Why can't I fit a curve to this plot?

I am trying to fit a curve to this plot, but am having trouble.

Code:

library(ggplot2)



data <- data.frame(pressure = c(400, 200, 100, 50, 25, 12.5, 6.25, 3.125),

               volume = c(.25, .5, 1, 2, 4, 8, 16, 32))





plot(data, ylim = c(min(data[,2]), max(data[,2])), xlim = c(min(data[,1]), max(data[,1])),

 pch = 19, col = "firebrick")



lm_fit <- lm(data$pressure ~ poly(data$volume, 2, raw = TRUE))



lines(data$volume, predict (lm_fit, data.frame(x = data$volume)), col = "red")

Result:

Curve does not fit data

edited Nov 13 '18 at 15:04

Ben Bolker

133k11223311

asked Nov 13 '18 at 14:30

Luís Telles

576212

Are data$volume and predict(...) ordered how you want them?

– mickey
Nov 13 '18 at 14:36

Yes, they're ordered.

– Luís Telles
Nov 13 '18 at 14:53

add a comment |

I am trying to fit a curve to this plot, but am having trouble.

Code:

library(ggplot2)



data <- data.frame(pressure = c(400, 200, 100, 50, 25, 12.5, 6.25, 3.125),

               volume = c(.25, .5, 1, 2, 4, 8, 16, 32))





plot(data, ylim = c(min(data[,2]), max(data[,2])), xlim = c(min(data[,1]), max(data[,1])),

 pch = 19, col = "firebrick")



lm_fit <- lm(data$pressure ~ poly(data$volume, 2, raw = TRUE))



lines(data$volume, predict (lm_fit, data.frame(x = data$volume)), col = "red")

Result:

Curve does not fit data

edited Nov 13 '18 at 15:04

Ben Bolker

133k11223311

asked Nov 13 '18 at 14:30

Luís Telles

576212

Are data$volume and predict(...) ordered how you want them?

– mickey
Nov 13 '18 at 14:36

Yes, they're ordered.

– Luís Telles
Nov 13 '18 at 14:53

add a comment |

I am trying to fit a curve to this plot, but am having trouble.

Code:

library(ggplot2)



data <- data.frame(pressure = c(400, 200, 100, 50, 25, 12.5, 6.25, 3.125),

               volume = c(.25, .5, 1, 2, 4, 8, 16, 32))





plot(data, ylim = c(min(data[,2]), max(data[,2])), xlim = c(min(data[,1]), max(data[,1])),

 pch = 19, col = "firebrick")



lm_fit <- lm(data$pressure ~ poly(data$volume, 2, raw = TRUE))



lines(data$volume, predict (lm_fit, data.frame(x = data$volume)), col = "red")

Result:

Curve does not fit data

edited Nov 13 '18 at 15:04

Ben Bolker

133k11223311

asked Nov 13 '18 at 14:30

Luís Telles

576212

I am trying to fit a curve to this plot, but am having trouble.

Code:

library(ggplot2)



data <- data.frame(pressure = c(400, 200, 100, 50, 25, 12.5, 6.25, 3.125),

               volume = c(.25, .5, 1, 2, 4, 8, 16, 32))





plot(data, ylim = c(min(data[,2]), max(data[,2])), xlim = c(min(data[,1]), max(data[,1])),

 pch = 19, col = "firebrick")



lm_fit <- lm(data$pressure ~ poly(data$volume, 2, raw = TRUE))



lines(data$volume, predict (lm_fit, data.frame(x = data$volume)), col = "red")

Result:

Curve does not fit data

r plot curve

edited Nov 13 '18 at 15:04

Ben Bolker

133k11223311

asked Nov 13 '18 at 14:30

Luís Telles

576212

edited Nov 13 '18 at 15:04

Ben Bolker

133k11223311

asked Nov 13 '18 at 14:30

Luís Telles

576212

edited Nov 13 '18 at 15:04

Ben Bolker

133k11223311

edited Nov 13 '18 at 15:04

Ben Bolker

133k11223311

edited Nov 13 '18 at 15:04

Ben Bolker

133k11223311

asked Nov 13 '18 at 14:30

Luís Telles

576212

asked Nov 13 '18 at 14:30

Luís Telles

576212

asked Nov 13 '18 at 14:30

Luís Telles

576212

Are data$volume and predict(...) ordered how you want them?

– mickey
Nov 13 '18 at 14:36

Yes, they're ordered.

– Luís Telles
Nov 13 '18 at 14:53

add a comment |

Are data$volume and predict(...) ordered how you want them?

– mickey
Nov 13 '18 at 14:36

Yes, they're ordered.

– Luís Telles
Nov 13 '18 at 14:53

Are data$volume and predict(...) ordered how you want them?

– mickey
Nov 13 '18 at 14:36

Yes, they're ordered.

– Luís Telles
Nov 13 '18 at 14:53

add a comment |

2 Answers
2

active

oldest

votes

Use ggplot2: geom_smooth lm method

library(ggplot2)



df <- data.frame(pressure = c(400, 200, 100, 50, 25, 12.5, 6.25, 3.125),

                   volume = c(.25, .5, 1, 2, 4, 8, 16, 32))



# Fit a regression line. Change the method if you want the exponential fitting

ggplot(data = df, aes(x = pressure, y = volume)) +

  geom_point() +

  geom_smooth(method = "lm")



# If you just want to connect the dots

ggplot(data = df, aes(x = pressure, y = volume)) +

  geom_point() +

  geom_line()

enter image description here

answered Nov 13 '18 at 14:41

kon_u

1966

add a comment |

Your main problem is using data$ inside your regression model: you should use just the names of the variables.

dd <- data.frame(pressure = c(400, 200, 100, 50, 25, 12.5, 6.25, 3.125),

                 volume = c(.25, .5, 1, 2, 4, 8, 16, 32))

lm_fit <- lm(pressure ~ poly(volume, 2, raw = TRUE),

             data=dd)

For a smoother curve, I computed the predicted value at a finer sequence of volume values:

pframe <- data.frame(volume=seq(0,30,length=51))

pframe$pressure <- predict(lm_fit,newdata=pframe)

Now the picture:

## png("SO_poly.png")

par(las=1,bty="l") ## cosmetic

plot(pressure~volume, data=dd, pch = 19, col = "firebrick",

     ylim=c(-100,500))



with(pframe, lines(volume, pressure, col="red"))

This doesn't look great, so I tried some other curve-fits.

Log-linear fit:

lm_fit2 <- lm(log(pressure) ~ poly(volume, 2, raw = TRUE),

             data=dd)

pframe$lpressure <- exp(predict(lm_fit2,newdata=pframe))

with(pframe, lines(volume, lpressure, col="purple"))

Exponential fit:

glm_fit <- glm(pressure ~ poly(volume,2),

               family=gaussian(link="log"),

               data=dd)

pframe$gpressure <- predict(glm_fit, newdata=pframe, type="response")

with(pframe, lines(volume, gpressure, col="blue"))

## dev.off()

enter image description here

You could also use ggplot2:

library(ggplot2)

ggplot(dd, aes(volume,pressure))+

    geom_point()+

    geom_smooth(method="lm",

                formula=y~poly(x,2))

answered Nov 13 '18 at 15:04

Ben Bolker

133k11223311

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');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53283261%2fwhy-cant-i-fit-a-curve-to-this-plot%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

Use ggplot2: geom_smooth lm method

library(ggplot2)



df <- data.frame(pressure = c(400, 200, 100, 50, 25, 12.5, 6.25, 3.125),

                   volume = c(.25, .5, 1, 2, 4, 8, 16, 32))



# Fit a regression line. Change the method if you want the exponential fitting

ggplot(data = df, aes(x = pressure, y = volume)) +

  geom_point() +

  geom_smooth(method = "lm")



# If you just want to connect the dots

ggplot(data = df, aes(x = pressure, y = volume)) +

  geom_point() +

  geom_line()

enter image description here

answered Nov 13 '18 at 14:41

kon_u

1966

add a comment |

Use ggplot2: geom_smooth lm method

library(ggplot2)



df <- data.frame(pressure = c(400, 200, 100, 50, 25, 12.5, 6.25, 3.125),

                   volume = c(.25, .5, 1, 2, 4, 8, 16, 32))



# Fit a regression line. Change the method if you want the exponential fitting

ggplot(data = df, aes(x = pressure, y = volume)) +

  geom_point() +

  geom_smooth(method = "lm")



# If you just want to connect the dots

ggplot(data = df, aes(x = pressure, y = volume)) +

  geom_point() +

  geom_line()

enter image description here

answered Nov 13 '18 at 14:41

kon_u

1966

add a comment |

Use ggplot2: geom_smooth lm method

library(ggplot2)



df <- data.frame(pressure = c(400, 200, 100, 50, 25, 12.5, 6.25, 3.125),

                   volume = c(.25, .5, 1, 2, 4, 8, 16, 32))



# Fit a regression line. Change the method if you want the exponential fitting

ggplot(data = df, aes(x = pressure, y = volume)) +

  geom_point() +

  geom_smooth(method = "lm")



# If you just want to connect the dots

ggplot(data = df, aes(x = pressure, y = volume)) +

  geom_point() +

  geom_line()

enter image description here

answered Nov 13 '18 at 14:41

kon_u

1966

Use ggplot2: geom_smooth lm method

library(ggplot2)



df <- data.frame(pressure = c(400, 200, 100, 50, 25, 12.5, 6.25, 3.125),

                   volume = c(.25, .5, 1, 2, 4, 8, 16, 32))



# Fit a regression line. Change the method if you want the exponential fitting

ggplot(data = df, aes(x = pressure, y = volume)) +

  geom_point() +

  geom_smooth(method = "lm")



# If you just want to connect the dots

ggplot(data = df, aes(x = pressure, y = volume)) +

  geom_point() +

  geom_line()

enter image description here

answered Nov 13 '18 at 14:41

kon_u

1966

answered Nov 13 '18 at 14:41

kon_u

1966

answered Nov 13 '18 at 14:41

kon_u

1966

answered Nov 13 '18 at 14:41

kon_u

1966

add a comment |

Your main problem is using data$ inside your regression model: you should use just the names of the variables.

dd <- data.frame(pressure = c(400, 200, 100, 50, 25, 12.5, 6.25, 3.125),

                 volume = c(.25, .5, 1, 2, 4, 8, 16, 32))

lm_fit <- lm(pressure ~ poly(volume, 2, raw = TRUE),

             data=dd)

For a smoother curve, I computed the predicted value at a finer sequence of volume values:

pframe <- data.frame(volume=seq(0,30,length=51))

pframe$pressure <- predict(lm_fit,newdata=pframe)

Now the picture:

## png("SO_poly.png")

par(las=1,bty="l") ## cosmetic

plot(pressure~volume, data=dd, pch = 19, col = "firebrick",

     ylim=c(-100,500))



with(pframe, lines(volume, pressure, col="red"))

This doesn't look great, so I tried some other curve-fits.

Log-linear fit:

lm_fit2 <- lm(log(pressure) ~ poly(volume, 2, raw = TRUE),

             data=dd)

pframe$lpressure <- exp(predict(lm_fit2,newdata=pframe))

with(pframe, lines(volume, lpressure, col="purple"))

Exponential fit:

glm_fit <- glm(pressure ~ poly(volume,2),

               family=gaussian(link="log"),

               data=dd)

pframe$gpressure <- predict(glm_fit, newdata=pframe, type="response")

with(pframe, lines(volume, gpressure, col="blue"))

## dev.off()

enter image description here

You could also use ggplot2:

library(ggplot2)

ggplot(dd, aes(volume,pressure))+

    geom_point()+

    geom_smooth(method="lm",

                formula=y~poly(x,2))

answered Nov 13 '18 at 15:04

Ben Bolker

133k11223311

add a comment |

Your main problem is using data$ inside your regression model: you should use just the names of the variables.

dd <- data.frame(pressure = c(400, 200, 100, 50, 25, 12.5, 6.25, 3.125),

                 volume = c(.25, .5, 1, 2, 4, 8, 16, 32))

lm_fit <- lm(pressure ~ poly(volume, 2, raw = TRUE),

             data=dd)

For a smoother curve, I computed the predicted value at a finer sequence of volume values:

pframe <- data.frame(volume=seq(0,30,length=51))

pframe$pressure <- predict(lm_fit,newdata=pframe)

Now the picture:

## png("SO_poly.png")

par(las=1,bty="l") ## cosmetic

plot(pressure~volume, data=dd, pch = 19, col = "firebrick",

     ylim=c(-100,500))



with(pframe, lines(volume, pressure, col="red"))

This doesn't look great, so I tried some other curve-fits.

Log-linear fit:

lm_fit2 <- lm(log(pressure) ~ poly(volume, 2, raw = TRUE),

             data=dd)

pframe$lpressure <- exp(predict(lm_fit2,newdata=pframe))

with(pframe, lines(volume, lpressure, col="purple"))

Exponential fit:

glm_fit <- glm(pressure ~ poly(volume,2),

               family=gaussian(link="log"),

               data=dd)

pframe$gpressure <- predict(glm_fit, newdata=pframe, type="response")

with(pframe, lines(volume, gpressure, col="blue"))

## dev.off()

enter image description here

You could also use ggplot2:

library(ggplot2)

ggplot(dd, aes(volume,pressure))+

    geom_point()+

    geom_smooth(method="lm",

                formula=y~poly(x,2))

answered Nov 13 '18 at 15:04

Ben Bolker

133k11223311

add a comment |

Your main problem is using data$ inside your regression model: you should use just the names of the variables.

dd <- data.frame(pressure = c(400, 200, 100, 50, 25, 12.5, 6.25, 3.125),

                 volume = c(.25, .5, 1, 2, 4, 8, 16, 32))

lm_fit <- lm(pressure ~ poly(volume, 2, raw = TRUE),

             data=dd)

For a smoother curve, I computed the predicted value at a finer sequence of volume values:

pframe <- data.frame(volume=seq(0,30,length=51))

pframe$pressure <- predict(lm_fit,newdata=pframe)

Now the picture:

## png("SO_poly.png")

par(las=1,bty="l") ## cosmetic

plot(pressure~volume, data=dd, pch = 19, col = "firebrick",

     ylim=c(-100,500))



with(pframe, lines(volume, pressure, col="red"))

This doesn't look great, so I tried some other curve-fits.

Log-linear fit:

lm_fit2 <- lm(log(pressure) ~ poly(volume, 2, raw = TRUE),

             data=dd)

pframe$lpressure <- exp(predict(lm_fit2,newdata=pframe))

with(pframe, lines(volume, lpressure, col="purple"))

Exponential fit:

glm_fit <- glm(pressure ~ poly(volume,2),

               family=gaussian(link="log"),

               data=dd)

pframe$gpressure <- predict(glm_fit, newdata=pframe, type="response")

with(pframe, lines(volume, gpressure, col="blue"))

## dev.off()

enter image description here

You could also use ggplot2:

library(ggplot2)

ggplot(dd, aes(volume,pressure))+

    geom_point()+

    geom_smooth(method="lm",

                formula=y~poly(x,2))

answered Nov 13 '18 at 15:04

Ben Bolker

133k11223311

Your main problem is using data$ inside your regression model: you should use just the names of the variables.

dd <- data.frame(pressure = c(400, 200, 100, 50, 25, 12.5, 6.25, 3.125),

                 volume = c(.25, .5, 1, 2, 4, 8, 16, 32))

lm_fit <- lm(pressure ~ poly(volume, 2, raw = TRUE),

             data=dd)

For a smoother curve, I computed the predicted value at a finer sequence of volume values:

pframe <- data.frame(volume=seq(0,30,length=51))

pframe$pressure <- predict(lm_fit,newdata=pframe)

Now the picture:

## png("SO_poly.png")

par(las=1,bty="l") ## cosmetic

plot(pressure~volume, data=dd, pch = 19, col = "firebrick",

     ylim=c(-100,500))



with(pframe, lines(volume, pressure, col="red"))

This doesn't look great, so I tried some other curve-fits.

Log-linear fit:

lm_fit2 <- lm(log(pressure) ~ poly(volume, 2, raw = TRUE),

             data=dd)

pframe$lpressure <- exp(predict(lm_fit2,newdata=pframe))

with(pframe, lines(volume, lpressure, col="purple"))

Exponential fit:

glm_fit <- glm(pressure ~ poly(volume,2),

               family=gaussian(link="log"),

               data=dd)

pframe$gpressure <- predict(glm_fit, newdata=pframe, type="response")

with(pframe, lines(volume, gpressure, col="blue"))

## dev.off()

enter image description here

You could also use ggplot2:

library(ggplot2)

ggplot(dd, aes(volume,pressure))+

    geom_point()+

    geom_smooth(method="lm",

                formula=y~poly(x,2))

answered Nov 13 '18 at 15:04

Ben Bolker

133k11223311

answered Nov 13 '18 at 15:04

Ben Bolker

133k11223311

answered Nov 13 '18 at 15:04

Ben Bolker

133k11223311

answered Nov 13 '18 at 15:04

Ben Bolker

133k11223311

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.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Vfrdtyky