Dividing columns by particular values using dplyr
0
I have a dataframe like this:
Setting q02_id c_school c_home c_work c_transport c_leisure Country
Rural 11900006 0 5 3 1 1 Vietnam
Rural 11900031 10 5 0 0 0 China
Rural 11900033 0 3 0 0 3 Vietnam
Rural 11900053 0 7 2 0 0 Vietnam
Rural 11900114 3 6 0 0 0 Malaysia
Rural 11900446 0 6 0 0 0 Vietnam
and I would like to divide columns 2, 3, 4, 5, 6 by the total for that particular country.
Doing it in base R is a bit clumsy:
df[df$Country=="Vietnam",][c(3, 4, 5, 6)] = df[df$Country=="Vietnam",][c(3, 4, 5, 6)] / sum(df[df$Country=="Vietnam",][c(3, 4, 5, 6)])
(I think that works).
I'm trying to convert as much of my code as possible to use tidyverse functions. Is there a way of doing the same thing more efficiently using, dplyr, for instance?
Thanks.
r dplyr
edited Nov 23 '18 at 7:31
Marcus Campbell
2,01921027
asked Mar 21 '18 at 10:55
sahwahn
669
add a comment |
0
I have a dataframe like this:
Setting q02_id c_school c_home c_work c_transport c_leisure Country
Rural 11900006 0 5 3 1 1 Vietnam
Rural 11900031 10 5 0 0 0 China
Rural 11900033 0 3 0 0 3 Vietnam
Rural 11900053 0 7 2 0 0 Vietnam
Rural 11900114 3 6 0 0 0 Malaysia
Rural 11900446 0 6 0 0 0 Vietnam
and I would like to divide columns 2, 3, 4, 5, 6 by the total for that particular country.
Doing it in base R is a bit clumsy:
df[df$Country=="Vietnam",][c(3, 4, 5, 6)] = df[df$Country=="Vietnam",][c(3, 4, 5, 6)] / sum(df[df$Country=="Vietnam",][c(3, 4, 5, 6)])
(I think that works).
I'm trying to convert as much of my code as possible to use tidyverse functions. Is there a way of doing the same thing more efficiently using, dplyr, for instance?
Thanks.
r dplyr
edited Nov 23 '18 at 7:31
Marcus Campbell
2,01921027
asked Mar 21 '18 at 10:55
sahwahn
669
divide with total (all columns) for the country, or total for the country per column?
– missuse
Mar 21 '18 at 11:11
add a comment |
0
0
0
I have a dataframe like this:
Setting q02_id c_school c_home c_work c_transport c_leisure Country
Rural 11900006 0 5 3 1 1 Vietnam
Rural 11900031 10 5 0 0 0 China
Rural 11900033 0 3 0 0 3 Vietnam
Rural 11900053 0 7 2 0 0 Vietnam
Rural 11900114 3 6 0 0 0 Malaysia
Rural 11900446 0 6 0 0 0 Vietnam
and I would like to divide columns 2, 3, 4, 5, 6 by the total for that particular country.
Doing it in base R is a bit clumsy:
df[df$Country=="Vietnam",][c(3, 4, 5, 6)] = df[df$Country=="Vietnam",][c(3, 4, 5, 6)] / sum(df[df$Country=="Vietnam",][c(3, 4, 5, 6)])
(I think that works).
I'm trying to convert as much of my code as possible to use tidyverse functions. Is there a way of doing the same thing more efficiently using, dplyr, for instance?
Thanks.
r dplyr
edited Nov 23 '18 at 7:31
Marcus Campbell
2,01921027
asked Mar 21 '18 at 10:55
sahwahn
669
I have a dataframe like this:
Setting q02_id c_school c_home c_work c_transport c_leisure Country
Rural 11900006 0 5 3 1 1 Vietnam
Rural 11900031 10 5 0 0 0 China
Rural 11900033 0 3 0 0 3 Vietnam
Rural 11900053 0 7 2 0 0 Vietnam
Rural 11900114 3 6 0 0 0 Malaysia
Rural 11900446 0 6 0 0 0 Vietnam
and I would like to divide columns 2, 3, 4, 5, 6 by the total for that particular country.
Doing it in base R is a bit clumsy:
df[df$Country=="Vietnam",][c(3, 4, 5, 6)] = df[df$Country=="Vietnam",][c(3, 4, 5, 6)] / sum(df[df$Country=="Vietnam",][c(3, 4, 5, 6)])
(I think that works).
I'm trying to convert as much of my code as possible to use tidyverse functions. Is there a way of doing the same thing more efficiently using, dplyr, for instance?
Thanks.
r dplyr
r dplyr
edited Nov 23 '18 at 7:31
Marcus Campbell
2,01921027
asked Mar 21 '18 at 10:55
sahwahn
669
edited Nov 23 '18 at 7:31
Marcus Campbell
2,01921027
asked Mar 21 '18 at 10:55
sahwahn
669
edited Nov 23 '18 at 7:31
Marcus Campbell
2,01921027
edited Nov 23 '18 at 7:31
Marcus Campbell
2,01921027
edited Nov 23 '18 at 7:31
Marcus Campbell
2,01921027
2,01921027
asked Mar 21 '18 at 10:55
sahwahn
669
asked Mar 21 '18 at 10:55
sahwahn
669
asked Mar 21 '18 at 10:55
sahwahn
669
669
divide with total (all columns) for the country, or total for the country per column?
– missuse
Mar 21 '18 at 11:11
add a comment |
divide with total (all columns) for the country, or total for the country per column?
– missuse
Mar 21 '18 at 11:11
divide with total (all columns) for the country, or total for the country per column?
– missuse
Mar 21 '18 at 11:11
divide with total (all columns) for the country, or total for the country per column?
– missuse
Mar 21 '18 at 11:11
add a comment |
2 Answers
2
active
oldest
votes
0
I trust this is what you are after:
Divide each column by the sum of that column - grouped by Country
library(tidyverse)
df1 %>%
group_by(Country) %>%
mutate_at(vars(c_school: c_leisure), funs(./ sum(.)))
#output
Setting q02_id c_school c_home c_work c_transport c_leisure Country
<fct> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <fct>
1 Rural 11900006 NaN 0.238 0.600 1.00 0.250 Vietnam
2 Rural 11900031 1.00 1.00 NaN NaN NaN China
3 Rural 11900033 NaN 0.143 0 0 0.750 Vietnam
4 Rural 11900053 NaN 0.333 0.400 0 0 Vietnam
5 Rural 11900114 1.00 1.00 NaN NaN NaN Malaysia
6 Rural 11900446 NaN 0.286 0 0 0 Vietnam
or alternatively divide each column by the total sum for each country as in your example (only difference is I used columns 3:7 as I trust you intended.
df1 %>%
mutate(sum = rowSums(.[,3:7])) %>%
group_by(Country) %>%
mutate_at(vars(c_school: c_leisure), funs(./ sum(sum))) %>%
select(-sum)
#output
Setting q02_id c_school c_home c_work c_transport c_leisure Country
<fct> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <fct>
1 Rural 11900006 0 0.161 0.0968 0.0323 0.0323 Vietnam
2 Rural 11900031 0.667 0.333 0 0 0 China
3 Rural 11900033 0 0.0968 0 0 0.0968 Vietnam
4 Rural 11900053 0 0.226 0.0645 0 0 Vietnam
5 Rural 11900114 0.333 0.667 0 0 0 Malaysia
6 Rural 11900446 0 0.194 0 0 0 Vietnam
data:
df1 = read.table(text ="Setting q02_id c_school c_home c_work c_transport c_leisure Country
Rural 11900006 0 5 3 1 1 Vietnam
Rural 11900031 10 5 0 0 0 China
Rural 11900033 0 3 0 0 3 Vietnam
Rural 11900053 0 7 2 0 0 Vietnam
Rural 11900114 3 6 0 0 0 Malaysia
Rural 11900446 0 6 0 0 0 Vietnam", header = T)
answered Mar 21 '18 at 11:06
missuse
11.6k2622
add a comment |
0
I know you ask for tidyverse functions, but this is also a task where the data.table package shines:
library(data.table)
setDT(df)
df[, lapply(.SD, function(x) x / sum(x)), by = Country, .SDcols = 3:7]
Country c_school c_home c_work c_transport c_leisure
1: Vietnam NaN 0.2380952 0.6 1 0.25
2: Vietnam NaN 0.1428571 0.0 0 0.75
3: Vietnam NaN 0.3333333 0.4 0 0.00
4: Vietnam NaN 0.2857143 0.0 0 0.00
5: China 1 1.0000000 NaN NaN NaN
6: Malaysia 1 1.0000000 NaN NaN NaN
answered Nov 23 '18 at 8:29
snoram
6,402831
add a comment |
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2 Answers
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active
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2 Answers
2
active
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oldest
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0
I trust this is what you are after:
Divide each column by the sum of that column - grouped by Country
library(tidyverse)
df1 %>%
group_by(Country) %>%
mutate_at(vars(c_school: c_leisure), funs(./ sum(.)))
#output
Setting q02_id c_school c_home c_work c_transport c_leisure Country
<fct> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <fct>
1 Rural 11900006 NaN 0.238 0.600 1.00 0.250 Vietnam
2 Rural 11900031 1.00 1.00 NaN NaN NaN China
3 Rural 11900033 NaN 0.143 0 0 0.750 Vietnam
4 Rural 11900053 NaN 0.333 0.400 0 0 Vietnam
5 Rural 11900114 1.00 1.00 NaN NaN NaN Malaysia
6 Rural 11900446 NaN 0.286 0 0 0 Vietnam
or alternatively divide each column by the total sum for each country as in your example (only difference is I used columns 3:7 as I trust you intended.
df1 %>%
mutate(sum = rowSums(.[,3:7])) %>%
group_by(Country) %>%
mutate_at(vars(c_school: c_leisure), funs(./ sum(sum))) %>%
select(-sum)
#output
Setting q02_id c_school c_home c_work c_transport c_leisure Country
<fct> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <fct>
1 Rural 11900006 0 0.161 0.0968 0.0323 0.0323 Vietnam
2 Rural 11900031 0.667 0.333 0 0 0 China
3 Rural 11900033 0 0.0968 0 0 0.0968 Vietnam
4 Rural 11900053 0 0.226 0.0645 0 0 Vietnam
5 Rural 11900114 0.333 0.667 0 0 0 Malaysia
6 Rural 11900446 0 0.194 0 0 0 Vietnam
data:
df1 = read.table(text ="Setting q02_id c_school c_home c_work c_transport c_leisure Country
Rural 11900006 0 5 3 1 1 Vietnam
Rural 11900031 10 5 0 0 0 China
Rural 11900033 0 3 0 0 3 Vietnam
Rural 11900053 0 7 2 0 0 Vietnam
Rural 11900114 3 6 0 0 0 Malaysia
Rural 11900446 0 6 0 0 0 Vietnam", header = T)
answered Mar 21 '18 at 11:06
missuse
11.6k2622
add a comment |
0
I trust this is what you are after:
Divide each column by the sum of that column - grouped by Country
library(tidyverse)
df1 %>%
group_by(Country) %>%
mutate_at(vars(c_school: c_leisure), funs(./ sum(.)))
#output
Setting q02_id c_school c_home c_work c_transport c_leisure Country
<fct> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <fct>
1 Rural 11900006 NaN 0.238 0.600 1.00 0.250 Vietnam
2 Rural 11900031 1.00 1.00 NaN NaN NaN China
3 Rural 11900033 NaN 0.143 0 0 0.750 Vietnam
4 Rural 11900053 NaN 0.333 0.400 0 0 Vietnam
5 Rural 11900114 1.00 1.00 NaN NaN NaN Malaysia
6 Rural 11900446 NaN 0.286 0 0 0 Vietnam
or alternatively divide each column by the total sum for each country as in your example (only difference is I used columns 3:7 as I trust you intended.
df1 %>%
mutate(sum = rowSums(.[,3:7])) %>%
group_by(Country) %>%
mutate_at(vars(c_school: c_leisure), funs(./ sum(sum))) %>%
select(-sum)
#output
Setting q02_id c_school c_home c_work c_transport c_leisure Country
<fct> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <fct>
1 Rural 11900006 0 0.161 0.0968 0.0323 0.0323 Vietnam
2 Rural 11900031 0.667 0.333 0 0 0 China
3 Rural 11900033 0 0.0968 0 0 0.0968 Vietnam
4 Rural 11900053 0 0.226 0.0645 0 0 Vietnam
5 Rural 11900114 0.333 0.667 0 0 0 Malaysia
6 Rural 11900446 0 0.194 0 0 0 Vietnam
data:
df1 = read.table(text ="Setting q02_id c_school c_home c_work c_transport c_leisure Country
Rural 11900006 0 5 3 1 1 Vietnam
Rural 11900031 10 5 0 0 0 China
Rural 11900033 0 3 0 0 3 Vietnam
Rural 11900053 0 7 2 0 0 Vietnam
Rural 11900114 3 6 0 0 0 Malaysia
Rural 11900446 0 6 0 0 0 Vietnam", header = T)
answered Mar 21 '18 at 11:06
missuse
11.6k2622
add a comment |
0
0
0
I trust this is what you are after:
Divide each column by the sum of that column - grouped by Country
library(tidyverse)
df1 %>%
group_by(Country) %>%
mutate_at(vars(c_school: c_leisure), funs(./ sum(.)))
#output
Setting q02_id c_school c_home c_work c_transport c_leisure Country
<fct> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <fct>
1 Rural 11900006 NaN 0.238 0.600 1.00 0.250 Vietnam
2 Rural 11900031 1.00 1.00 NaN NaN NaN China
3 Rural 11900033 NaN 0.143 0 0 0.750 Vietnam
4 Rural 11900053 NaN 0.333 0.400 0 0 Vietnam
5 Rural 11900114 1.00 1.00 NaN NaN NaN Malaysia
6 Rural 11900446 NaN 0.286 0 0 0 Vietnam
or alternatively divide each column by the total sum for each country as in your example (only difference is I used columns 3:7 as I trust you intended.
df1 %>%
mutate(sum = rowSums(.[,3:7])) %>%
group_by(Country) %>%
mutate_at(vars(c_school: c_leisure), funs(./ sum(sum))) %>%
select(-sum)
#output
Setting q02_id c_school c_home c_work c_transport c_leisure Country
<fct> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <fct>
1 Rural 11900006 0 0.161 0.0968 0.0323 0.0323 Vietnam
2 Rural 11900031 0.667 0.333 0 0 0 China
3 Rural 11900033 0 0.0968 0 0 0.0968 Vietnam
4 Rural 11900053 0 0.226 0.0645 0 0 Vietnam
5 Rural 11900114 0.333 0.667 0 0 0 Malaysia
6 Rural 11900446 0 0.194 0 0 0 Vietnam
data:
df1 = read.table(text ="Setting q02_id c_school c_home c_work c_transport c_leisure Country
Rural 11900006 0 5 3 1 1 Vietnam
Rural 11900031 10 5 0 0 0 China
Rural 11900033 0 3 0 0 3 Vietnam
Rural 11900053 0 7 2 0 0 Vietnam
Rural 11900114 3 6 0 0 0 Malaysia
Rural 11900446 0 6 0 0 0 Vietnam", header = T)
answered Mar 21 '18 at 11:06
missuse
11.6k2622
I trust this is what you are after:
Divide each column by the sum of that column - grouped by Country
library(tidyverse)
df1 %>%
group_by(Country) %>%
mutate_at(vars(c_school: c_leisure), funs(./ sum(.)))
#output
Setting q02_id c_school c_home c_work c_transport c_leisure Country
<fct> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <fct>
1 Rural 11900006 NaN 0.238 0.600 1.00 0.250 Vietnam
2 Rural 11900031 1.00 1.00 NaN NaN NaN China
3 Rural 11900033 NaN 0.143 0 0 0.750 Vietnam
4 Rural 11900053 NaN 0.333 0.400 0 0 Vietnam
5 Rural 11900114 1.00 1.00 NaN NaN NaN Malaysia
6 Rural 11900446 NaN 0.286 0 0 0 Vietnam
or alternatively divide each column by the total sum for each country as in your example (only difference is I used columns 3:7 as I trust you intended.
df1 %>%
mutate(sum = rowSums(.[,3:7])) %>%
group_by(Country) %>%
mutate_at(vars(c_school: c_leisure), funs(./ sum(sum))) %>%
select(-sum)
#output
Setting q02_id c_school c_home c_work c_transport c_leisure Country
<fct> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <fct>
1 Rural 11900006 0 0.161 0.0968 0.0323 0.0323 Vietnam
2 Rural 11900031 0.667 0.333 0 0 0 China
3 Rural 11900033 0 0.0968 0 0 0.0968 Vietnam
4 Rural 11900053 0 0.226 0.0645 0 0 Vietnam
5 Rural 11900114 0.333 0.667 0 0 0 Malaysia
6 Rural 11900446 0 0.194 0 0 0 Vietnam
data:
df1 = read.table(text ="Setting q02_id c_school c_home c_work c_transport c_leisure Country
Rural 11900006 0 5 3 1 1 Vietnam
Rural 11900031 10 5 0 0 0 China
Rural 11900033 0 3 0 0 3 Vietnam
Rural 11900053 0 7 2 0 0 Vietnam
Rural 11900114 3 6 0 0 0 Malaysia
Rural 11900446 0 6 0 0 0 Vietnam", header = T)
answered Mar 21 '18 at 11:06
missuse
11.6k2622
edited Mar 21 '18 at 11:16
answered Mar 21 '18 at 11:06
missuse
11.6k2622
answered Mar 21 '18 at 11:06
missuse
11.6k2622
answered Mar 21 '18 at 11:06
missuse
11.6k2622
11.6k2622
add a comment |
add a comment |
0
I know you ask for tidyverse functions, but this is also a task where the data.table package shines:
library(data.table)
setDT(df)
df[, lapply(.SD, function(x) x / sum(x)), by = Country, .SDcols = 3:7]
Country c_school c_home c_work c_transport c_leisure
1: Vietnam NaN 0.2380952 0.6 1 0.25
2: Vietnam NaN 0.1428571 0.0 0 0.75
3: Vietnam NaN 0.3333333 0.4 0 0.00
4: Vietnam NaN 0.2857143 0.0 0 0.00
5: China 1 1.0000000 NaN NaN NaN
6: Malaysia 1 1.0000000 NaN NaN NaN
answered Nov 23 '18 at 8:29
snoram
6,402831
add a comment |
0
I know you ask for tidyverse functions, but this is also a task where the data.table package shines:
library(data.table)
setDT(df)
df[, lapply(.SD, function(x) x / sum(x)), by = Country, .SDcols = 3:7]
Country c_school c_home c_work c_transport c_leisure
1: Vietnam NaN 0.2380952 0.6 1 0.25
2: Vietnam NaN 0.1428571 0.0 0 0.75
3: Vietnam NaN 0.3333333 0.4 0 0.00
4: Vietnam NaN 0.2857143 0.0 0 0.00
5: China 1 1.0000000 NaN NaN NaN
6: Malaysia 1 1.0000000 NaN NaN NaN
answered Nov 23 '18 at 8:29
snoram
6,402831
add a comment |
0
0
0
I know you ask for tidyverse functions, but this is also a task where the data.table package shines:
library(data.table)
setDT(df)
df[, lapply(.SD, function(x) x / sum(x)), by = Country, .SDcols = 3:7]
Country c_school c_home c_work c_transport c_leisure
1: Vietnam NaN 0.2380952 0.6 1 0.25
2: Vietnam NaN 0.1428571 0.0 0 0.75
3: Vietnam NaN 0.3333333 0.4 0 0.00
4: Vietnam NaN 0.2857143 0.0 0 0.00
5: China 1 1.0000000 NaN NaN NaN
6: Malaysia 1 1.0000000 NaN NaN NaN
answered Nov 23 '18 at 8:29
snoram
6,402831
I know you ask for tidyverse functions, but this is also a task where the data.table package shines:
library(data.table)
setDT(df)
df[, lapply(.SD, function(x) x / sum(x)), by = Country, .SDcols = 3:7]
Country c_school c_home c_work c_transport c_leisure
1: Vietnam NaN 0.2380952 0.6 1 0.25
2: Vietnam NaN 0.1428571 0.0 0 0.75
3: Vietnam NaN 0.3333333 0.4 0 0.00
4: Vietnam NaN 0.2857143 0.0 0 0.00
5: China 1 1.0000000 NaN NaN NaN
6: Malaysia 1 1.0000000 NaN NaN NaN
answered Nov 23 '18 at 8:29
snoram
6,402831
answered Nov 23 '18 at 8:29
snoram
6,402831
answered Nov 23 '18 at 8:29
snoram
6,402831
answered Nov 23 '18 at 8:29
snoram
6,402831
6,402831
add a comment |
add a comment |
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divide with total (all columns) for the country, or total for the country per column?
– missuse
Mar 21 '18 at 11:11