_present_5 is_present_6
1 1 0 0 0 0 0
2 0 0 0 0 0 0
3 0 0 0 0 0 0
4 0 0 0 0 0 1
5 1 0 0 0 0 0
6 0 0 0 0 0 0


Sample data:



 df <- fread("
id
1: 1
2: 29
3: 26
4: 6
5: 1
6: 14") %>%
select(2) %>%
rownames_to_column("row")





share|improve this answer




















  • Thank you for your help :)
    – Jenny
    Nov 13 at 10:41












up vote
0
down vote



accepted







up vote
0
down vote



accepted






Not sure if I understand the question. But i think you are looking for a map function from purr with dynamic columns names. If the logic is wrong you can just adjust inside the function.



library(tidyverse)
library(data.table)

map_dfc(df$row, function(x)
nm <- paste("is_present_", x, sep = "")
df %>%
mutate(!!nm := ifelse(id == x, 1, 0))) %>%
select(contains("is_present_"))


results in:



 is_present_1 is_present_2 is_present_3 is_present_4 is_present_5 is_present_6
1 1 0 0 0 0 0
2 0 0 0 0 0 0
3 0 0 0 0 0 0
4 0 0 0 0 0 1
5 1 0 0 0 0 0
6 0 0 0 0 0 0


Sample data:



 df <- fread("
id
1: 1
2: 29
3: 26
4: 6
5: 1
6: 14") %>%
select(2) %>%
rownames_to_column("row")





share|improve this answer












Not sure if I understand the question. But i think you are looking for a map function from purr with dynamic columns names. If the logic is wrong you can just adjust inside the function.



library(tidyverse)
library(data.table)

map_dfc(df$row, function(x)
nm <- paste("is_present_", x, sep = "")
df %>%
mutate(!!nm := ifelse(id == x, 1, 0))) %>%
select(contains("is_present_"))


results in:



 is_present_1 is_present_2 is_present_3 is_present_4 is_present_5 is_present_6
1 1 0 0 0 0 0
2 0 0 0 0 0 0
3 0 0 0 0 0 0
4 0 0 0 0 0 1
5 1 0 0 0 0 0
6 0 0 0 0 0 0


Sample data:



 df <- fread("
id
1: 1
2: 29
3: 26
4: 6
5: 1
6: 14") %>%
select(2) %>%
rownames_to_column("row")






share|improve this answer












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answered Nov 12 at 18:59









davsjob

49726




49726











  • Thank you for your help :)
    – Jenny
    Nov 13 at 10:41
















  • Thank you for your help :)
    – Jenny
    Nov 13 at 10:41















Thank you for your help :)
– Jenny
Nov 13 at 10:41




Thank you for your help :)
– Jenny
Nov 13 at 10:41

















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