Pass multiple columns in dataframe into function at once in R
0
After much searching, I can't seem to figure this out.
Trying to write a function that:
- takes a data frame, db
- groups the data frame by var1
- returns the mean and sd by group on several different columns
Here is my function,
myfun <- function(db,var1, ...) {
var1 <- enquo(var1)
var2 <- quos(...)
for (i in var2) {
db %>%
group_by(!!var1) %>%
summarise(mean_var = mean(!!!var2))
}}
when I pass the following, nothing returns
myfun(data, group, age, bmi)
Ideally, I would like to group both age and bmi by group and return the mean and sd for each. In the future, I would like to pass many more columns from data into the function...
The output would be similar to summaryBy from doby package, but on many columns at once and would look like:
Group age.mean age.sd
0
1
bmi.mean bmi.sd
0
1
r function for-loop dplyr summarytools
asked Nov 17 '18 at 13:47
mdb_ftlmdb_ftl
52
add a comment |
0
After much searching, I can't seem to figure this out.
Trying to write a function that:
- takes a data frame, db
- groups the data frame by var1
- returns the mean and sd by group on several different columns
Here is my function,
myfun <- function(db,var1, ...) {
var1 <- enquo(var1)
var2 <- quos(...)
for (i in var2) {
db %>%
group_by(!!var1) %>%
summarise(mean_var = mean(!!!var2))
}}
when I pass the following, nothing returns
myfun(data, group, age, bmi)
Ideally, I would like to group both age and bmi by group and return the mean and sd for each. In the future, I would like to pass many more columns from data into the function...
The output would be similar to summaryBy from doby package, but on many columns at once and would look like:
Group age.mean age.sd
0
1
bmi.mean bmi.sd
0
1
r function for-loop dplyr summarytools
asked Nov 17 '18 at 13:47
mdb_ftlmdb_ftl
52
add a comment |
0
0
0
After much searching, I can't seem to figure this out.
Trying to write a function that:
- takes a data frame, db
- groups the data frame by var1
- returns the mean and sd by group on several different columns
Here is my function,
myfun <- function(db,var1, ...) {
var1 <- enquo(var1)
var2 <- quos(...)
for (i in var2) {
db %>%
group_by(!!var1) %>%
summarise(mean_var = mean(!!!var2))
}}
when I pass the following, nothing returns
myfun(data, group, age, bmi)
Ideally, I would like to group both age and bmi by group and return the mean and sd for each. In the future, I would like to pass many more columns from data into the function...
The output would be similar to summaryBy from doby package, but on many columns at once and would look like:
Group age.mean age.sd
0
1
bmi.mean bmi.sd
0
1
r function for-loop dplyr summarytools
asked Nov 17 '18 at 13:47
mdb_ftlmdb_ftl
52
After much searching, I can't seem to figure this out.
Trying to write a function that:
- takes a data frame, db
- groups the data frame by var1
- returns the mean and sd by group on several different columns
Here is my function,
myfun <- function(db,var1, ...) {
var1 <- enquo(var1)
var2 <- quos(...)
for (i in var2) {
db %>%
group_by(!!var1) %>%
summarise(mean_var = mean(!!!var2))
}}
when I pass the following, nothing returns
myfun(data, group, age, bmi)
Ideally, I would like to group both age and bmi by group and return the mean and sd for each. In the future, I would like to pass many more columns from data into the function...
The output would be similar to summaryBy from doby package, but on many columns at once and would look like:
Group age.mean age.sd
0
1
bmi.mean bmi.sd
0
1
r function for-loop dplyr summarytools
r function for-loop dplyr summarytools
asked Nov 17 '18 at 13:47
mdb_ftlmdb_ftl
52
asked Nov 17 '18 at 13:47
mdb_ftlmdb_ftl
52
asked Nov 17 '18 at 13:47
mdb_ftlmdb_ftl
52
asked Nov 17 '18 at 13:47
mdb_ftlmdb_ftl
52
asked Nov 17 '18 at 13:47
mdb_ftlmdb_ftl
52
52
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
2
Your loop appears to be unnecessary (you aren't doing anything with i). Instead, you could use summarize_at to achieve the effect you want:
myfun <- function(db,var1, ...) {
var1 <- enquo(var1)
var2 <- quos(...)
db %>%
group_by(!!var1) %>%
summarise_at(vars(!!!var2), c(mean = mean, sd = sd))
}
And if we test it out with diamonds dataset:
myfun(diamonds, cut, x, z)
cut x_mean z_mean x_sd z_sd
<ord> <dbl> <dbl> <dbl> <dbl>
1 Fair 6.25 3.98 0.964 0.652
2 Good 5.84 3.64 1.06 0.655
3 Very Good 5.74 3.56 1.10 0.730
4 Premium 5.97 3.65 1.19 0.731
5 Ideal 5.51 3.40 1.06 0.658
To get the formatting closer to what you had in mind in your original post, we can use a bit of tidyr magic:
myfun <- function(db,var1, ...) {
var1 <- enquo(var1)
var2 <- quos(...)
db %>%
group_by(!!var1) %>%
summarise_at(vars(!!!var2), c(mean = mean, sd = sd)) %>%
gather(variable, value, -(!!var1)) %>%
separate(variable, c('variable', 'measure'), sep = '_') %>%
spread(measure, value) %>%
arrange(variable, !!var1)
}
cut variable mean sd
<ord> <chr> <dbl> <dbl>
1 Fair x 6.25 0.964
2 Good x 5.84 1.06
3 Very Good x 5.74 1.10
4 Premium x 5.97 1.19
5 Ideal x 5.51 1.06
6 Fair z 3.98 0.652
7 Good z 3.64 0.655
8 Very Good z 3.56 0.730
9 Premium z 3.65 0.731
10 Ideal z 3.40 0.658
answered Nov 17 '18 at 14:05
jdobresjdobres
4,8561522
This solution worked well for me. Can you explain how I might output the mean and sd by group for each column into a format which is similar to the original post? (more vertically aligned?, perhaps each result is outputted as a matrix?). If I input many more columns (x, y, z, a, b, c, etc...) the result is going to very difficult to read horizontally
– mdb_ftl
Nov 17 '18 at 14:14
See updated answer.
– jdobres
Nov 17 '18 at 14:28
This is precisely what I was looking for and works very well. I learned a lot from this answer.
– mdb_ftl
Nov 17 '18 at 14:48
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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2
Your loop appears to be unnecessary (you aren't doing anything with i). Instead, you could use summarize_at to achieve the effect you want:
myfun <- function(db,var1, ...) {
var1 <- enquo(var1)
var2 <- quos(...)
db %>%
group_by(!!var1) %>%
summarise_at(vars(!!!var2), c(mean = mean, sd = sd))
}
And if we test it out with diamonds dataset:
myfun(diamonds, cut, x, z)
cut x_mean z_mean x_sd z_sd
<ord> <dbl> <dbl> <dbl> <dbl>
1 Fair 6.25 3.98 0.964 0.652
2 Good 5.84 3.64 1.06 0.655
3 Very Good 5.74 3.56 1.10 0.730
4 Premium 5.97 3.65 1.19 0.731
5 Ideal 5.51 3.40 1.06 0.658
To get the formatting closer to what you had in mind in your original post, we can use a bit of tidyr magic:
myfun <- function(db,var1, ...) {
var1 <- enquo(var1)
var2 <- quos(...)
db %>%
group_by(!!var1) %>%
summarise_at(vars(!!!var2), c(mean = mean, sd = sd)) %>%
gather(variable, value, -(!!var1)) %>%
separate(variable, c('variable', 'measure'), sep = '_') %>%
spread(measure, value) %>%
arrange(variable, !!var1)
}
cut variable mean sd
<ord> <chr> <dbl> <dbl>
1 Fair x 6.25 0.964
2 Good x 5.84 1.06
3 Very Good x 5.74 1.10
4 Premium x 5.97 1.19
5 Ideal x 5.51 1.06
6 Fair z 3.98 0.652
7 Good z 3.64 0.655
8 Very Good z 3.56 0.730
9 Premium z 3.65 0.731
10 Ideal z 3.40 0.658
answered Nov 17 '18 at 14:05
jdobresjdobres
4,8561522
This solution worked well for me. Can you explain how I might output the mean and sd by group for each column into a format which is similar to the original post? (more vertically aligned?, perhaps each result is outputted as a matrix?). If I input many more columns (x, y, z, a, b, c, etc...) the result is going to very difficult to read horizontally
– mdb_ftl
Nov 17 '18 at 14:14
See updated answer.
– jdobres
Nov 17 '18 at 14:28
This is precisely what I was looking for and works very well. I learned a lot from this answer.
– mdb_ftl
Nov 17 '18 at 14:48
add a comment |
2
Your loop appears to be unnecessary (you aren't doing anything with i). Instead, you could use summarize_at to achieve the effect you want:
myfun <- function(db,var1, ...) {
var1 <- enquo(var1)
var2 <- quos(...)
db %>%
group_by(!!var1) %>%
summarise_at(vars(!!!var2), c(mean = mean, sd = sd))
}
And if we test it out with diamonds dataset:
myfun(diamonds, cut, x, z)
cut x_mean z_mean x_sd z_sd
<ord> <dbl> <dbl> <dbl> <dbl>
1 Fair 6.25 3.98 0.964 0.652
2 Good 5.84 3.64 1.06 0.655
3 Very Good 5.74 3.56 1.10 0.730
4 Premium 5.97 3.65 1.19 0.731
5 Ideal 5.51 3.40 1.06 0.658
To get the formatting closer to what you had in mind in your original post, we can use a bit of tidyr magic:
myfun <- function(db,var1, ...) {
var1 <- enquo(var1)
var2 <- quos(...)
db %>%
group_by(!!var1) %>%
summarise_at(vars(!!!var2), c(mean = mean, sd = sd)) %>%
gather(variable, value, -(!!var1)) %>%
separate(variable, c('variable', 'measure'), sep = '_') %>%
spread(measure, value) %>%
arrange(variable, !!var1)
}
cut variable mean sd
<ord> <chr> <dbl> <dbl>
1 Fair x 6.25 0.964
2 Good x 5.84 1.06
3 Very Good x 5.74 1.10
4 Premium x 5.97 1.19
5 Ideal x 5.51 1.06
6 Fair z 3.98 0.652
7 Good z 3.64 0.655
8 Very Good z 3.56 0.730
9 Premium z 3.65 0.731
10 Ideal z 3.40 0.658
answered Nov 17 '18 at 14:05
jdobresjdobres
4,8561522
This solution worked well for me. Can you explain how I might output the mean and sd by group for each column into a format which is similar to the original post? (more vertically aligned?, perhaps each result is outputted as a matrix?). If I input many more columns (x, y, z, a, b, c, etc...) the result is going to very difficult to read horizontally
– mdb_ftl
Nov 17 '18 at 14:14
See updated answer.
– jdobres
Nov 17 '18 at 14:28
This is precisely what I was looking for and works very well. I learned a lot from this answer.
– mdb_ftl
Nov 17 '18 at 14:48
add a comment |
2
2
2
Your loop appears to be unnecessary (you aren't doing anything with i). Instead, you could use summarize_at to achieve the effect you want:
myfun <- function(db,var1, ...) {
var1 <- enquo(var1)
var2 <- quos(...)
db %>%
group_by(!!var1) %>%
summarise_at(vars(!!!var2), c(mean = mean, sd = sd))
}
And if we test it out with diamonds dataset:
myfun(diamonds, cut, x, z)
cut x_mean z_mean x_sd z_sd
<ord> <dbl> <dbl> <dbl> <dbl>
1 Fair 6.25 3.98 0.964 0.652
2 Good 5.84 3.64 1.06 0.655
3 Very Good 5.74 3.56 1.10 0.730
4 Premium 5.97 3.65 1.19 0.731
5 Ideal 5.51 3.40 1.06 0.658
To get the formatting closer to what you had in mind in your original post, we can use a bit of tidyr magic:
myfun <- function(db,var1, ...) {
var1 <- enquo(var1)
var2 <- quos(...)
db %>%
group_by(!!var1) %>%
summarise_at(vars(!!!var2), c(mean = mean, sd = sd)) %>%
gather(variable, value, -(!!var1)) %>%
separate(variable, c('variable', 'measure'), sep = '_') %>%
spread(measure, value) %>%
arrange(variable, !!var1)
}
cut variable mean sd
<ord> <chr> <dbl> <dbl>
1 Fair x 6.25 0.964
2 Good x 5.84 1.06
3 Very Good x 5.74 1.10
4 Premium x 5.97 1.19
5 Ideal x 5.51 1.06
6 Fair z 3.98 0.652
7 Good z 3.64 0.655
8 Very Good z 3.56 0.730
9 Premium z 3.65 0.731
10 Ideal z 3.40 0.658
answered Nov 17 '18 at 14:05
jdobresjdobres
4,8561522
Your loop appears to be unnecessary (you aren't doing anything with i). Instead, you could use summarize_at to achieve the effect you want:
myfun <- function(db,var1, ...) {
var1 <- enquo(var1)
var2 <- quos(...)
db %>%
group_by(!!var1) %>%
summarise_at(vars(!!!var2), c(mean = mean, sd = sd))
}
And if we test it out with diamonds dataset:
myfun(diamonds, cut, x, z)
cut x_mean z_mean x_sd z_sd
<ord> <dbl> <dbl> <dbl> <dbl>
1 Fair 6.25 3.98 0.964 0.652
2 Good 5.84 3.64 1.06 0.655
3 Very Good 5.74 3.56 1.10 0.730
4 Premium 5.97 3.65 1.19 0.731
5 Ideal 5.51 3.40 1.06 0.658
To get the formatting closer to what you had in mind in your original post, we can use a bit of tidyr magic:
myfun <- function(db,var1, ...) {
var1 <- enquo(var1)
var2 <- quos(...)
db %>%
group_by(!!var1) %>%
summarise_at(vars(!!!var2), c(mean = mean, sd = sd)) %>%
gather(variable, value, -(!!var1)) %>%
separate(variable, c('variable', 'measure'), sep = '_') %>%
spread(measure, value) %>%
arrange(variable, !!var1)
}
cut variable mean sd
<ord> <chr> <dbl> <dbl>
1 Fair x 6.25 0.964
2 Good x 5.84 1.06
3 Very Good x 5.74 1.10
4 Premium x 5.97 1.19
5 Ideal x 5.51 1.06
6 Fair z 3.98 0.652
7 Good z 3.64 0.655
8 Very Good z 3.56 0.730
9 Premium z 3.65 0.731
10 Ideal z 3.40 0.658
answered Nov 17 '18 at 14:05
jdobresjdobres
4,8561522
edited Nov 17 '18 at 14:28
answered Nov 17 '18 at 14:05
jdobresjdobres
4,8561522
answered Nov 17 '18 at 14:05
jdobresjdobres
4,8561522
answered Nov 17 '18 at 14:05
jdobresjdobres
4,8561522
4,8561522
This solution worked well for me. Can you explain how I might output the mean and sd by group for each column into a format which is similar to the original post? (more vertically aligned?, perhaps each result is outputted as a matrix?). If I input many more columns (x, y, z, a, b, c, etc...) the result is going to very difficult to read horizontally
– mdb_ftl
Nov 17 '18 at 14:14
See updated answer.
– jdobres
Nov 17 '18 at 14:28
This is precisely what I was looking for and works very well. I learned a lot from this answer.
– mdb_ftl
Nov 17 '18 at 14:48
add a comment |
This solution worked well for me. Can you explain how I might output the mean and sd by group for each column into a format which is similar to the original post? (more vertically aligned?, perhaps each result is outputted as a matrix?). If I input many more columns (x, y, z, a, b, c, etc...) the result is going to very difficult to read horizontally
– mdb_ftl
Nov 17 '18 at 14:14
See updated answer.
– jdobres
Nov 17 '18 at 14:28
This is precisely what I was looking for and works very well. I learned a lot from this answer.
– mdb_ftl
Nov 17 '18 at 14:48
This solution worked well for me. Can you explain how I might output the mean and sd by group for each column into a format which is similar to the original post? (more vertically aligned?, perhaps each result is outputted as a matrix?). If I input many more columns (x, y, z, a, b, c, etc...) the result is going to very difficult to read horizontally
– mdb_ftl
Nov 17 '18 at 14:14
This solution worked well for me. Can you explain how I might output the mean and sd by group for each column into a format which is similar to the original post? (more vertically aligned?, perhaps each result is outputted as a matrix?). If I input many more columns (x, y, z, a, b, c, etc...) the result is going to very difficult to read horizontally
– mdb_ftl
Nov 17 '18 at 14:14
See updated answer.
– jdobres
Nov 17 '18 at 14:28
See updated answer.
– jdobres
Nov 17 '18 at 14:28
This is precisely what I was looking for and works very well. I learned a lot from this answer.
– mdb_ftl
Nov 17 '18 at 14:48
This is precisely what I was looking for and works very well. I learned a lot from this answer.
– mdb_ftl
Nov 17 '18 at 14:48
add a comment |
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