Reputation: 67
I have a df in the following format:
name | other_info | revenues_2015 | ebitda_2015 | ebitda_2016 | revenues_2015 | other_2017 |
---|---|---|---|---|---|---|
A | Info1 | 1 | 2 | 3 | 4 | 5 |
B | Info2 | 6 | 7 | 8 | 9 | 10 |
C | Info3 | 11 | 12 | 13 | 14 | 15 |
I would like to change it to long format where I have it structured in the following way:
Name | Info | Year | Metric name | Value
Can you show me how to do that in R? Since the real dataframe has more than 300 columns, is there a way to automate the creation of the year column?
Data:
structure(list(name = structure(1:3, .Label = c("A", "B", "C"
), class = "factor"), other_info = structure(1:3, .Label = c("Info1",
"Info2", "Info3"), class = "factor"), revenues_2015 = structure(c(1L,
3L, 2L), .Label = c("1", "11", "6"), class = "factor"), ebitda_2015 = structure(c(2L,
3L, 1L), .Label = c("12", "2", "7"), class = "factor"), ebitda_2016 = structure(c(2L,
3L, 1L), .Label = c("13", "3", "8"), class = "factor"), revenues_2015 = structure(c(2L,
3L, 1L), .Label = c("14", "4", "9"), class = "factor"), other_2017 = structure(c(3L,
1L, 2L), .Label = c("10", "15", "5"), class = "factor")), class = "data.frame", row.names = c(NA,
-3L))
Upvotes: 1
Views: 63
Reputation: 124
You have two option, you can use the utils package (base-r functions, you do not have to call it using library()) or the melt function from reshape2 package.
With the function reshape():
data = structure(list(name = structure(1:3, .Label = c("A", "B", "C"
), class = "factor"), other_info = structure(1:3, .Label = c("Info1",
"Info2", "Info3"), class = "factor"), revenues_2015 = structure(c(1L,
3L, 2L), .Label = c("1", "11", "6"), class = "factor"), ebitda_2015 = structure(c(2L,
3L, 1L), .Label = c("12", "2", "7"), class = "factor"), ebitda_2016 = structure(c(2L,
3L, 1L), .Label = c("13", "3", "8"), class = "factor"), revenues_2015 = structure(c(2L,
3L, 1L), .Label = c("14", "4", "9"), class = "factor"), other_2017 = structure(c(3L,
1L, 2L), .Label = c("10", "15", "5"), class = "factor")), class = "data.frame", row.names = c(NA,
-3L))
LF_data = reshape(data=data, idvar = c("name","other_info"), varying =c("revenues_2015","ebitda_2015","ebitda_2016","revenues_2015","other_2017"),
v.names = c("Value"),times=c("revenues_2015","ebitda_2015","ebitda_2016","revenues_2015","other_2017"), direction = "long")
Using the package reshape2 melt() function:
data=data.frame(structure(list(name = structure(1:3, .Label = c("A", "B", "C"
), class = "factor"), other_info = structure(1:3, .Label = c("Info1",
"Info2", "Info3"), class = "factor"), revenues_2015 = structure(c(1L,
3L, 2L), .Label = c("1", "11", "6"), class = "factor"), ebitda_2015 = structure(c(2L,
3L, 1L), .Label = c("12", "2", "7"), class = "factor"), ebitda_2016 = structure(c(2L,
3L, 1L), .Label = c("13", "3", "8"), class = "factor"), revenues_2015 = structure(c(2L,
3L, 1L), .Label = c("14", "4", "9"), class = "factor"), other_2017 = structure(c(3L,
1L, 2L), .Label = c("10", "15", "5"), class = "factor")), class = "data.frame", row.names = c(NA,
-3L)),stringsAsFactors=False)
2. Then:
LF_data=reshape2::melt(data,id.vars=c("name","other_info"), mesure.vars=c("revenues_2015","ebitda_2015","ebitda_2016","revenues_2015","other_2017"))
melt wont let you have a combination of "name","other_info" and "variable" unless they are unique. In your example it would change the second triplets of revenues_2015 to revenues_2015.1
Upvotes: 1
Reputation: 78907
A little too late: Similar to the-mad-statter solution. Slightly different using mutate:
library(tidyr)
library(dplyr)
df <- structure(list(name = structure(1:3, .Label = c("A", "B", "C"
), class = "factor"), other_info = structure(1:3, .Label = c("Info1",
"Info2", "Info3"), class = "factor"), revenues_2015 = structure(c(1L,
3L, 2L), .Label = c("1", "11", "6"), class = "factor"), ebitda_2015 = structure(c(2L,
3L, 1L), .Label = c("12", "2", "7"), class = "factor"), ebitda_2016 = structure(c(2L,
3L, 1L), .Label = c("13", "3", "8"), class = "factor"), revenues_2015 = structure(c(2L,
3L, 1L), .Label = c("14", "4", "9"), class = "factor"), other_2017 = structure(c(3L,
1L, 2L), .Label = c("10", "15", "5"), class = "factor")), class = "data.frame", row.names = c(NA, -3L)) %>%
pivot_longer(revenues_2015:other_2017, names_to = c("Metric name", "Year"),
names_sep ="_", values_to = "Value") %>%
dplyr::mutate(Year = stringr::str_remove(Year, "\\D")) %>%
rename(Name=name, Info = other_info)
Upvotes: 0
Reputation: 8676
Does this work for you?
library(dplyr)
library(tidyr)
structure(list(name = structure(1:3, .Label = c("A", "B", "C"
), class = "factor"), other_info = structure(1:3, .Label = c("Info1",
"Info2", "Info3"), class = "factor"), revenues_2015 = structure(c(1L,
3L, 2L), .Label = c("1", "11", "6"), class = "factor"), ebitda_2015 = structure(c(2L,
3L, 1L), .Label = c("12", "2", "7"), class = "factor"), ebitda_2016 = structure(c(2L,
3L, 1L), .Label = c("13", "3", "8"), class = "factor"), revenues_2015 = structure(c(2L,
3L, 1L), .Label = c("14", "4", "9"), class = "factor"), other_2017 = structure(c(3L,
1L, 2L), .Label = c("10", "15", "5"), class = "factor")), class = "data.frame", row.names = c(NA,
-3L)) %>%
pivot_longer(revenues_2015:other_2017, names_pattern = "(.+)_(\\d{4})", names_to = c("metric", "year"))
Upvotes: 1