Plot polynomial regression curve in R
I have a simple polynomial regression which I do as follows
attach(mtcars)
fit <- lm(mpg ~ hp + I(hp^2))
Now, I plot as follows
> plot(mpg~hp)
> points(hp, fitted(fit), col='red', pch=20)
This gives me the following
I want to connect these points into a smooth curve, using lines gives me the following
> lines(hp, fitted(fit), col='red', type='b')
What am I missing here. I want the output to be a smooth curve which connects the points
r plot lm
edited Apr 28 '14 at 7:08
Davide Passaretti
2,0941227
asked Apr 28 '14 at 6:51
psteelkpsteelk
5002820
add a comment |
I have a simple polynomial regression which I do as follows
attach(mtcars)
fit <- lm(mpg ~ hp + I(hp^2))
Now, I plot as follows
> plot(mpg~hp)
> points(hp, fitted(fit), col='red', pch=20)
This gives me the following
I want to connect these points into a smooth curve, using lines gives me the following
> lines(hp, fitted(fit), col='red', type='b')
What am I missing here. I want the output to be a smooth curve which connects the points
r plot lm
edited Apr 28 '14 at 7:08
Davide Passaretti
2,0941227
asked Apr 28 '14 at 6:51
psteelkpsteelk
5002820
1
You really shouldn't useattach, it can cause many bugs.
– Gregor
Apr 9 '16 at 17:10
add a comment |
I have a simple polynomial regression which I do as follows
attach(mtcars)
fit <- lm(mpg ~ hp + I(hp^2))
Now, I plot as follows
> plot(mpg~hp)
> points(hp, fitted(fit), col='red', pch=20)
This gives me the following
I want to connect these points into a smooth curve, using lines gives me the following
> lines(hp, fitted(fit), col='red', type='b')
What am I missing here. I want the output to be a smooth curve which connects the points
r plot lm
edited Apr 28 '14 at 7:08
Davide Passaretti
2,0941227
asked Apr 28 '14 at 6:51
psteelkpsteelk
5002820
I have a simple polynomial regression which I do as follows
attach(mtcars)
fit <- lm(mpg ~ hp + I(hp^2))
Now, I plot as follows
> plot(mpg~hp)
> points(hp, fitted(fit), col='red', pch=20)
This gives me the following
I want to connect these points into a smooth curve, using lines gives me the following
> lines(hp, fitted(fit), col='red', type='b')
What am I missing here. I want the output to be a smooth curve which connects the points
r plot lm
r plot lm
edited Apr 28 '14 at 7:08
Davide Passaretti
2,0941227
asked Apr 28 '14 at 6:51
psteelkpsteelk
5002820
edited Apr 28 '14 at 7:08
Davide Passaretti
2,0941227
asked Apr 28 '14 at 6:51
psteelkpsteelk
5002820
edited Apr 28 '14 at 7:08
Davide Passaretti
2,0941227
edited Apr 28 '14 at 7:08
Davide Passaretti
2,0941227
edited Apr 28 '14 at 7:08
Davide Passaretti
2,0941227
2,0941227
asked Apr 28 '14 at 6:51
psteelkpsteelk
5002820
asked Apr 28 '14 at 6:51
psteelkpsteelk
5002820
asked Apr 28 '14 at 6:51
psteelkpsteelk
5002820
5002820
1
You really shouldn't useattach, it can cause many bugs.
– Gregor
Apr 9 '16 at 17:10
add a comment |
1
You really shouldn't useattach, it can cause many bugs.
– Gregor
Apr 9 '16 at 17:10
1
1
You really shouldn't use
attach, it can cause many bugs.– Gregor
Apr 9 '16 at 17:10
You really shouldn't use
attach, it can cause many bugs.– Gregor
Apr 9 '16 at 17:10
add a comment |
3 Answers
3
active
oldest
votes
Try:
lines(sort(hp), fitted(fit)[order(hp)], col='red', type='b')
Because your statistical units in the dataset are not ordered, thus, when you use lines it's a mess.
answered Apr 28 '14 at 6:59
Davide PassarettiDavide Passaretti
2,0941227
Unless you have evenly spaced values or many observations, using thisfitted()approach is not going to produce a smooth realisation of the fitted polynomial/function
– Gavin Simpson
Apr 9 '16 at 17:17
@GavinSimpson of course, generating a sequence of close and evenly spaced points, and fitting the function on it would produce a smoother curve. But I think the aim of the question was to find a way to connect the existing fitted points by a line, not the curve itself.
– Davide Passaretti
May 9 '16 at 7:04
add a comment |
I like to use ggplot2 for this because it's usually very intuitive to add layers of data.
library(ggplot2)
fit <- lm(mpg ~ hp + I(hp^2), data = mtcars)
prd <- data.frame(hp = seq(from = range(mtcars$hp)[1], to = range(mtcars$hp)[2], length.out = 100))
err <- predict(fit, newdata = prd, se.fit = TRUE)
prd$lci <- err$fit - 1.96 * err$se.fit
prd$fit <- err$fit
prd$uci <- err$fit + 1.96 * err$se.fit
ggplot(prd, aes(x = hp, y = fit)) +
theme_bw() +
geom_line() +
geom_smooth(aes(ymin = lci, ymax = uci), stat = "identity") +
geom_point(data = mtcars, aes(x = hp, y = mpg))
answered Apr 28 '14 at 7:32
Roman LuštrikRoman Luštrik
49.7k19109160
When using your code (with R 3.3.3 and ggplot2_2.2.1 sp_1.2-4) I get the Warning: Ignoring unknown aesthetics: ymin, ymax
– Pertinax
Nov 15 '17 at 11:44
1
@TheThunderChimp they appear to be there... ggplot2.tidyverse.org/reference/geom_smooth.html
– Roman Luštrik
Nov 15 '17 at 19:53
1
This is apparently a bug on recent versions of ggplot2: github.com/tidyverse/ggplot2/issues/1939
– Pertinax
Nov 16 '17 at 9:45
add a comment |
Generally a good way to go is to use the predict() function. Pick some x values, use predict() to generate corresponding y values, and plot them. It can look something like this:
newdat = data.frame(hp = seq(min(mtcars$hp), max(mtcars$hp), length.out = 100))
newdat$pred = predict(fit, newdata = newdat)
plot(mpg ~ hp, data = mtcars)
with(newdat, lines(x = hp, y = pred))
See Roman's answer for a fancier version of this method, where confidence intervals are calculated too. In both cases the actual plotting of the solution is incidental - you can use base graphics or ggplot2 or anything else you'd like - the key is just use the predict function to generate the proper y values. It's a good method because it extends to all sorts of fits, not just polynomial linear models. You can use it with non-linear models, GLMs, smoothing splines, etc. - anything with a predict method.
answered Apr 9 '16 at 17:16
GregorGregor
64.2k990171
Whilst not explained as such, Romain's answer already shows thispredict()approach, does it not?
– Gavin Simpson
Apr 9 '16 at 17:18
1
Yes it does, but as you say it's not explained as such. This seems to be a standard source for this info with many linked duplicates - I think having an explanation of the general method is valuable, and I also think thatggplotcan be a barrier for new R users so it's nice to demo the method using base. But I will edit to acknowledge Roman's efforts.
– Gregor
Apr 9 '16 at 17:44
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Try:
lines(sort(hp), fitted(fit)[order(hp)], col='red', type='b')
Because your statistical units in the dataset are not ordered, thus, when you use lines it's a mess.
answered Apr 28 '14 at 6:59
Davide PassarettiDavide Passaretti
2,0941227
Unless you have evenly spaced values or many observations, using thisfitted()approach is not going to produce a smooth realisation of the fitted polynomial/function
– Gavin Simpson
Apr 9 '16 at 17:17
@GavinSimpson of course, generating a sequence of close and evenly spaced points, and fitting the function on it would produce a smoother curve. But I think the aim of the question was to find a way to connect the existing fitted points by a line, not the curve itself.
– Davide Passaretti
May 9 '16 at 7:04
add a comment |
Try:
lines(sort(hp), fitted(fit)[order(hp)], col='red', type='b')
Because your statistical units in the dataset are not ordered, thus, when you use lines it's a mess.
answered Apr 28 '14 at 6:59
Davide PassarettiDavide Passaretti
2,0941227
Unless you have evenly spaced values or many observations, using thisfitted()approach is not going to produce a smooth realisation of the fitted polynomial/function
– Gavin Simpson
Apr 9 '16 at 17:17
@GavinSimpson of course, generating a sequence of close and evenly spaced points, and fitting the function on it would produce a smoother curve. But I think the aim of the question was to find a way to connect the existing fitted points by a line, not the curve itself.
– Davide Passaretti
May 9 '16 at 7:04
add a comment |
Try:
lines(sort(hp), fitted(fit)[order(hp)], col='red', type='b')
Because your statistical units in the dataset are not ordered, thus, when you use lines it's a mess.
answered Apr 28 '14 at 6:59
Davide PassarettiDavide Passaretti
2,0941227
Try:
lines(sort(hp), fitted(fit)[order(hp)], col='red', type='b')
Because your statistical units in the dataset are not ordered, thus, when you use lines it's a mess.
answered Apr 28 '14 at 6:59
Davide PassarettiDavide Passaretti
2,0941227
answered Apr 28 '14 at 6:59
Davide PassarettiDavide Passaretti
2,0941227
answered Apr 28 '14 at 6:59
Davide PassarettiDavide Passaretti
2,0941227
answered Apr 28 '14 at 6:59
Davide PassarettiDavide Passaretti
2,0941227
2,0941227
Unless you have evenly spaced values or many observations, using thisfitted()approach is not going to produce a smooth realisation of the fitted polynomial/function
– Gavin Simpson
Apr 9 '16 at 17:17
@GavinSimpson of course, generating a sequence of close and evenly spaced points, and fitting the function on it would produce a smoother curve. But I think the aim of the question was to find a way to connect the existing fitted points by a line, not the curve itself.
– Davide Passaretti
May 9 '16 at 7:04
add a comment |
Unless you have evenly spaced values or many observations, using thisfitted()approach is not going to produce a smooth realisation of the fitted polynomial/function
– Gavin Simpson
Apr 9 '16 at 17:17
@GavinSimpson of course, generating a sequence of close and evenly spaced points, and fitting the function on it would produce a smoother curve. But I think the aim of the question was to find a way to connect the existing fitted points by a line, not the curve itself.
– Davide Passaretti
May 9 '16 at 7:04
Unless you have evenly spaced values or many observations, using this
fitted() approach is not going to produce a smooth realisation of the fitted polynomial/function– Gavin Simpson
Apr 9 '16 at 17:17
Unless you have evenly spaced values or many observations, using this
fitted() approach is not going to produce a smooth realisation of the fitted polynomial/function– Gavin Simpson
Apr 9 '16 at 17:17
@GavinSimpson of course, generating a sequence of close and evenly spaced points, and fitting the function on it would produce a smoother curve. But I think the aim of the question was to find a way to connect the existing fitted points by a line, not the curve itself.
– Davide Passaretti
May 9 '16 at 7:04
@GavinSimpson of course, generating a sequence of close and evenly spaced points, and fitting the function on it would produce a smoother curve. But I think the aim of the question was to find a way to connect the existing fitted points by a line, not the curve itself.
– Davide Passaretti
May 9 '16 at 7:04
add a comment |
I like to use ggplot2 for this because it's usually very intuitive to add layers of data.
library(ggplot2)
fit <- lm(mpg ~ hp + I(hp^2), data = mtcars)
prd <- data.frame(hp = seq(from = range(mtcars$hp)[1], to = range(mtcars$hp)[2], length.out = 100))
err <- predict(fit, newdata = prd, se.fit = TRUE)
prd$lci <- err$fit - 1.96 * err$se.fit
prd$fit <- err$fit
prd$uci <- err$fit + 1.96 * err$se.fit
ggplot(prd, aes(x = hp, y = fit)) +
theme_bw() +
geom_line() +
geom_smooth(aes(ymin = lci, ymax = uci), stat = "identity") +
geom_point(data = mtcars, aes(x = hp, y = mpg))
answered Apr 28 '14 at 7:32
Roman LuštrikRoman Luštrik
49.7k19109160
When using your code (with R 3.3.3 and ggplot2_2.2.1 sp_1.2-4) I get the Warning: Ignoring unknown aesthetics: ymin, ymax
– Pertinax
Nov 15 '17 at 11:44
1
@TheThunderChimp they appear to be there... ggplot2.tidyverse.org/reference/geom_smooth.html
– Roman Luštrik
Nov 15 '17 at 19:53
1
This is apparently a bug on recent versions of ggplot2: github.com/tidyverse/ggplot2/issues/1939
– Pertinax
Nov 16 '17 at 9:45
add a comment |
I like to use ggplot2 for this because it's usually very intuitive to add layers of data.
library(ggplot2)
fit <- lm(mpg ~ hp + I(hp^2), data = mtcars)
prd <- data.frame(hp = seq(from = range(mtcars$hp)[1], to = range(mtcars$hp)[2], length.out = 100))
err <- predict(fit, newdata = prd, se.fit = TRUE)
prd$lci <- err$fit - 1.96 * err$se.fit
prd$fit <- err$fit
prd$uci <- err$fit + 1.96 * err$se.fit
ggplot(prd, aes(x = hp, y = fit)) +
theme_bw() +
geom_line() +
geom_smooth(aes(ymin = lci, ymax = uci), stat = "identity") +
geom_point(data = mtcars, aes(x = hp, y = mpg))
answered Apr 28 '14 at 7:32
Roman LuštrikRoman Luštrik
49.7k19109160
When using your code (with R 3.3.3 and ggplot2_2.2.1 sp_1.2-4) I get the Warning: Ignoring unknown aesthetics: ymin, ymax
– Pertinax
Nov 15 '17 at 11:44
1
@TheThunderChimp they appear to be there... ggplot2.tidyverse.org/reference/geom_smooth.html
– Roman Luštrik
Nov 15 '17 at 19:53
1
This is apparently a bug on recent versions of ggplot2: github.com/tidyverse/ggplot2/issues/1939
– Pertinax
Nov 16 '17 at 9:45
add a comment |
I like to use ggplot2 for this because it's usually very intuitive to add layers of data.
library(ggplot2)
fit <- lm(mpg ~ hp + I(hp^2), data = mtcars)
prd <- data.frame(hp = seq(from = range(mtcars$hp)[1], to = range(mtcars$hp)[2], length.out = 100))
err <- predict(fit, newdata = prd, se.fit = TRUE)
prd$lci <- err$fit - 1.96 * err$se.fit
prd$fit <- err$fit
prd$uci <- err$fit + 1.96 * err$se.fit
ggplot(prd, aes(x = hp, y = fit)) +
theme_bw() +
geom_line() +
geom_smooth(aes(ymin = lci, ymax = uci), stat = "identity") +
geom_point(data = mtcars, aes(x = hp, y = mpg))
answered Apr 28 '14 at 7:32
Roman LuštrikRoman Luštrik
49.7k19109160
I like to use ggplot2 for this because it's usually very intuitive to add layers of data.
library(ggplot2)
fit <- lm(mpg ~ hp + I(hp^2), data = mtcars)
prd <- data.frame(hp = seq(from = range(mtcars$hp)[1], to = range(mtcars$hp)[2], length.out = 100))
err <- predict(fit, newdata = prd, se.fit = TRUE)
prd$lci <- err$fit - 1.96 * err$se.fit
prd$fit <- err$fit
prd$uci <- err$fit + 1.96 * err$se.fit
ggplot(prd, aes(x = hp, y = fit)) +
theme_bw() +
geom_line() +
geom_smooth(aes(ymin = lci, ymax = uci), stat = "identity") +
geom_point(data = mtcars, aes(x = hp, y = mpg))
answered Apr 28 '14 at 7:32
Roman LuštrikRoman Luštrik
49.7k19109160
answered Apr 28 '14 at 7:32
Roman LuštrikRoman Luštrik
49.7k19109160
answered Apr 28 '14 at 7:32
Roman LuštrikRoman Luštrik
49.7k19109160
answered Apr 28 '14 at 7:32
Roman LuštrikRoman Luštrik
49.7k19109160
49.7k19109160
When using your code (with R 3.3.3 and ggplot2_2.2.1 sp_1.2-4) I get the Warning: Ignoring unknown aesthetics: ymin, ymax
– Pertinax
Nov 15 '17 at 11:44
1
@TheThunderChimp they appear to be there... ggplot2.tidyverse.org/reference/geom_smooth.html
– Roman Luštrik
Nov 15 '17 at 19:53
1
This is apparently a bug on recent versions of ggplot2: github.com/tidyverse/ggplot2/issues/1939
– Pertinax
Nov 16 '17 at 9:45
add a comment |
When using your code (with R 3.3.3 and ggplot2_2.2.1 sp_1.2-4) I get the Warning: Ignoring unknown aesthetics: ymin, ymax
– Pertinax
Nov 15 '17 at 11:44
1
@TheThunderChimp they appear to be there... ggplot2.tidyverse.org/reference/geom_smooth.html
– Roman Luštrik
Nov 15 '17 at 19:53
1
This is apparently a bug on recent versions of ggplot2: github.com/tidyverse/ggplot2/issues/1939
– Pertinax
Nov 16 '17 at 9:45
When using your code (with R 3.3.3 and ggplot2_2.2.1 sp_1.2-4) I get the Warning: Ignoring unknown aesthetics: ymin, ymax
– Pertinax
Nov 15 '17 at 11:44
When using your code (with R 3.3.3 and ggplot2_2.2.1 sp_1.2-4) I get the Warning: Ignoring unknown aesthetics: ymin, ymax
– Pertinax
Nov 15 '17 at 11:44
1
1
@TheThunderChimp they appear to be there... ggplot2.tidyverse.org/reference/geom_smooth.html
– Roman Luštrik
Nov 15 '17 at 19:53
@TheThunderChimp they appear to be there... ggplot2.tidyverse.org/reference/geom_smooth.html
– Roman Luštrik
Nov 15 '17 at 19:53
1
1
This is apparently a bug on recent versions of ggplot2: github.com/tidyverse/ggplot2/issues/1939
– Pertinax
Nov 16 '17 at 9:45
This is apparently a bug on recent versions of ggplot2: github.com/tidyverse/ggplot2/issues/1939
– Pertinax
Nov 16 '17 at 9:45
add a comment |
Generally a good way to go is to use the predict() function. Pick some x values, use predict() to generate corresponding y values, and plot them. It can look something like this:
newdat = data.frame(hp = seq(min(mtcars$hp), max(mtcars$hp), length.out = 100))
newdat$pred = predict(fit, newdata = newdat)
plot(mpg ~ hp, data = mtcars)
with(newdat, lines(x = hp, y = pred))
See Roman's answer for a fancier version of this method, where confidence intervals are calculated too. In both cases the actual plotting of the solution is incidental - you can use base graphics or ggplot2 or anything else you'd like - the key is just use the predict function to generate the proper y values. It's a good method because it extends to all sorts of fits, not just polynomial linear models. You can use it with non-linear models, GLMs, smoothing splines, etc. - anything with a predict method.
answered Apr 9 '16 at 17:16
GregorGregor
64.2k990171
Whilst not explained as such, Romain's answer already shows thispredict()approach, does it not?
– Gavin Simpson
Apr 9 '16 at 17:18
1
Yes it does, but as you say it's not explained as such. This seems to be a standard source for this info with many linked duplicates - I think having an explanation of the general method is valuable, and I also think thatggplotcan be a barrier for new R users so it's nice to demo the method using base. But I will edit to acknowledge Roman's efforts.
– Gregor
Apr 9 '16 at 17:44
add a comment |
Generally a good way to go is to use the predict() function. Pick some x values, use predict() to generate corresponding y values, and plot them. It can look something like this:
newdat = data.frame(hp = seq(min(mtcars$hp), max(mtcars$hp), length.out = 100))
newdat$pred = predict(fit, newdata = newdat)
plot(mpg ~ hp, data = mtcars)
with(newdat, lines(x = hp, y = pred))
See Roman's answer for a fancier version of this method, where confidence intervals are calculated too. In both cases the actual plotting of the solution is incidental - you can use base graphics or ggplot2 or anything else you'd like - the key is just use the predict function to generate the proper y values. It's a good method because it extends to all sorts of fits, not just polynomial linear models. You can use it with non-linear models, GLMs, smoothing splines, etc. - anything with a predict method.
answered Apr 9 '16 at 17:16
GregorGregor
64.2k990171
Whilst not explained as such, Romain's answer already shows thispredict()approach, does it not?
– Gavin Simpson
Apr 9 '16 at 17:18
1
Yes it does, but as you say it's not explained as such. This seems to be a standard source for this info with many linked duplicates - I think having an explanation of the general method is valuable, and I also think thatggplotcan be a barrier for new R users so it's nice to demo the method using base. But I will edit to acknowledge Roman's efforts.
– Gregor
Apr 9 '16 at 17:44
add a comment |
Generally a good way to go is to use the predict() function. Pick some x values, use predict() to generate corresponding y values, and plot them. It can look something like this:
newdat = data.frame(hp = seq(min(mtcars$hp), max(mtcars$hp), length.out = 100))
newdat$pred = predict(fit, newdata = newdat)
plot(mpg ~ hp, data = mtcars)
with(newdat, lines(x = hp, y = pred))
See Roman's answer for a fancier version of this method, where confidence intervals are calculated too. In both cases the actual plotting of the solution is incidental - you can use base graphics or ggplot2 or anything else you'd like - the key is just use the predict function to generate the proper y values. It's a good method because it extends to all sorts of fits, not just polynomial linear models. You can use it with non-linear models, GLMs, smoothing splines, etc. - anything with a predict method.
answered Apr 9 '16 at 17:16
GregorGregor
64.2k990171
Generally a good way to go is to use the predict() function. Pick some x values, use predict() to generate corresponding y values, and plot them. It can look something like this:
newdat = data.frame(hp = seq(min(mtcars$hp), max(mtcars$hp), length.out = 100))
newdat$pred = predict(fit, newdata = newdat)
plot(mpg ~ hp, data = mtcars)
with(newdat, lines(x = hp, y = pred))
See Roman's answer for a fancier version of this method, where confidence intervals are calculated too. In both cases the actual plotting of the solution is incidental - you can use base graphics or ggplot2 or anything else you'd like - the key is just use the predict function to generate the proper y values. It's a good method because it extends to all sorts of fits, not just polynomial linear models. You can use it with non-linear models, GLMs, smoothing splines, etc. - anything with a predict method.
answered Apr 9 '16 at 17:16
GregorGregor
64.2k990171
edited Apr 9 '16 at 17:47
answered Apr 9 '16 at 17:16
GregorGregor
64.2k990171
answered Apr 9 '16 at 17:16
GregorGregor
64.2k990171
answered Apr 9 '16 at 17:16
GregorGregor
64.2k990171
64.2k990171
Whilst not explained as such, Romain's answer already shows thispredict()approach, does it not?
– Gavin Simpson
Apr 9 '16 at 17:18
1
Yes it does, but as you say it's not explained as such. This seems to be a standard source for this info with many linked duplicates - I think having an explanation of the general method is valuable, and I also think thatggplotcan be a barrier for new R users so it's nice to demo the method using base. But I will edit to acknowledge Roman's efforts.
– Gregor
Apr 9 '16 at 17:44
add a comment |
Whilst not explained as such, Romain's answer already shows thispredict()approach, does it not?
– Gavin Simpson
Apr 9 '16 at 17:18
1
Yes it does, but as you say it's not explained as such. This seems to be a standard source for this info with many linked duplicates - I think having an explanation of the general method is valuable, and I also think thatggplotcan be a barrier for new R users so it's nice to demo the method using base. But I will edit to acknowledge Roman's efforts.
– Gregor
Apr 9 '16 at 17:44
Whilst not explained as such, Romain's answer already shows this
predict() approach, does it not?– Gavin Simpson
Apr 9 '16 at 17:18
Whilst not explained as such, Romain's answer already shows this
predict() approach, does it not?– Gavin Simpson
Apr 9 '16 at 17:18
1
1
Yes it does, but as you say it's not explained as such. This seems to be a standard source for this info with many linked duplicates - I think having an explanation of the general method is valuable, and I also think that
ggplot can be a barrier for new R users so it's nice to demo the method using base. But I will edit to acknowledge Roman's efforts.– Gregor
Apr 9 '16 at 17:44
Yes it does, but as you say it's not explained as such. This seems to be a standard source for this info with many linked duplicates - I think having an explanation of the general method is valuable, and I also think that
ggplot can be a barrier for new R users so it's nice to demo the method using base. But I will edit to acknowledge Roman's efforts.– Gregor
Apr 9 '16 at 17:44
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You really shouldn't use
attach, it can cause many bugs.– Gregor
Apr 9 '16 at 17:10