Struggling to adapt code from stackoverflow to data set - Adding Regression Line Equation and R2 on graph
0
I found a function that adds a regression line equation on graph.
Adding Regression Line Equation and R2 on graph
If I copy the function and data sample, everything works fine. But when I try to adapt the code, I get the message, that Y can't be found.
The Function:
lm_eqn <- function(df){
m <- lm(y ~ x, df);
eq <- substitute(italic(y) == a + b %.% italic(x)*","~~italic(r)^2~"="~r2,
list(a = format(coef(m)[1], digits = 2),
b = format(coef(m)[2], digits = 2),
r2 = format(summary(m)$r.squared, digits = 3)))
as.character(as.expression(eq));
}
My data sample and adapted code:
library(tidyverse)
set.seed(1)
id <- 1:5000
zip <- sample(100:200, 5000, replace = TRUE)
outcome <- rbinom(5000, 1, 0.23)
df <- data.frame(id, outcome, zip) %>% as_tibble()
new_df <- df %>% group_by(zip) %>% summarise(ratio = mean(outcome))
library(ggplot2)
t <- ggplot(data = new_df, aes(x = zip, y = ratio)) +
geom_smooth(method = "lm", se=FALSE, color="black", formula = y ~ x) +
geom_point()
t1 <- t + geom_text(x = 25, y = 300, label = lm_eqn(df), parse = TRUE)
Error in eval(predvars, data, env) : object 'y' not found
Where did I make the mistake? Thanks in advance!
r
edited Nov 28 '18 at 21:22
MrFlick
124k11141173
asked Nov 28 '18 at 21:01
SchillerlockeSchillerlocke
246
add a comment |
0
I found a function that adds a regression line equation on graph.
Adding Regression Line Equation and R2 on graph
If I copy the function and data sample, everything works fine. But when I try to adapt the code, I get the message, that Y can't be found.
The Function:
lm_eqn <- function(df){
m <- lm(y ~ x, df);
eq <- substitute(italic(y) == a + b %.% italic(x)*","~~italic(r)^2~"="~r2,
list(a = format(coef(m)[1], digits = 2),
b = format(coef(m)[2], digits = 2),
r2 = format(summary(m)$r.squared, digits = 3)))
as.character(as.expression(eq));
}
My data sample and adapted code:
library(tidyverse)
set.seed(1)
id <- 1:5000
zip <- sample(100:200, 5000, replace = TRUE)
outcome <- rbinom(5000, 1, 0.23)
df <- data.frame(id, outcome, zip) %>% as_tibble()
new_df <- df %>% group_by(zip) %>% summarise(ratio = mean(outcome))
library(ggplot2)
t <- ggplot(data = new_df, aes(x = zip, y = ratio)) +
geom_smooth(method = "lm", se=FALSE, color="black", formula = y ~ x) +
geom_point()
t1 <- t + geom_text(x = 25, y = 300, label = lm_eqn(df), parse = TRUE)
Error in eval(predvars, data, env) : object 'y' not found
Where did I make the mistake? Thanks in advance!
r
edited Nov 28 '18 at 21:22
MrFlick
124k11141173
asked Nov 28 '18 at 21:01
SchillerlockeSchillerlocke
246
2
Your function has the linem <- lm(y ~ x, df)but there is no variable calledxoryin yourdfvariable.
– MrFlick
Nov 28 '18 at 21:04
I have edited the function and I don't get an error anymore but no equation on the graph. I have changed the variables names from the example so I understand which operator does what but clearly I can't the see right answer. And I'm fairly new to R and functions especially.
– Schillerlocke
Nov 28 '18 at 22:25
1
Well, you are plotting the label at zip=25 and ratio=300. those values are outside where all your points are. Maybe choose a different position.
– MrFlick
Nov 28 '18 at 22:27
Thank you, it works now and I understand a bit more.
– Schillerlocke
Nov 28 '18 at 22:45
add a comment |
0
0
0
I found a function that adds a regression line equation on graph.
Adding Regression Line Equation and R2 on graph
If I copy the function and data sample, everything works fine. But when I try to adapt the code, I get the message, that Y can't be found.
The Function:
lm_eqn <- function(df){
m <- lm(y ~ x, df);
eq <- substitute(italic(y) == a + b %.% italic(x)*","~~italic(r)^2~"="~r2,
list(a = format(coef(m)[1], digits = 2),
b = format(coef(m)[2], digits = 2),
r2 = format(summary(m)$r.squared, digits = 3)))
as.character(as.expression(eq));
}
My data sample and adapted code:
library(tidyverse)
set.seed(1)
id <- 1:5000
zip <- sample(100:200, 5000, replace = TRUE)
outcome <- rbinom(5000, 1, 0.23)
df <- data.frame(id, outcome, zip) %>% as_tibble()
new_df <- df %>% group_by(zip) %>% summarise(ratio = mean(outcome))
library(ggplot2)
t <- ggplot(data = new_df, aes(x = zip, y = ratio)) +
geom_smooth(method = "lm", se=FALSE, color="black", formula = y ~ x) +
geom_point()
t1 <- t + geom_text(x = 25, y = 300, label = lm_eqn(df), parse = TRUE)
Error in eval(predvars, data, env) : object 'y' not found
Where did I make the mistake? Thanks in advance!
r
edited Nov 28 '18 at 21:22
MrFlick
124k11141173
asked Nov 28 '18 at 21:01
SchillerlockeSchillerlocke
246
I found a function that adds a regression line equation on graph.
Adding Regression Line Equation and R2 on graph
If I copy the function and data sample, everything works fine. But when I try to adapt the code, I get the message, that Y can't be found.
The Function:
lm_eqn <- function(df){
m <- lm(y ~ x, df);
eq <- substitute(italic(y) == a + b %.% italic(x)*","~~italic(r)^2~"="~r2,
list(a = format(coef(m)[1], digits = 2),
b = format(coef(m)[2], digits = 2),
r2 = format(summary(m)$r.squared, digits = 3)))
as.character(as.expression(eq));
}
My data sample and adapted code:
library(tidyverse)
set.seed(1)
id <- 1:5000
zip <- sample(100:200, 5000, replace = TRUE)
outcome <- rbinom(5000, 1, 0.23)
df <- data.frame(id, outcome, zip) %>% as_tibble()
new_df <- df %>% group_by(zip) %>% summarise(ratio = mean(outcome))
library(ggplot2)
t <- ggplot(data = new_df, aes(x = zip, y = ratio)) +
geom_smooth(method = "lm", se=FALSE, color="black", formula = y ~ x) +
geom_point()
t1 <- t + geom_text(x = 25, y = 300, label = lm_eqn(df), parse = TRUE)
Error in eval(predvars, data, env) : object 'y' not found
Where did I make the mistake? Thanks in advance!
r
r
edited Nov 28 '18 at 21:22
MrFlick
124k11141173
asked Nov 28 '18 at 21:01
SchillerlockeSchillerlocke
246
edited Nov 28 '18 at 21:22
MrFlick
124k11141173
asked Nov 28 '18 at 21:01
SchillerlockeSchillerlocke
246
edited Nov 28 '18 at 21:22
MrFlick
124k11141173
edited Nov 28 '18 at 21:22
MrFlick
124k11141173
edited Nov 28 '18 at 21:22
MrFlick
124k11141173
124k11141173
asked Nov 28 '18 at 21:01
SchillerlockeSchillerlocke
246
asked Nov 28 '18 at 21:01
SchillerlockeSchillerlocke
246
asked Nov 28 '18 at 21:01
SchillerlockeSchillerlocke
246
246
2
Your function has the linem <- lm(y ~ x, df)but there is no variable calledxoryin yourdfvariable.
– MrFlick
Nov 28 '18 at 21:04
I have edited the function and I don't get an error anymore but no equation on the graph. I have changed the variables names from the example so I understand which operator does what but clearly I can't the see right answer. And I'm fairly new to R and functions especially.
– Schillerlocke
Nov 28 '18 at 22:25
1
Well, you are plotting the label at zip=25 and ratio=300. those values are outside where all your points are. Maybe choose a different position.
– MrFlick
Nov 28 '18 at 22:27
Thank you, it works now and I understand a bit more.
– Schillerlocke
Nov 28 '18 at 22:45
add a comment |
2
Your function has the linem <- lm(y ~ x, df)but there is no variable calledxoryin yourdfvariable.
– MrFlick
Nov 28 '18 at 21:04
I have edited the function and I don't get an error anymore but no equation on the graph. I have changed the variables names from the example so I understand which operator does what but clearly I can't the see right answer. And I'm fairly new to R and functions especially.
– Schillerlocke
Nov 28 '18 at 22:25
1
Well, you are plotting the label at zip=25 and ratio=300. those values are outside where all your points are. Maybe choose a different position.
– MrFlick
Nov 28 '18 at 22:27
Thank you, it works now and I understand a bit more.
– Schillerlocke
Nov 28 '18 at 22:45
2
2
Your function has the line
m <- lm(y ~ x, df) but there is no variable called x or y in your df variable.– MrFlick
Nov 28 '18 at 21:04
Your function has the line
m <- lm(y ~ x, df) but there is no variable called x or y in your df variable.– MrFlick
Nov 28 '18 at 21:04
I have edited the function and I don't get an error anymore but no equation on the graph. I have changed the variables names from the example so I understand which operator does what but clearly I can't the see right answer. And I'm fairly new to R and functions especially.
– Schillerlocke
Nov 28 '18 at 22:25
I have edited the function and I don't get an error anymore but no equation on the graph. I have changed the variables names from the example so I understand which operator does what but clearly I can't the see right answer. And I'm fairly new to R and functions especially.
– Schillerlocke
Nov 28 '18 at 22:25
1
1
Well, you are plotting the label at zip=25 and ratio=300. those values are outside where all your points are. Maybe choose a different position.
– MrFlick
Nov 28 '18 at 22:27
Well, you are plotting the label at zip=25 and ratio=300. those values are outside where all your points are. Maybe choose a different position.
– MrFlick
Nov 28 '18 at 22:27
Thank you, it works now and I understand a bit more.
– Schillerlocke
Nov 28 '18 at 22:45
Thank you, it works now and I understand a bit more.
– Schillerlocke
Nov 28 '18 at 22:45
add a comment |
0
active
oldest
votes
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%2f53528032%2fstruggling-to-adapt-code-from-stackoverflow-to-data-set-adding-regression-line%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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%2f53528032%2fstruggling-to-adapt-code-from-stackoverflow-to-data-set-adding-regression-line%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
2
Your function has the line
m <- lm(y ~ x, df)but there is no variable calledxoryin yourdfvariable.– MrFlick
Nov 28 '18 at 21:04
I have edited the function and I don't get an error anymore but no equation on the graph. I have changed the variables names from the example so I understand which operator does what but clearly I can't the see right answer. And I'm fairly new to R and functions especially.
– Schillerlocke
Nov 28 '18 at 22:25
1
Well, you are plotting the label at zip=25 and ratio=300. those values are outside where all your points are. Maybe choose a different position.
– MrFlick
Nov 28 '18 at 22:27
Thank you, it works now and I understand a bit more.
– Schillerlocke
Nov 28 '18 at 22:45