Replacing values across a data frame's variables from a list of nested data frames
Let's presume I have the following dataframe:
df <- data.frame(x = rnorm(10), y = rnorm(10), z = rnorm(10))
And I would like to replace the values in the variables by their corresponding data frame and variable names in the following list:
replace_df <- list(x = data.frame(x = 1:10),
y = data.frame(y = 11:20),
z = data.frame(z = 21:30))
How would I do that using dplyr?
I feel like my issue is related to this Q&A, but I haven't been able to implement the answers to that question correctly to my situation.
I've attempted the below, among others, without success:
library(tidyverse)
variables <- c("x", "y", "z")
df %>%
mutate_at(vars(variables), funs(replace_df[[.]][[.]]))
The "dumb" way would be the following:
df %>%
mutate(x = replace_df[["x"]][["x"]],
y = replace_df[["y"]][["y"]],
z = replace_df[["z"]][["z"]])
r dplyr rlang
asked Nov 21 '18 at 21:35
PhilPhil
1,774628
add a comment |
Let's presume I have the following dataframe:
df <- data.frame(x = rnorm(10), y = rnorm(10), z = rnorm(10))
And I would like to replace the values in the variables by their corresponding data frame and variable names in the following list:
replace_df <- list(x = data.frame(x = 1:10),
y = data.frame(y = 11:20),
z = data.frame(z = 21:30))
How would I do that using dplyr?
I feel like my issue is related to this Q&A, but I haven't been able to implement the answers to that question correctly to my situation.
I've attempted the below, among others, without success:
library(tidyverse)
variables <- c("x", "y", "z")
df %>%
mutate_at(vars(variables), funs(replace_df[[.]][[.]]))
The "dumb" way would be the following:
df %>%
mutate(x = replace_df[["x"]][["x"]],
y = replace_df[["y"]][["y"]],
z = replace_df[["z"]][["z"]])
r dplyr rlang
asked Nov 21 '18 at 21:35
PhilPhil
1,774628
add a comment |
Let's presume I have the following dataframe:
df <- data.frame(x = rnorm(10), y = rnorm(10), z = rnorm(10))
And I would like to replace the values in the variables by their corresponding data frame and variable names in the following list:
replace_df <- list(x = data.frame(x = 1:10),
y = data.frame(y = 11:20),
z = data.frame(z = 21:30))
How would I do that using dplyr?
I feel like my issue is related to this Q&A, but I haven't been able to implement the answers to that question correctly to my situation.
I've attempted the below, among others, without success:
library(tidyverse)
variables <- c("x", "y", "z")
df %>%
mutate_at(vars(variables), funs(replace_df[[.]][[.]]))
The "dumb" way would be the following:
df %>%
mutate(x = replace_df[["x"]][["x"]],
y = replace_df[["y"]][["y"]],
z = replace_df[["z"]][["z"]])
r dplyr rlang
asked Nov 21 '18 at 21:35
PhilPhil
1,774628
Let's presume I have the following dataframe:
df <- data.frame(x = rnorm(10), y = rnorm(10), z = rnorm(10))
And I would like to replace the values in the variables by their corresponding data frame and variable names in the following list:
replace_df <- list(x = data.frame(x = 1:10),
y = data.frame(y = 11:20),
z = data.frame(z = 21:30))
How would I do that using dplyr?
I feel like my issue is related to this Q&A, but I haven't been able to implement the answers to that question correctly to my situation.
I've attempted the below, among others, without success:
library(tidyverse)
variables <- c("x", "y", "z")
df %>%
mutate_at(vars(variables), funs(replace_df[[.]][[.]]))
The "dumb" way would be the following:
df %>%
mutate(x = replace_df[["x"]][["x"]],
y = replace_df[["y"]][["y"]],
z = replace_df[["z"]][["z"]])
r dplyr rlang
r dplyr rlang
asked Nov 21 '18 at 21:35
PhilPhil
1,774628
asked Nov 21 '18 at 21:35
PhilPhil
1,774628
asked Nov 21 '18 at 21:35
PhilPhil
1,774628
asked Nov 21 '18 at 21:35
PhilPhil
1,774628
asked Nov 21 '18 at 21:35
PhilPhil
1,774628
1,774628
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
You need to use expr! I am not sure if the subsetting will work as you tried above, but I was able to get the correct output by making a simple function and passing in an argument that was wrapped in expr()
df <- data.frame(x = rnorm(10), y = rnorm(10), z = rnorm(10))
replace_df <- list(x = data.frame(x = 1:10),
y = data.frame(y = 11:20),
z = data.frame(z = 21:30))
my_func <- function(string) {
return(
replace_df[[string]][[string]]
)
}
df %>%
mutate_at(vars(x, y, z), funs(my_func(expr(.))))
answered Nov 21 '18 at 22:31
buchmaynebuchmayne
10917
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
});
}
});
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%2f53420793%2freplacing-values-across-a-data-frames-variables-from-a-list-of-nested-data-fram%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
You need to use expr! I am not sure if the subsetting will work as you tried above, but I was able to get the correct output by making a simple function and passing in an argument that was wrapped in expr()
df <- data.frame(x = rnorm(10), y = rnorm(10), z = rnorm(10))
replace_df <- list(x = data.frame(x = 1:10),
y = data.frame(y = 11:20),
z = data.frame(z = 21:30))
my_func <- function(string) {
return(
replace_df[[string]][[string]]
)
}
df %>%
mutate_at(vars(x, y, z), funs(my_func(expr(.))))
answered Nov 21 '18 at 22:31
buchmaynebuchmayne
10917
add a comment |
You need to use expr! I am not sure if the subsetting will work as you tried above, but I was able to get the correct output by making a simple function and passing in an argument that was wrapped in expr()
df <- data.frame(x = rnorm(10), y = rnorm(10), z = rnorm(10))
replace_df <- list(x = data.frame(x = 1:10),
y = data.frame(y = 11:20),
z = data.frame(z = 21:30))
my_func <- function(string) {
return(
replace_df[[string]][[string]]
)
}
df %>%
mutate_at(vars(x, y, z), funs(my_func(expr(.))))
answered Nov 21 '18 at 22:31
buchmaynebuchmayne
10917
add a comment |
You need to use expr! I am not sure if the subsetting will work as you tried above, but I was able to get the correct output by making a simple function and passing in an argument that was wrapped in expr()
df <- data.frame(x = rnorm(10), y = rnorm(10), z = rnorm(10))
replace_df <- list(x = data.frame(x = 1:10),
y = data.frame(y = 11:20),
z = data.frame(z = 21:30))
my_func <- function(string) {
return(
replace_df[[string]][[string]]
)
}
df %>%
mutate_at(vars(x, y, z), funs(my_func(expr(.))))
answered Nov 21 '18 at 22:31
buchmaynebuchmayne
10917
You need to use expr! I am not sure if the subsetting will work as you tried above, but I was able to get the correct output by making a simple function and passing in an argument that was wrapped in expr()
df <- data.frame(x = rnorm(10), y = rnorm(10), z = rnorm(10))
replace_df <- list(x = data.frame(x = 1:10),
y = data.frame(y = 11:20),
z = data.frame(z = 21:30))
my_func <- function(string) {
return(
replace_df[[string]][[string]]
)
}
df %>%
mutate_at(vars(x, y, z), funs(my_func(expr(.))))
answered Nov 21 '18 at 22:31
buchmaynebuchmayne
10917
answered Nov 21 '18 at 22:31
buchmaynebuchmayne
10917
answered Nov 21 '18 at 22:31
buchmaynebuchmayne
10917
answered Nov 21 '18 at 22:31
buchmaynebuchmayne
10917
10917
add a comment |
add a comment |
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.
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%2f53420793%2freplacing-values-across-a-data-frames-variables-from-a-list-of-nested-data-fram%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
