calculate new column in dataframe or list using a function with column as param
1
I'm trying to calculate a new column with a user defined function that needs data from same row and a fixed value valid for all rows:
myfunc <- function(ds,colname,val1,col1,col2){
# content of new column <colname> should be computed from:
ds[colname] = val1 + ds[col1] * ds[col2] # for each row of ds
return(ds)
}
v1 = 2
data(mtcars)
mt = head(mtcars)
mt
mpg cyl disp hp drat wt qsec vs am gear
carb
Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
apply(mt,'newcol',v1,mt$wt,mt$qsec)
mt
What I would like to see in mt$newcol in first row is: 2 + 2.620 * 16.46 (-> 45.12) and all other rows similiar.
So, how can I send a fixed value (v1) and two values from each row to my function and store returned value in this row in a new column?
Thanks
r
asked Nov 10 at 13:32
Tom
102
add a comment |
1
I'm trying to calculate a new column with a user defined function that needs data from same row and a fixed value valid for all rows:
myfunc <- function(ds,colname,val1,col1,col2){
# content of new column <colname> should be computed from:
ds[colname] = val1 + ds[col1] * ds[col2] # for each row of ds
return(ds)
}
v1 = 2
data(mtcars)
mt = head(mtcars)
mt
mpg cyl disp hp drat wt qsec vs am gear
carb
Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
apply(mt,'newcol',v1,mt$wt,mt$qsec)
mt
What I would like to see in mt$newcol in first row is: 2 + 2.620 * 16.46 (-> 45.12) and all other rows similiar.
So, how can I send a fixed value (v1) and two values from each row to my function and store returned value in this row in a new column?
Thanks
r
asked Nov 10 at 13:32
Tom
102
1
Why don't you simply saymtcars$newcol <- 2 + mtcars$weight * mtcars$qsec?
– vaettchen
Nov 10 at 14:31
@vaettchen: This is just an example. I have a function that gets more information from a server depending on content of cells in each row.
– Tom
Nov 10 at 16:40
add a comment |
1
1
1
I'm trying to calculate a new column with a user defined function that needs data from same row and a fixed value valid for all rows:
myfunc <- function(ds,colname,val1,col1,col2){
# content of new column <colname> should be computed from:
ds[colname] = val1 + ds[col1] * ds[col2] # for each row of ds
return(ds)
}
v1 = 2
data(mtcars)
mt = head(mtcars)
mt
mpg cyl disp hp drat wt qsec vs am gear
carb
Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
apply(mt,'newcol',v1,mt$wt,mt$qsec)
mt
What I would like to see in mt$newcol in first row is: 2 + 2.620 * 16.46 (-> 45.12) and all other rows similiar.
So, how can I send a fixed value (v1) and two values from each row to my function and store returned value in this row in a new column?
Thanks
r
asked Nov 10 at 13:32
Tom
102
I'm trying to calculate a new column with a user defined function that needs data from same row and a fixed value valid for all rows:
myfunc <- function(ds,colname,val1,col1,col2){
# content of new column <colname> should be computed from:
ds[colname] = val1 + ds[col1] * ds[col2] # for each row of ds
return(ds)
}
v1 = 2
data(mtcars)
mt = head(mtcars)
mt
mpg cyl disp hp drat wt qsec vs am gear
carb
Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
apply(mt,'newcol',v1,mt$wt,mt$qsec)
mt
What I would like to see in mt$newcol in first row is: 2 + 2.620 * 16.46 (-> 45.12) and all other rows similiar.
So, how can I send a fixed value (v1) and two values from each row to my function and store returned value in this row in a new column?
Thanks
r
r
asked Nov 10 at 13:32
Tom
102
asked Nov 10 at 13:32
Tom
102
asked Nov 10 at 13:32
Tom
102
asked Nov 10 at 13:32
Tom
102
asked Nov 10 at 13:32
Tom
102
102
1
Why don't you simply saymtcars$newcol <- 2 + mtcars$weight * mtcars$qsec?
– vaettchen
Nov 10 at 14:31
@vaettchen: This is just an example. I have a function that gets more information from a server depending on content of cells in each row.
– Tom
Nov 10 at 16:40
add a comment |
1
Why don't you simply saymtcars$newcol <- 2 + mtcars$weight * mtcars$qsec?
– vaettchen
Nov 10 at 14:31
@vaettchen: This is just an example. I have a function that gets more information from a server depending on content of cells in each row.
– Tom
Nov 10 at 16:40
1
1
Why don't you simply say
mtcars$newcol <- 2 + mtcars$weight * mtcars$qsec ?– vaettchen
Nov 10 at 14:31
Why don't you simply say
mtcars$newcol <- 2 + mtcars$weight * mtcars$qsec ?– vaettchen
Nov 10 at 14:31
@vaettchen: This is just an example. I have a function that gets more information from a server depending on content of cells in each row.
– Tom
Nov 10 at 16:40
@vaettchen: This is just an example. I have a function that gets more information from a server depending on content of cells in each row.
– Tom
Nov 10 at 16:40
add a comment |
1 Answer
1
active
oldest
votes
1
dplyr approach:
library(dplyr)
data(mtcars)
myfunc <- function(ds, new_column, val1, col1, col2){
name <- rownames(ds)
ds <- ds %>%
mutate(!!as.name(new_column) := val1 + !!as.name(col1) + !!as.name(col2),
car_name = name) %>%
select(car_name, mpg:!!as.name(new_column))
return(ds)
}
head(
myfunc(ds = mtcars,
new_column = "new_column",
val1 = 2,
col1 = "hp",
col2 = "vs")
)
output
car_name mpg cyl disp hp drat wt qsec vs am gear carb new_column
1 Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4 112
2 Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4 112
3 Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1 96
4 Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1 113
5 Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2 177
6 Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1 108
answered Nov 10 at 13:43
Aleksandr
1,376716
Great, it almost works. I changed your function to ... := val1 + ... because val1 is not a column but a fixed value. But why do we loose our first column (car names)?
– Tom
Nov 10 at 14:21
@Tom car names are not column. It isrownames(), and tidy data structure does not treat it.dplyrwould erase it. If you want it, there is a function likerownames_to_column()
– Blended
Nov 10 at 14:45
@Tom , I updated the code. Now the car names in a separate column. HTH
– Aleksandr
Nov 10 at 15:57
1
Thanks, works as expected. I think I should learn more about dplyr.
– Tom
Nov 10 at 16:41
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53239473%2fcalculate-new-column-in-dataframe-or-list-using-a-function-with-column-as-param%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
dplyr approach:
library(dplyr)
data(mtcars)
myfunc <- function(ds, new_column, val1, col1, col2){
name <- rownames(ds)
ds <- ds %>%
mutate(!!as.name(new_column) := val1 + !!as.name(col1) + !!as.name(col2),
car_name = name) %>%
select(car_name, mpg:!!as.name(new_column))
return(ds)
}
head(
myfunc(ds = mtcars,
new_column = "new_column",
val1 = 2,
col1 = "hp",
col2 = "vs")
)
output
car_name mpg cyl disp hp drat wt qsec vs am gear carb new_column
1 Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4 112
2 Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4 112
3 Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1 96
4 Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1 113
5 Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2 177
6 Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1 108
answered Nov 10 at 13:43
Aleksandr
1,376716
Great, it almost works. I changed your function to ... := val1 + ... because val1 is not a column but a fixed value. But why do we loose our first column (car names)?
– Tom
Nov 10 at 14:21
@Tom car names are not column. It isrownames(), and tidy data structure does not treat it.dplyrwould erase it. If you want it, there is a function likerownames_to_column()
– Blended
Nov 10 at 14:45
@Tom , I updated the code. Now the car names in a separate column. HTH
– Aleksandr
Nov 10 at 15:57
1
Thanks, works as expected. I think I should learn more about dplyr.
– Tom
Nov 10 at 16:41
add a comment |
1
dplyr approach:
library(dplyr)
data(mtcars)
myfunc <- function(ds, new_column, val1, col1, col2){
name <- rownames(ds)
ds <- ds %>%
mutate(!!as.name(new_column) := val1 + !!as.name(col1) + !!as.name(col2),
car_name = name) %>%
select(car_name, mpg:!!as.name(new_column))
return(ds)
}
head(
myfunc(ds = mtcars,
new_column = "new_column",
val1 = 2,
col1 = "hp",
col2 = "vs")
)
output
car_name mpg cyl disp hp drat wt qsec vs am gear carb new_column
1 Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4 112
2 Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4 112
3 Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1 96
4 Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1 113
5 Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2 177
6 Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1 108
answered Nov 10 at 13:43
Aleksandr
1,376716
Great, it almost works. I changed your function to ... := val1 + ... because val1 is not a column but a fixed value. But why do we loose our first column (car names)?
– Tom
Nov 10 at 14:21
@Tom car names are not column. It isrownames(), and tidy data structure does not treat it.dplyrwould erase it. If you want it, there is a function likerownames_to_column()
– Blended
Nov 10 at 14:45
@Tom , I updated the code. Now the car names in a separate column. HTH
– Aleksandr
Nov 10 at 15:57
1
Thanks, works as expected. I think I should learn more about dplyr.
– Tom
Nov 10 at 16:41
add a comment |
1
1
1
dplyr approach:
library(dplyr)
data(mtcars)
myfunc <- function(ds, new_column, val1, col1, col2){
name <- rownames(ds)
ds <- ds %>%
mutate(!!as.name(new_column) := val1 + !!as.name(col1) + !!as.name(col2),
car_name = name) %>%
select(car_name, mpg:!!as.name(new_column))
return(ds)
}
head(
myfunc(ds = mtcars,
new_column = "new_column",
val1 = 2,
col1 = "hp",
col2 = "vs")
)
output
car_name mpg cyl disp hp drat wt qsec vs am gear carb new_column
1 Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4 112
2 Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4 112
3 Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1 96
4 Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1 113
5 Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2 177
6 Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1 108
answered Nov 10 at 13:43
Aleksandr
1,376716
dplyr approach:
library(dplyr)
data(mtcars)
myfunc <- function(ds, new_column, val1, col1, col2){
name <- rownames(ds)
ds <- ds %>%
mutate(!!as.name(new_column) := val1 + !!as.name(col1) + !!as.name(col2),
car_name = name) %>%
select(car_name, mpg:!!as.name(new_column))
return(ds)
}
head(
myfunc(ds = mtcars,
new_column = "new_column",
val1 = 2,
col1 = "hp",
col2 = "vs")
)
output
car_name mpg cyl disp hp drat wt qsec vs am gear carb new_column
1 Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4 112
2 Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4 112
3 Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1 96
4 Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1 113
5 Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2 177
6 Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1 108
answered Nov 10 at 13:43
Aleksandr
1,376716
edited Nov 10 at 15:56
answered Nov 10 at 13:43
Aleksandr
1,376716
answered Nov 10 at 13:43
Aleksandr
1,376716
answered Nov 10 at 13:43
Aleksandr
1,376716
1,376716
Great, it almost works. I changed your function to ... := val1 + ... because val1 is not a column but a fixed value. But why do we loose our first column (car names)?
– Tom
Nov 10 at 14:21
@Tom car names are not column. It isrownames(), and tidy data structure does not treat it.dplyrwould erase it. If you want it, there is a function likerownames_to_column()
– Blended
Nov 10 at 14:45
@Tom , I updated the code. Now the car names in a separate column. HTH
– Aleksandr
Nov 10 at 15:57
1
Thanks, works as expected. I think I should learn more about dplyr.
– Tom
Nov 10 at 16:41
add a comment |
Great, it almost works. I changed your function to ... := val1 + ... because val1 is not a column but a fixed value. But why do we loose our first column (car names)?
– Tom
Nov 10 at 14:21
@Tom car names are not column. It isrownames(), and tidy data structure does not treat it.dplyrwould erase it. If you want it, there is a function likerownames_to_column()
– Blended
Nov 10 at 14:45
@Tom , I updated the code. Now the car names in a separate column. HTH
– Aleksandr
Nov 10 at 15:57
1
Thanks, works as expected. I think I should learn more about dplyr.
– Tom
Nov 10 at 16:41
Great, it almost works. I changed your function to ... := val1 + ... because val1 is not a column but a fixed value. But why do we loose our first column (car names)?
– Tom
Nov 10 at 14:21
Great, it almost works. I changed your function to ... := val1 + ... because val1 is not a column but a fixed value. But why do we loose our first column (car names)?
– Tom
Nov 10 at 14:21
@Tom car names are not column. It is
rownames(), and tidy data structure does not treat it. dplyr would erase it. If you want it, there is a function like rownames_to_column()– Blended
Nov 10 at 14:45
@Tom car names are not column. It is
rownames(), and tidy data structure does not treat it. dplyr would erase it. If you want it, there is a function like rownames_to_column()– Blended
Nov 10 at 14:45
@Tom , I updated the code. Now the car names in a separate column. HTH
– Aleksandr
Nov 10 at 15:57
@Tom , I updated the code. Now the car names in a separate column. HTH
– Aleksandr
Nov 10 at 15:57
1
1
Thanks, works as expected. I think I should learn more about dplyr.
– Tom
Nov 10 at 16:41
Thanks, works as expected. I think I should learn more about dplyr.
– Tom
Nov 10 at 16:41
add a comment |
draft saved
draft discarded
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53239473%2fcalculate-new-column-in-dataframe-or-list-using-a-function-with-column-as-param%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
Why don't you simply say
mtcars$newcol <- 2 + mtcars$weight * mtcars$qsec?– vaettchen
Nov 10 at 14:31
@vaettchen: This is just an example. I have a function that gets more information from a server depending on content of cells in each row.
– Tom
Nov 10 at 16:40