How can I create this graphic?
The picture shows a normal linear regression.
Although it is easy to compute on paper, I have no idea how to code a linear regression in order to get an output shown as in the picture.
I would be very pleased if anyone could help me.
I have already tried {tikzpicture} etc. but it does not work out that good as pleased.
diagrams plot
|
show 6 more comments
The picture shows a normal linear regression.
Although it is easy to compute on paper, I have no idea how to code a linear regression in order to get an output shown as in the picture.
I would be very pleased if anyone could help me.
I have already tried {tikzpicture} etc. but it does not work out that good as pleased.
diagrams plot
8
Welcome to TeX.SE! Please edit your question and add a minimal example of what you tried.
– CarLaTeX
Dec 1 at 13:07
1
pgfplots
allows you to do this, see p 396Fitting Lines - Regression
: ctan.org/pkg/pgfplots
– AndréC
Dec 1 at 13:15
1
@CarLaTeX think a little before you get carried away! Rebecca wants to knowhow she can create this graph
. If she knew, she wouldn't have asked that question.
– AndréC
Dec 1 at 13:39
1
@AndréC Rebecca wrote "I have already tried...". Did you read the entire post?
– CarLaTeX
Dec 1 at 14:38
2
Pleqee, edit the question title to reflect the inquiry of drawing a linear regression graph out of raw data. This will help future readers coming from relevant Google search results.
– Diaa
Dec 2 at 15:43
|
show 6 more comments
The picture shows a normal linear regression.
Although it is easy to compute on paper, I have no idea how to code a linear regression in order to get an output shown as in the picture.
I would be very pleased if anyone could help me.
I have already tried {tikzpicture} etc. but it does not work out that good as pleased.
diagrams plot
The picture shows a normal linear regression.
Although it is easy to compute on paper, I have no idea how to code a linear regression in order to get an output shown as in the picture.
I would be very pleased if anyone could help me.
I have already tried {tikzpicture} etc. but it does not work out that good as pleased.
diagrams plot
diagrams plot
edited Dec 1 at 15:50
Torbjørn T.
155k13245435
155k13245435
asked Dec 1 at 13:05
Rebecca
513
513
8
Welcome to TeX.SE! Please edit your question and add a minimal example of what you tried.
– CarLaTeX
Dec 1 at 13:07
1
pgfplots
allows you to do this, see p 396Fitting Lines - Regression
: ctan.org/pkg/pgfplots
– AndréC
Dec 1 at 13:15
1
@CarLaTeX think a little before you get carried away! Rebecca wants to knowhow she can create this graph
. If she knew, she wouldn't have asked that question.
– AndréC
Dec 1 at 13:39
1
@AndréC Rebecca wrote "I have already tried...". Did you read the entire post?
– CarLaTeX
Dec 1 at 14:38
2
Pleqee, edit the question title to reflect the inquiry of drawing a linear regression graph out of raw data. This will help future readers coming from relevant Google search results.
– Diaa
Dec 2 at 15:43
|
show 6 more comments
8
Welcome to TeX.SE! Please edit your question and add a minimal example of what you tried.
– CarLaTeX
Dec 1 at 13:07
1
pgfplots
allows you to do this, see p 396Fitting Lines - Regression
: ctan.org/pkg/pgfplots
– AndréC
Dec 1 at 13:15
1
@CarLaTeX think a little before you get carried away! Rebecca wants to knowhow she can create this graph
. If she knew, she wouldn't have asked that question.
– AndréC
Dec 1 at 13:39
1
@AndréC Rebecca wrote "I have already tried...". Did you read the entire post?
– CarLaTeX
Dec 1 at 14:38
2
Pleqee, edit the question title to reflect the inquiry of drawing a linear regression graph out of raw data. This will help future readers coming from relevant Google search results.
– Diaa
Dec 2 at 15:43
8
8
Welcome to TeX.SE! Please edit your question and add a minimal example of what you tried.
– CarLaTeX
Dec 1 at 13:07
Welcome to TeX.SE! Please edit your question and add a minimal example of what you tried.
– CarLaTeX
Dec 1 at 13:07
1
1
pgfplots
allows you to do this, see p 396 Fitting Lines - Regression
: ctan.org/pkg/pgfplots– AndréC
Dec 1 at 13:15
pgfplots
allows you to do this, see p 396 Fitting Lines - Regression
: ctan.org/pkg/pgfplots– AndréC
Dec 1 at 13:15
1
1
@CarLaTeX think a little before you get carried away! Rebecca wants to know
how she can create this graph
. If she knew, she wouldn't have asked that question.– AndréC
Dec 1 at 13:39
@CarLaTeX think a little before you get carried away! Rebecca wants to know
how she can create this graph
. If she knew, she wouldn't have asked that question.– AndréC
Dec 1 at 13:39
1
1
@AndréC Rebecca wrote "I have already tried...". Did you read the entire post?
– CarLaTeX
Dec 1 at 14:38
@AndréC Rebecca wrote "I have already tried...". Did you read the entire post?
– CarLaTeX
Dec 1 at 14:38
2
2
Pleqee, edit the question title to reflect the inquiry of drawing a linear regression graph out of raw data. This will help future readers coming from relevant Google search results.
– Diaa
Dec 2 at 15:43
Pleqee, edit the question title to reflect the inquiry of drawing a linear regression graph out of raw data. This will help future readers coming from relevant Google search results.
– Diaa
Dec 2 at 15:43
|
show 6 more comments
2 Answers
2
active
oldest
votes
Motivated by AndréC's comments... ;-)
documentclass{article}
usepackage{tikzlings}
usepackage{pgfplots, pgfplotstable}
pgfplotsset{compat=1.16}
pgfplotstableread{
X Y
1 2
2 2.5
3 6
3 6.5
4 10
5 8
}datatable
begin{document}
pgfplotsset{every axis legend/.append style={
cells={anchor=west}}}
begin{tikzpicture}
begin{axis}[legend pos=north west,xmin=0,xmax=7,
ymin=0,ymax=15,enlargelimits=0.1]
addplot[only marks, mark=*] table[x=X,y=Y] {datatable};
addlegendentry{$y_i$}
addplot[draw=none,color=red] table [
x=X,
y={create col/linear regression={y=Y}},
] {datatable};
xdefslope{pgfplotstableregressiona}
xdefoffset{pgfplotstableregressionb}
addplot[no marks,color=red,domain=-2:9] {slope*x+offset};
addlegendentry{$f(x_i)=beta_0+beta_1x_i$}
coordinate (aux1) at (2,{slope*2+offset});
coordinate (aux2) at (2,2.5);
end{axis}
draw[latex-latex,red] (aux1) -- (aux2)
node[midway,right,text=black,font=sffamily]{St"orterm:
$varepsilon_i=y_i-f(x_i)$};
marmot[xshift=8cm,whiskers,teeth,crystal ball]
end{tikzpicture}
end{document}
And this is motivated by Sebastiano's comment.
documentclass{article}
usepackage{tikzlings}
usepackage{pgfplots, pgfplotstable}
pgfplotsset{compat=1.16}
pgfplotstableread{
X Y
1 2
2 2.5
3 6
3 6.5
4 10
5 8
}datatable
% pgfmanual p. 1087
pgfdeclareradialshading{ballshading}{pgfpoint{-10bp}{10bp}}
{color(0bp)=(cyan!15!white); color(9bp)=(cyan!75!white);
color(18bp)=(cyan!70!black); color(25bp)=(cyan!50!black); color(50bp)=(black)}
pgfdeclareplotmark{crystal ball}{pgfpathcircle{pgfpoint{0ex}{0ex}}{1ex}
pgfshadepath{ballshading}{0}
pgfusepath{}}
begin{document}
pgfplotsset{every axis legend/.append style={
cells={anchor=west}}}
begin{tikzpicture}
begin{axis}[legend pos=north west,xmin=0,xmax=7,
ymin=0,ymax=15,enlargelimits=0.1]
addplot[only marks, mark=crystal ball,opacity=0.7] table[x=X,y=Y] {datatable};
addlegendentry{$y_i$}
addplot[draw=none,color=red] table [
x=X,
y={create col/linear regression={y=Y}},
] {datatable};
xdefslope{pgfplotstableregressiona}
xdefoffset{pgfplotstableregressionb}
addplot[no marks,color=red,domain=-2:9] {slope*x+offset};
addlegendentry{$f(x_i)=beta_0+beta_1x_i$}
coordinate (aux1) at (2,{slope*2+offset});
coordinate (aux2) at (2,2.5);
end{axis}
draw[latex-latex,red] (aux1) -- (aux2)
node[midway,right,text=black,font=sffamily]{St"orterm:
$varepsilon_i=y_i-f(x_i)$};
marmot[xshift=8cm,whiskers,teeth,crystal ball]
end{tikzpicture}
end{document}
Off topic, but how did you make the little bear on the right hand side? It is very cute! :) Just anothertikzpicture
? In particular, how did roughly you make that ball with varying translucency?
– Dan Hoynoski
Dec 1 at 18:31
2
@DanHoynoski Which bear??? Do you see a bear here? You can find bears here, even with crystal balls (and indeed this is just a ball shading with some nontrivial opacity). But I insist that the "little bear" is a "cute marmot". ;-)
– marmot
Dec 1 at 18:40
It would be good to improve the title of the question to say that it is a question of building a normal linear regression.
– AndréC
Dec 1 at 23:03
@AndréC I guess that is more a request to the OP. You may refer to this question to motivate the request. (I won't do that as long as the OP does not confirm that this is what she's after.)
– marmot
Dec 1 at 23:07
3
@AndréC The OP can always comment on their own question
– Joseph Wright♦
Dec 2 at 9:34
|
show 5 more comments
With R and knitr
this plot is relatively simple. However, the MWE is a bit complex to show automatically the actual coefficients (intercept, slope and error) as well as to place legend, arrow and label automatically, so one can change the values at some range (for instance, the second y from 2 to -3) and still have a correct output in all aspects, even in the text out of the figure.
documentclass{article}
usepackage{lipsum}
usepackage[german]{babel}
usepackage[utf8]{inputenc}
<<Daten,echo=F>>=
df <- data.frame(x=c(1,2,3,3,4,5),y=c(1,2,6,7,10,8))
@
begin{document}
lipsum[2]
<<Streudiagramm,echo=F,dev="tikz", fig.cap="Regressionszeile zeigt den Fehler $\varepsilon_2$", fig.width=4.2, fig.height=3.5,fig.align='center',fig.pos="h">>=
par(mar=c(4,4,1,4)) # optional, just to crop
mod <- lm(df$y~df$x)
with(df,plot(x,y, pch=21, col="red",bg="yellow",ylim=c(min(df$y-.1),max(df$y+.1))))
abline(mod,col="blue",lwd=3)
legend(1, max(df$y), legend=c("$f(x_i)=\beta_0+\beta_1x_i$",
paste("$y=",
signif(mod$coefficients[1],3),"+",
signif(mod$coefficients[2],3),"x$")),
col=c("blue","white"), lty=1:2, cex=0.8)
arrows(df$x[2],df$y[2],df$x[2],predict(mod)[2], length=0.05, col =2, code=3)
text(df$x[2]+.1,mean(c(df$y[2],predict(mod)[2])),paste('Str\"{o}erm: $\varepsilon_i=y_i-f(x_i)$ =',signif(df$y[2]-predict(mod)[2],3)),adj=0)
@
Die Abbildung ref{fig:Streudiagramm} zeigt das $varepsilon_2 =
Sexpr{signif(df$y[2],3)} -
Sexpr{signif(predict(mod)[2],3)} =
Sexpr{signif(df$y[2]-(mod$coefficients[1]+(mod$coefficients[2]*df$x[2])),3)} $.
lipsum[3]
end{document}
thank you so much guys!
– Rebecca
Dec 2 at 13:22
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Motivated by AndréC's comments... ;-)
documentclass{article}
usepackage{tikzlings}
usepackage{pgfplots, pgfplotstable}
pgfplotsset{compat=1.16}
pgfplotstableread{
X Y
1 2
2 2.5
3 6
3 6.5
4 10
5 8
}datatable
begin{document}
pgfplotsset{every axis legend/.append style={
cells={anchor=west}}}
begin{tikzpicture}
begin{axis}[legend pos=north west,xmin=0,xmax=7,
ymin=0,ymax=15,enlargelimits=0.1]
addplot[only marks, mark=*] table[x=X,y=Y] {datatable};
addlegendentry{$y_i$}
addplot[draw=none,color=red] table [
x=X,
y={create col/linear regression={y=Y}},
] {datatable};
xdefslope{pgfplotstableregressiona}
xdefoffset{pgfplotstableregressionb}
addplot[no marks,color=red,domain=-2:9] {slope*x+offset};
addlegendentry{$f(x_i)=beta_0+beta_1x_i$}
coordinate (aux1) at (2,{slope*2+offset});
coordinate (aux2) at (2,2.5);
end{axis}
draw[latex-latex,red] (aux1) -- (aux2)
node[midway,right,text=black,font=sffamily]{St"orterm:
$varepsilon_i=y_i-f(x_i)$};
marmot[xshift=8cm,whiskers,teeth,crystal ball]
end{tikzpicture}
end{document}
And this is motivated by Sebastiano's comment.
documentclass{article}
usepackage{tikzlings}
usepackage{pgfplots, pgfplotstable}
pgfplotsset{compat=1.16}
pgfplotstableread{
X Y
1 2
2 2.5
3 6
3 6.5
4 10
5 8
}datatable
% pgfmanual p. 1087
pgfdeclareradialshading{ballshading}{pgfpoint{-10bp}{10bp}}
{color(0bp)=(cyan!15!white); color(9bp)=(cyan!75!white);
color(18bp)=(cyan!70!black); color(25bp)=(cyan!50!black); color(50bp)=(black)}
pgfdeclareplotmark{crystal ball}{pgfpathcircle{pgfpoint{0ex}{0ex}}{1ex}
pgfshadepath{ballshading}{0}
pgfusepath{}}
begin{document}
pgfplotsset{every axis legend/.append style={
cells={anchor=west}}}
begin{tikzpicture}
begin{axis}[legend pos=north west,xmin=0,xmax=7,
ymin=0,ymax=15,enlargelimits=0.1]
addplot[only marks, mark=crystal ball,opacity=0.7] table[x=X,y=Y] {datatable};
addlegendentry{$y_i$}
addplot[draw=none,color=red] table [
x=X,
y={create col/linear regression={y=Y}},
] {datatable};
xdefslope{pgfplotstableregressiona}
xdefoffset{pgfplotstableregressionb}
addplot[no marks,color=red,domain=-2:9] {slope*x+offset};
addlegendentry{$f(x_i)=beta_0+beta_1x_i$}
coordinate (aux1) at (2,{slope*2+offset});
coordinate (aux2) at (2,2.5);
end{axis}
draw[latex-latex,red] (aux1) -- (aux2)
node[midway,right,text=black,font=sffamily]{St"orterm:
$varepsilon_i=y_i-f(x_i)$};
marmot[xshift=8cm,whiskers,teeth,crystal ball]
end{tikzpicture}
end{document}
Off topic, but how did you make the little bear on the right hand side? It is very cute! :) Just anothertikzpicture
? In particular, how did roughly you make that ball with varying translucency?
– Dan Hoynoski
Dec 1 at 18:31
2
@DanHoynoski Which bear??? Do you see a bear here? You can find bears here, even with crystal balls (and indeed this is just a ball shading with some nontrivial opacity). But I insist that the "little bear" is a "cute marmot". ;-)
– marmot
Dec 1 at 18:40
It would be good to improve the title of the question to say that it is a question of building a normal linear regression.
– AndréC
Dec 1 at 23:03
@AndréC I guess that is more a request to the OP. You may refer to this question to motivate the request. (I won't do that as long as the OP does not confirm that this is what she's after.)
– marmot
Dec 1 at 23:07
3
@AndréC The OP can always comment on their own question
– Joseph Wright♦
Dec 2 at 9:34
|
show 5 more comments
Motivated by AndréC's comments... ;-)
documentclass{article}
usepackage{tikzlings}
usepackage{pgfplots, pgfplotstable}
pgfplotsset{compat=1.16}
pgfplotstableread{
X Y
1 2
2 2.5
3 6
3 6.5
4 10
5 8
}datatable
begin{document}
pgfplotsset{every axis legend/.append style={
cells={anchor=west}}}
begin{tikzpicture}
begin{axis}[legend pos=north west,xmin=0,xmax=7,
ymin=0,ymax=15,enlargelimits=0.1]
addplot[only marks, mark=*] table[x=X,y=Y] {datatable};
addlegendentry{$y_i$}
addplot[draw=none,color=red] table [
x=X,
y={create col/linear regression={y=Y}},
] {datatable};
xdefslope{pgfplotstableregressiona}
xdefoffset{pgfplotstableregressionb}
addplot[no marks,color=red,domain=-2:9] {slope*x+offset};
addlegendentry{$f(x_i)=beta_0+beta_1x_i$}
coordinate (aux1) at (2,{slope*2+offset});
coordinate (aux2) at (2,2.5);
end{axis}
draw[latex-latex,red] (aux1) -- (aux2)
node[midway,right,text=black,font=sffamily]{St"orterm:
$varepsilon_i=y_i-f(x_i)$};
marmot[xshift=8cm,whiskers,teeth,crystal ball]
end{tikzpicture}
end{document}
And this is motivated by Sebastiano's comment.
documentclass{article}
usepackage{tikzlings}
usepackage{pgfplots, pgfplotstable}
pgfplotsset{compat=1.16}
pgfplotstableread{
X Y
1 2
2 2.5
3 6
3 6.5
4 10
5 8
}datatable
% pgfmanual p. 1087
pgfdeclareradialshading{ballshading}{pgfpoint{-10bp}{10bp}}
{color(0bp)=(cyan!15!white); color(9bp)=(cyan!75!white);
color(18bp)=(cyan!70!black); color(25bp)=(cyan!50!black); color(50bp)=(black)}
pgfdeclareplotmark{crystal ball}{pgfpathcircle{pgfpoint{0ex}{0ex}}{1ex}
pgfshadepath{ballshading}{0}
pgfusepath{}}
begin{document}
pgfplotsset{every axis legend/.append style={
cells={anchor=west}}}
begin{tikzpicture}
begin{axis}[legend pos=north west,xmin=0,xmax=7,
ymin=0,ymax=15,enlargelimits=0.1]
addplot[only marks, mark=crystal ball,opacity=0.7] table[x=X,y=Y] {datatable};
addlegendentry{$y_i$}
addplot[draw=none,color=red] table [
x=X,
y={create col/linear regression={y=Y}},
] {datatable};
xdefslope{pgfplotstableregressiona}
xdefoffset{pgfplotstableregressionb}
addplot[no marks,color=red,domain=-2:9] {slope*x+offset};
addlegendentry{$f(x_i)=beta_0+beta_1x_i$}
coordinate (aux1) at (2,{slope*2+offset});
coordinate (aux2) at (2,2.5);
end{axis}
draw[latex-latex,red] (aux1) -- (aux2)
node[midway,right,text=black,font=sffamily]{St"orterm:
$varepsilon_i=y_i-f(x_i)$};
marmot[xshift=8cm,whiskers,teeth,crystal ball]
end{tikzpicture}
end{document}
Off topic, but how did you make the little bear on the right hand side? It is very cute! :) Just anothertikzpicture
? In particular, how did roughly you make that ball with varying translucency?
– Dan Hoynoski
Dec 1 at 18:31
2
@DanHoynoski Which bear??? Do you see a bear here? You can find bears here, even with crystal balls (and indeed this is just a ball shading with some nontrivial opacity). But I insist that the "little bear" is a "cute marmot". ;-)
– marmot
Dec 1 at 18:40
It would be good to improve the title of the question to say that it is a question of building a normal linear regression.
– AndréC
Dec 1 at 23:03
@AndréC I guess that is more a request to the OP. You may refer to this question to motivate the request. (I won't do that as long as the OP does not confirm that this is what she's after.)
– marmot
Dec 1 at 23:07
3
@AndréC The OP can always comment on their own question
– Joseph Wright♦
Dec 2 at 9:34
|
show 5 more comments
Motivated by AndréC's comments... ;-)
documentclass{article}
usepackage{tikzlings}
usepackage{pgfplots, pgfplotstable}
pgfplotsset{compat=1.16}
pgfplotstableread{
X Y
1 2
2 2.5
3 6
3 6.5
4 10
5 8
}datatable
begin{document}
pgfplotsset{every axis legend/.append style={
cells={anchor=west}}}
begin{tikzpicture}
begin{axis}[legend pos=north west,xmin=0,xmax=7,
ymin=0,ymax=15,enlargelimits=0.1]
addplot[only marks, mark=*] table[x=X,y=Y] {datatable};
addlegendentry{$y_i$}
addplot[draw=none,color=red] table [
x=X,
y={create col/linear regression={y=Y}},
] {datatable};
xdefslope{pgfplotstableregressiona}
xdefoffset{pgfplotstableregressionb}
addplot[no marks,color=red,domain=-2:9] {slope*x+offset};
addlegendentry{$f(x_i)=beta_0+beta_1x_i$}
coordinate (aux1) at (2,{slope*2+offset});
coordinate (aux2) at (2,2.5);
end{axis}
draw[latex-latex,red] (aux1) -- (aux2)
node[midway,right,text=black,font=sffamily]{St"orterm:
$varepsilon_i=y_i-f(x_i)$};
marmot[xshift=8cm,whiskers,teeth,crystal ball]
end{tikzpicture}
end{document}
And this is motivated by Sebastiano's comment.
documentclass{article}
usepackage{tikzlings}
usepackage{pgfplots, pgfplotstable}
pgfplotsset{compat=1.16}
pgfplotstableread{
X Y
1 2
2 2.5
3 6
3 6.5
4 10
5 8
}datatable
% pgfmanual p. 1087
pgfdeclareradialshading{ballshading}{pgfpoint{-10bp}{10bp}}
{color(0bp)=(cyan!15!white); color(9bp)=(cyan!75!white);
color(18bp)=(cyan!70!black); color(25bp)=(cyan!50!black); color(50bp)=(black)}
pgfdeclareplotmark{crystal ball}{pgfpathcircle{pgfpoint{0ex}{0ex}}{1ex}
pgfshadepath{ballshading}{0}
pgfusepath{}}
begin{document}
pgfplotsset{every axis legend/.append style={
cells={anchor=west}}}
begin{tikzpicture}
begin{axis}[legend pos=north west,xmin=0,xmax=7,
ymin=0,ymax=15,enlargelimits=0.1]
addplot[only marks, mark=crystal ball,opacity=0.7] table[x=X,y=Y] {datatable};
addlegendentry{$y_i$}
addplot[draw=none,color=red] table [
x=X,
y={create col/linear regression={y=Y}},
] {datatable};
xdefslope{pgfplotstableregressiona}
xdefoffset{pgfplotstableregressionb}
addplot[no marks,color=red,domain=-2:9] {slope*x+offset};
addlegendentry{$f(x_i)=beta_0+beta_1x_i$}
coordinate (aux1) at (2,{slope*2+offset});
coordinate (aux2) at (2,2.5);
end{axis}
draw[latex-latex,red] (aux1) -- (aux2)
node[midway,right,text=black,font=sffamily]{St"orterm:
$varepsilon_i=y_i-f(x_i)$};
marmot[xshift=8cm,whiskers,teeth,crystal ball]
end{tikzpicture}
end{document}
Motivated by AndréC's comments... ;-)
documentclass{article}
usepackage{tikzlings}
usepackage{pgfplots, pgfplotstable}
pgfplotsset{compat=1.16}
pgfplotstableread{
X Y
1 2
2 2.5
3 6
3 6.5
4 10
5 8
}datatable
begin{document}
pgfplotsset{every axis legend/.append style={
cells={anchor=west}}}
begin{tikzpicture}
begin{axis}[legend pos=north west,xmin=0,xmax=7,
ymin=0,ymax=15,enlargelimits=0.1]
addplot[only marks, mark=*] table[x=X,y=Y] {datatable};
addlegendentry{$y_i$}
addplot[draw=none,color=red] table [
x=X,
y={create col/linear regression={y=Y}},
] {datatable};
xdefslope{pgfplotstableregressiona}
xdefoffset{pgfplotstableregressionb}
addplot[no marks,color=red,domain=-2:9] {slope*x+offset};
addlegendentry{$f(x_i)=beta_0+beta_1x_i$}
coordinate (aux1) at (2,{slope*2+offset});
coordinate (aux2) at (2,2.5);
end{axis}
draw[latex-latex,red] (aux1) -- (aux2)
node[midway,right,text=black,font=sffamily]{St"orterm:
$varepsilon_i=y_i-f(x_i)$};
marmot[xshift=8cm,whiskers,teeth,crystal ball]
end{tikzpicture}
end{document}
And this is motivated by Sebastiano's comment.
documentclass{article}
usepackage{tikzlings}
usepackage{pgfplots, pgfplotstable}
pgfplotsset{compat=1.16}
pgfplotstableread{
X Y
1 2
2 2.5
3 6
3 6.5
4 10
5 8
}datatable
% pgfmanual p. 1087
pgfdeclareradialshading{ballshading}{pgfpoint{-10bp}{10bp}}
{color(0bp)=(cyan!15!white); color(9bp)=(cyan!75!white);
color(18bp)=(cyan!70!black); color(25bp)=(cyan!50!black); color(50bp)=(black)}
pgfdeclareplotmark{crystal ball}{pgfpathcircle{pgfpoint{0ex}{0ex}}{1ex}
pgfshadepath{ballshading}{0}
pgfusepath{}}
begin{document}
pgfplotsset{every axis legend/.append style={
cells={anchor=west}}}
begin{tikzpicture}
begin{axis}[legend pos=north west,xmin=0,xmax=7,
ymin=0,ymax=15,enlargelimits=0.1]
addplot[only marks, mark=crystal ball,opacity=0.7] table[x=X,y=Y] {datatable};
addlegendentry{$y_i$}
addplot[draw=none,color=red] table [
x=X,
y={create col/linear regression={y=Y}},
] {datatable};
xdefslope{pgfplotstableregressiona}
xdefoffset{pgfplotstableregressionb}
addplot[no marks,color=red,domain=-2:9] {slope*x+offset};
addlegendentry{$f(x_i)=beta_0+beta_1x_i$}
coordinate (aux1) at (2,{slope*2+offset});
coordinate (aux2) at (2,2.5);
end{axis}
draw[latex-latex,red] (aux1) -- (aux2)
node[midway,right,text=black,font=sffamily]{St"orterm:
$varepsilon_i=y_i-f(x_i)$};
marmot[xshift=8cm,whiskers,teeth,crystal ball]
end{tikzpicture}
end{document}
edited Dec 1 at 22:50
answered Dec 1 at 16:07
marmot
87.2k4100187
87.2k4100187
Off topic, but how did you make the little bear on the right hand side? It is very cute! :) Just anothertikzpicture
? In particular, how did roughly you make that ball with varying translucency?
– Dan Hoynoski
Dec 1 at 18:31
2
@DanHoynoski Which bear??? Do you see a bear here? You can find bears here, even with crystal balls (and indeed this is just a ball shading with some nontrivial opacity). But I insist that the "little bear" is a "cute marmot". ;-)
– marmot
Dec 1 at 18:40
It would be good to improve the title of the question to say that it is a question of building a normal linear regression.
– AndréC
Dec 1 at 23:03
@AndréC I guess that is more a request to the OP. You may refer to this question to motivate the request. (I won't do that as long as the OP does not confirm that this is what she's after.)
– marmot
Dec 1 at 23:07
3
@AndréC The OP can always comment on their own question
– Joseph Wright♦
Dec 2 at 9:34
|
show 5 more comments
Off topic, but how did you make the little bear on the right hand side? It is very cute! :) Just anothertikzpicture
? In particular, how did roughly you make that ball with varying translucency?
– Dan Hoynoski
Dec 1 at 18:31
2
@DanHoynoski Which bear??? Do you see a bear here? You can find bears here, even with crystal balls (and indeed this is just a ball shading with some nontrivial opacity). But I insist that the "little bear" is a "cute marmot". ;-)
– marmot
Dec 1 at 18:40
It would be good to improve the title of the question to say that it is a question of building a normal linear regression.
– AndréC
Dec 1 at 23:03
@AndréC I guess that is more a request to the OP. You may refer to this question to motivate the request. (I won't do that as long as the OP does not confirm that this is what she's after.)
– marmot
Dec 1 at 23:07
3
@AndréC The OP can always comment on their own question
– Joseph Wright♦
Dec 2 at 9:34
Off topic, but how did you make the little bear on the right hand side? It is very cute! :) Just another
tikzpicture
? In particular, how did roughly you make that ball with varying translucency?– Dan Hoynoski
Dec 1 at 18:31
Off topic, but how did you make the little bear on the right hand side? It is very cute! :) Just another
tikzpicture
? In particular, how did roughly you make that ball with varying translucency?– Dan Hoynoski
Dec 1 at 18:31
2
2
@DanHoynoski Which bear??? Do you see a bear here? You can find bears here, even with crystal balls (and indeed this is just a ball shading with some nontrivial opacity). But I insist that the "little bear" is a "cute marmot". ;-)
– marmot
Dec 1 at 18:40
@DanHoynoski Which bear??? Do you see a bear here? You can find bears here, even with crystal balls (and indeed this is just a ball shading with some nontrivial opacity). But I insist that the "little bear" is a "cute marmot". ;-)
– marmot
Dec 1 at 18:40
It would be good to improve the title of the question to say that it is a question of building a normal linear regression.
– AndréC
Dec 1 at 23:03
It would be good to improve the title of the question to say that it is a question of building a normal linear regression.
– AndréC
Dec 1 at 23:03
@AndréC I guess that is more a request to the OP. You may refer to this question to motivate the request. (I won't do that as long as the OP does not confirm that this is what she's after.)
– marmot
Dec 1 at 23:07
@AndréC I guess that is more a request to the OP. You may refer to this question to motivate the request. (I won't do that as long as the OP does not confirm that this is what she's after.)
– marmot
Dec 1 at 23:07
3
3
@AndréC The OP can always comment on their own question
– Joseph Wright♦
Dec 2 at 9:34
@AndréC The OP can always comment on their own question
– Joseph Wright♦
Dec 2 at 9:34
|
show 5 more comments
With R and knitr
this plot is relatively simple. However, the MWE is a bit complex to show automatically the actual coefficients (intercept, slope and error) as well as to place legend, arrow and label automatically, so one can change the values at some range (for instance, the second y from 2 to -3) and still have a correct output in all aspects, even in the text out of the figure.
documentclass{article}
usepackage{lipsum}
usepackage[german]{babel}
usepackage[utf8]{inputenc}
<<Daten,echo=F>>=
df <- data.frame(x=c(1,2,3,3,4,5),y=c(1,2,6,7,10,8))
@
begin{document}
lipsum[2]
<<Streudiagramm,echo=F,dev="tikz", fig.cap="Regressionszeile zeigt den Fehler $\varepsilon_2$", fig.width=4.2, fig.height=3.5,fig.align='center',fig.pos="h">>=
par(mar=c(4,4,1,4)) # optional, just to crop
mod <- lm(df$y~df$x)
with(df,plot(x,y, pch=21, col="red",bg="yellow",ylim=c(min(df$y-.1),max(df$y+.1))))
abline(mod,col="blue",lwd=3)
legend(1, max(df$y), legend=c("$f(x_i)=\beta_0+\beta_1x_i$",
paste("$y=",
signif(mod$coefficients[1],3),"+",
signif(mod$coefficients[2],3),"x$")),
col=c("blue","white"), lty=1:2, cex=0.8)
arrows(df$x[2],df$y[2],df$x[2],predict(mod)[2], length=0.05, col =2, code=3)
text(df$x[2]+.1,mean(c(df$y[2],predict(mod)[2])),paste('Str\"{o}erm: $\varepsilon_i=y_i-f(x_i)$ =',signif(df$y[2]-predict(mod)[2],3)),adj=0)
@
Die Abbildung ref{fig:Streudiagramm} zeigt das $varepsilon_2 =
Sexpr{signif(df$y[2],3)} -
Sexpr{signif(predict(mod)[2],3)} =
Sexpr{signif(df$y[2]-(mod$coefficients[1]+(mod$coefficients[2]*df$x[2])),3)} $.
lipsum[3]
end{document}
thank you so much guys!
– Rebecca
Dec 2 at 13:22
add a comment |
With R and knitr
this plot is relatively simple. However, the MWE is a bit complex to show automatically the actual coefficients (intercept, slope and error) as well as to place legend, arrow and label automatically, so one can change the values at some range (for instance, the second y from 2 to -3) and still have a correct output in all aspects, even in the text out of the figure.
documentclass{article}
usepackage{lipsum}
usepackage[german]{babel}
usepackage[utf8]{inputenc}
<<Daten,echo=F>>=
df <- data.frame(x=c(1,2,3,3,4,5),y=c(1,2,6,7,10,8))
@
begin{document}
lipsum[2]
<<Streudiagramm,echo=F,dev="tikz", fig.cap="Regressionszeile zeigt den Fehler $\varepsilon_2$", fig.width=4.2, fig.height=3.5,fig.align='center',fig.pos="h">>=
par(mar=c(4,4,1,4)) # optional, just to crop
mod <- lm(df$y~df$x)
with(df,plot(x,y, pch=21, col="red",bg="yellow",ylim=c(min(df$y-.1),max(df$y+.1))))
abline(mod,col="blue",lwd=3)
legend(1, max(df$y), legend=c("$f(x_i)=\beta_0+\beta_1x_i$",
paste("$y=",
signif(mod$coefficients[1],3),"+",
signif(mod$coefficients[2],3),"x$")),
col=c("blue","white"), lty=1:2, cex=0.8)
arrows(df$x[2],df$y[2],df$x[2],predict(mod)[2], length=0.05, col =2, code=3)
text(df$x[2]+.1,mean(c(df$y[2],predict(mod)[2])),paste('Str\"{o}erm: $\varepsilon_i=y_i-f(x_i)$ =',signif(df$y[2]-predict(mod)[2],3)),adj=0)
@
Die Abbildung ref{fig:Streudiagramm} zeigt das $varepsilon_2 =
Sexpr{signif(df$y[2],3)} -
Sexpr{signif(predict(mod)[2],3)} =
Sexpr{signif(df$y[2]-(mod$coefficients[1]+(mod$coefficients[2]*df$x[2])),3)} $.
lipsum[3]
end{document}
thank you so much guys!
– Rebecca
Dec 2 at 13:22
add a comment |
With R and knitr
this plot is relatively simple. However, the MWE is a bit complex to show automatically the actual coefficients (intercept, slope and error) as well as to place legend, arrow and label automatically, so one can change the values at some range (for instance, the second y from 2 to -3) and still have a correct output in all aspects, even in the text out of the figure.
documentclass{article}
usepackage{lipsum}
usepackage[german]{babel}
usepackage[utf8]{inputenc}
<<Daten,echo=F>>=
df <- data.frame(x=c(1,2,3,3,4,5),y=c(1,2,6,7,10,8))
@
begin{document}
lipsum[2]
<<Streudiagramm,echo=F,dev="tikz", fig.cap="Regressionszeile zeigt den Fehler $\varepsilon_2$", fig.width=4.2, fig.height=3.5,fig.align='center',fig.pos="h">>=
par(mar=c(4,4,1,4)) # optional, just to crop
mod <- lm(df$y~df$x)
with(df,plot(x,y, pch=21, col="red",bg="yellow",ylim=c(min(df$y-.1),max(df$y+.1))))
abline(mod,col="blue",lwd=3)
legend(1, max(df$y), legend=c("$f(x_i)=\beta_0+\beta_1x_i$",
paste("$y=",
signif(mod$coefficients[1],3),"+",
signif(mod$coefficients[2],3),"x$")),
col=c("blue","white"), lty=1:2, cex=0.8)
arrows(df$x[2],df$y[2],df$x[2],predict(mod)[2], length=0.05, col =2, code=3)
text(df$x[2]+.1,mean(c(df$y[2],predict(mod)[2])),paste('Str\"{o}erm: $\varepsilon_i=y_i-f(x_i)$ =',signif(df$y[2]-predict(mod)[2],3)),adj=0)
@
Die Abbildung ref{fig:Streudiagramm} zeigt das $varepsilon_2 =
Sexpr{signif(df$y[2],3)} -
Sexpr{signif(predict(mod)[2],3)} =
Sexpr{signif(df$y[2]-(mod$coefficients[1]+(mod$coefficients[2]*df$x[2])),3)} $.
lipsum[3]
end{document}
With R and knitr
this plot is relatively simple. However, the MWE is a bit complex to show automatically the actual coefficients (intercept, slope and error) as well as to place legend, arrow and label automatically, so one can change the values at some range (for instance, the second y from 2 to -3) and still have a correct output in all aspects, even in the text out of the figure.
documentclass{article}
usepackage{lipsum}
usepackage[german]{babel}
usepackage[utf8]{inputenc}
<<Daten,echo=F>>=
df <- data.frame(x=c(1,2,3,3,4,5),y=c(1,2,6,7,10,8))
@
begin{document}
lipsum[2]
<<Streudiagramm,echo=F,dev="tikz", fig.cap="Regressionszeile zeigt den Fehler $\varepsilon_2$", fig.width=4.2, fig.height=3.5,fig.align='center',fig.pos="h">>=
par(mar=c(4,4,1,4)) # optional, just to crop
mod <- lm(df$y~df$x)
with(df,plot(x,y, pch=21, col="red",bg="yellow",ylim=c(min(df$y-.1),max(df$y+.1))))
abline(mod,col="blue",lwd=3)
legend(1, max(df$y), legend=c("$f(x_i)=\beta_0+\beta_1x_i$",
paste("$y=",
signif(mod$coefficients[1],3),"+",
signif(mod$coefficients[2],3),"x$")),
col=c("blue","white"), lty=1:2, cex=0.8)
arrows(df$x[2],df$y[2],df$x[2],predict(mod)[2], length=0.05, col =2, code=3)
text(df$x[2]+.1,mean(c(df$y[2],predict(mod)[2])),paste('Str\"{o}erm: $\varepsilon_i=y_i-f(x_i)$ =',signif(df$y[2]-predict(mod)[2],3)),adj=0)
@
Die Abbildung ref{fig:Streudiagramm} zeigt das $varepsilon_2 =
Sexpr{signif(df$y[2],3)} -
Sexpr{signif(predict(mod)[2],3)} =
Sexpr{signif(df$y[2]-(mod$coefficients[1]+(mod$coefficients[2]*df$x[2])),3)} $.
lipsum[3]
end{document}
edited Dec 2 at 10:31
answered Dec 2 at 2:54
Fran
51.3k6112175
51.3k6112175
thank you so much guys!
– Rebecca
Dec 2 at 13:22
add a comment |
thank you so much guys!
– Rebecca
Dec 2 at 13:22
thank you so much guys!
– Rebecca
Dec 2 at 13:22
thank you so much guys!
– Rebecca
Dec 2 at 13:22
add a comment |
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8
Welcome to TeX.SE! Please edit your question and add a minimal example of what you tried.
– CarLaTeX
Dec 1 at 13:07
1
pgfplots
allows you to do this, see p 396Fitting Lines - Regression
: ctan.org/pkg/pgfplots– AndréC
Dec 1 at 13:15
1
@CarLaTeX think a little before you get carried away! Rebecca wants to know
how she can create this graph
. If she knew, she wouldn't have asked that question.– AndréC
Dec 1 at 13:39
1
@AndréC Rebecca wrote "I have already tried...". Did you read the entire post?
– CarLaTeX
Dec 1 at 14:38
2
Pleqee, edit the question title to reflect the inquiry of drawing a linear regression graph out of raw data. This will help future readers coming from relevant Google search results.
– Diaa
Dec 2 at 15:43