blocks|key|1170791|text|我非常喜欢loess()的平滑效果：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1170792|x+<-+1:10
y+<-+c(2,4,6,8,7,12,14,16,18,20)
lo+<-+loess(y~x)
plot(x,y)
lines(predict(lo),+col='red',+lwd=2)|code-block|syntax|javascript|1170793|Venables和Ripley的质量书有一整节关于平滑的内容，还涵盖了样条和多项式--但loess()几乎是每个人的最爱。|1170794|entityMap^0|5|7|0|0|18|7|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

I like <code>loess()</code> a lot for smoothing:

<pre><code>x &lt;- 1:10
y &lt;- c(2,4,6,8,7,12,14,16,18,20)
lo &lt;- loess(y~x)
plot(x,y)
lines(predict(lo), col='red', lwd=2)
</code></pre>

Venables and Ripley's MASS book has an entire section on smoothing that also covers splines and polynomials -- but <code>loess()</code> is just about everybody's favourite.

blocks|key|1170898|text|也许smooth.spline是一个选项，你可以在这里设置一个平滑参数(通常在0和1之间|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1170899|smoothingSpline+=+smooth.spline(x,+y,+spar=0.35)
plot(x,y)
lines(smoothingSpline)|code-block|syntax|javascript|1170900|您还可以对smooth.spline对象使用预测。该函数以R为基数，详情请参见?smooth.spline。|1170901|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Maybe smooth.spline is an option, You can set a smoothing parameter (typically between 0 and 1) here

<pre><code>smoothingSpline = smooth.spline(x, y, spar=0.35)
plot(x,y)
lines(smoothingSpline)
</code></pre>

you can also use predict on smooth.spline objects. The function comes with base R, see 
?smooth.spline for details.

blocks|key|1170969|text|为了让它变得更流畅。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1170970|x+<-+1:10
y+<-+c(2,4,6,8,7,8,14,16,18,20)
lo+<-+loess(y~x)
plot(x,y)
xl+<-+seq(min(x),max(x),+(max(x)+-+min(x))/1000)
lines(xl,+predict(lo,xl),+col='red',+lwd=2)|code-block|syntax|javascript|1170971|这种风格插入了很多额外的点，并得到了一条非常平滑的曲线。这似乎也是ggplot所采用的方法。如果标准的平滑度很好，你可以直接使用。|1170972|scatter.smooth(x,+y)|1170973|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

In order to get it REALLY smoooth...

<pre><code>x &lt;- 1:10
y &lt;- c(2,4,6,8,7,8,14,16,18,20)
lo &lt;- loess(y~x)
plot(x,y)
xl &lt;- seq(min(x),max(x), (max(x) - min(x))/1000)
lines(xl, predict(lo,xl), col='red', lwd=2)
</code></pre>

This style interpolates lots of extra points and gets you a curve that is very smooth. It also appears to be the the approach that ggplot takes. If the standard level of smoothness is fine you can just use.

<pre><code>scatter.smooth(x, y)
</code></pre>

blocks|key|956552|text|正如德克所说，黄土是一种非常好的方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|956553|另一种选择是使用Bezier样条，在某些情况下，如果数据点不多，这种方法可能比LOESS更好。|956554|在这里你可以找到一个例子：http://rosettacode.org/wiki/Cubic_bezier_curves#R|offset|length|956555|#+x,+y:+the+x+and+y+coordinates+of+the+hull+points
#+n:+the+number+of+points+in+the+curve.
bezierCurve+<-+function(x,+y,+n=10)
++++{
++++outx+<-+NULL
++++outy+<-+NULL

++++i+<-+1
++++for+(t+in+seq(0,+1,+length.out=n))
++++++++{
++++++++b+<-+bez(x,+y,+t)
++++++++outx[i]+<-+b$x
++++++++outy[i]+<-+b$y

++++++++i+<-+i%2B1
++++++++}

++++return+(list(x=outx,+y=outy))
++++}

bez+<-+function(x,+y,+t)
++++{
++++outx+<-+0
++++outy+<-+0
++++n+<-+length(x)-1
++++for+(i+in+0:n)
++++++++{
++++++++outx+<-+outx+%2B+choose(n,+i)*((1-t)%5E(n-i))*t%5Ei*x[i%2B1]
++++++++outy+<-+outy+%2B+choose(n,+i)*((1-t)%5E(n-i))*t%5Ei*y[i%2B1]
++++++++}

++++return+(list(x=outx,+y=outy))
++++}

#+Example+usage
x+<-+c(4,6,4,5,6,7)
y+<-+1:6
plot(x,+y,+"o",+pch=20)
points(bezierCurve(x,y,20),+type="l",+col="red")|code-block|syntax|javascript|956556|entityMap|0|LINK|mutability|MUTABLE|url|http://rosettacode.org/wiki/Cubic_bezier_curves#R^0|0|0|D|1D|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@$F|X|G|Y|1|Z]]|A|$]]|$1|H|3|I|5|J|7|10|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

LOESS is a very good approach, as Dirk said.

Another option is using Bezier splines, which may in some cases work better than LOESS if you don't have many data points.

Here you'll find an example: <a href="http://rosettacode.org/wiki/Cubic_bezier_curves#R" rel="noreferrer">http://rosettacode.org/wiki/Cubic_bezier_curves#R</a>

<pre><code># x, y: the x and y coordinates of the hull points
# n: the number of points in the curve.
bezierCurve &lt;- function(x, y, n=10)
 {
 outx &lt;- NULL
 outy &lt;- NULL

 i &lt;- 1
 for (t in seq(0, 1, length.out=n))
 {
 b &lt;- bez(x, y, t)
 outx[i] &lt;- b$x
 outy[i] &lt;- b$y

 i &lt;- i+1
 }

 return (list(x=outx, y=outy))
 }

bez &lt;- function(x, y, t)
 {
 outx &lt;- 0
 outy &lt;- 0
 n &lt;- length(x)-1
 for (i in 0:n)
 {
 outx &lt;- outx + choose(n, i)*((1-t)^(n-i))*t^i*x[i+1]
 outy &lt;- outy + choose(n, i)*((1-t)^(n-i))*t^i*y[i+1]
 }

 return (list(x=outx, y=outy))
 }

# Example usage
x &lt;- c(4,6,4,5,6,7)
y &lt;- 1:6
plot(x, y, "o", pch=20)
points(bezierCurve(x,y,20), type="l", col="red")
</code></pre>

blocks|key|1171138|text|其他答案都是很好的方法。然而，在R中还有一些其他选项没有被提及，包括lowess和approx，它们可能会提供更好的拟合或更快的性能。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1171139|使用备用数据集可以更容易地展示其优势：|1171140|sigmoid+<-+function(x)
{
++y<-1/(1%2Bexp(-.15*(x-100)))
++return(y)
}

dat<-data.frame(x=rnorm(5000)*30%2B100)
dat$y<-as.numeric(as.logical(round(sigmoid(dat$x)%2Brnorm(5000)*.3,0)))|code-block|syntax|javascript|1171141|下面是生成它的sigmoid曲线覆盖的数据：|1171142|​|1171143|📷|atomic|1171144|1171145|在观察群体中的二进制行为时，这种类型的数据很常见。例如，这可能是客户是否购买了某种东西(y轴上的二进制1/0+)与他们在网站上花费的时间(x轴)的关系图。|1171146|为了更好地演示这些函数的性能差异，我们使用了大量的点。|1171147|Smooth、spline和smooth.spline都会使用我尝试过的任何一组参数在这样的数据集上产生乱码，这可能是因为它们倾向于映射到每个点，这对噪声数据不起作用。|1171148|loess、lowess和approx函数都会产生有用的结果，尽管只对approx有用。这是使用轻微优化的参数的每一个的代码：|1171149|loessFit+<-+loess(y~x,+dat,+span+=+0.6)
loessFit+<-+data.frame(x=loessFit$x,y=loessFit$fitted)
loessFit+<-+loessFit[order(loessFit$x),]

approxFit+<-+approx(dat,n+=+15)

lowessFit+<-data.frame(lowess(dat,f+=+.6,iter=1))|1171150|结果是：|1171151|plot(dat,col='gray')
curve(sigmoid,0,200,add=TRUE,col='blue',)
lines(lowessFit,col='red')
lines(loessFit,col='green')
lines(approxFit,col='purple')
legend(150,.6,
+++++++legend=c("Sigmoid","Loess","Lowess",'Approx'),
+++++++lty=c(1,1),
+++++++lwd=c(2.5,2.5),col=c("blue","green","red","purple"))|1171152|1171153|1171154|1171155|如您所见，lowess生成了与原始生成曲线近乎完美的拟合。Loess很接近，但在两端都出现了奇怪的偏差。|1171156|尽管您的数据集将非常不同，但我发现其他数据集的性能相似，loess和lowess都能够产生良好的结果。当您查看基准时，差异会变得更加显著：|1171157|>+microbenchmark::microbenchmark(loess(y~x,+dat,+span+=+0.6),approx(dat,n+=+20),lowess(dat,f+=+.6,iter=1),times=20)
Unit:+milliseconds
+++++++++++++++++++++++++++expr++++++++min+++++++++lq+++++++mean+++++median++++++++uq++++++++max+neval+cld
++loess(y+~+x,+dat,+span+=+0.6)+153.034810+154.450750+156.794257+156.004357+159.23183+163.117746++++20+++c
++++++++++++approx(dat,+n+=+20)+++1.297685+++1.346773+++1.689133+++1.441823+++1.86018+++4.281735++++20+a++
+lowess(dat,+f+=+0.6,+iter+=+1)+++9.637583++10.085613++11.270911++11.350722++12.33046++12.495343++++20++b+|1171158|Loess非常慢，耗时是approx的100倍。Lowess比approx产生更好的结果，同时运行速度仍然相当快(比loess快15倍)。|1171159|随着点数的增加，Loess也变得越来越陷入困境，在50,000左右变得不可用。|1171160|编辑:其他研究表明，loess对某些数据集具有更好的拟合能力。如果您处理的是小型数据集或性能不是考虑因素，请尝试这两个函数并比较结果。|1171161|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/13d0c4ae98867c6fb4adfc36cdf892bd.png|imageAlt|1|https://ask.qcloudimg.com/http-save/yehe-900000/3520502b41bc969430305dc1f59c9a9d.png^0|Y|6|15|6|0|0|0|0|0|0|1|0|0|0|0|0|0|6|7|6|E|D|0|0|5|6|6|D|6|Z|6|0|0|0|0|0|0|1|1|0|0|5|6|T|5|0|S|5|Y|6|0|0|0|5|C|6|O|6|V|6|0|8|5|0|A|5|0^^$0|@$1|2|3|4|5|6|7|1Y|8|@$9|1Z|A|20|B|C]|$9|21|A|22|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|23|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|24|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|25|8|@]|D|@]|E|$]]|$1|O|3|P|5|6|7|26|8|@]|D|@]|E|$]]|$1|Q|3|R|5|S|7|27|8|@]|D|@$9|28|A|29|1|2A]]|E|$]]|$1|T|3|P|5|6|7|2B|8|@]|D|@]|E|$]]|$1|U|3|V|5|6|7|2C|8|@]|D|@]|E|$]]|$1|W|3|X|5|6|7|2D|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|6|7|2E|8|@$9|2F|A|2G|B|C]|$9|2H|A|2I|B|C]|$9|2J|A|2K|B|C]]|D|@]|E|$]]|$1|10|3|11|5|6|7|2L|8|@$9|2M|A|2N|B|C]|$9|2O|A|2P|B|C]|$9|2Q|A|2R|B|C]|$9|2S|A|2T|B|C]]|D|@]|E|$]]|$1|12|3|13|5|J|7|2U|8|@]|D|@]|E|$K|L]]|$1|14|3|15|5|6|7|2V|8|@]|D|@]|E|$]]|$1|16|3|17|5|J|7|2W|8|@]|D|@]|E|$K|L]]|$1|18|3|P|5|6|7|2X|8|@]|D|@]|E|$]]|$1|19|3|R|5|S|7|2Y|8|@]|D|@$9|2Z|A|30|1|31]]|E|$]]|$1|1A|3|P|5|6|7|32|8|@]|D|@]|E|$]]|$1|1B|3|1C|5|6|7|33|8|@$9|34|A|35|B|C]|$9|36|A|37|B|C]]|D|@]|E|$]]|$1|1D|3|1E|5|6|7|38|8|@$9|39|A|3A|B|C]|$9|3B|A|3C|B|C]]|D|@]|E|$]]|$1|1F|3|1G|5|J|7|3D|8|@]|D|@]|E|$K|L]]|$1|1H|3|1I|5|6|7|3E|8|@$9|3F|A|3G|B|C]|$9|3H|A|3I|B|C]|$9|3J|A|3K|B|C]|$9|3L|A|3M|B|C]]|D|@]|E|$]]|$1|1J|3|1K|5|6|7|3N|8|@$9|3O|A|3P|B|C]]|D|@]|E|$]]|$1|1L|3|1M|5|6|7|3Q|8|@$9|3R|A|3S|B|C]]|D|@]|E|$]]|$1|1N|3|-4|5|6|7|3T|8|@]|D|@]|E|$]]]|1O|$1P|$5|1Q|1R|1S|E|$1T|1U|1V|-4]]|1W|$5|1Q|1R|1S|E|$1T|1X|1V|-4]]]]

The other answers are all good approaches. However, there are a few other options in R that haven't been mentioned, including <code>lowess</code> and <code>approx</code>, which may give better fits or faster performance.

The advantages are more easily demonstrated with an alternate dataset:

<pre><code>sigmoid &lt;- function(x)
{
 y&lt;-1/(1+exp(-.15*(x-100)))
 return(y)
}

dat&lt;-data.frame(x=rnorm(5000)*30+100)
dat$y&lt;-as.numeric(as.logical(round(sigmoid(dat$x)+rnorm(5000)*.3,0)))
</code></pre>

Here is the data overlaid with the sigmoid curve that generated it:

<a href="https://i.stack.imgur.com/N70sI.png" rel="noreferrer"><img src="https://i.stack.imgur.com/N70sI.png" alt="Data"></a>

This sort of data is common when looking at a binary behavior among a population. For example, this might be a plot of whether or not a customer purchased something (a binary 1/0 on the y-axis) versus the amount of time they spent on the site (x-axis). 

A large number of points are used to better demonstrate the performance differences of these functions.

<code>Smooth</code>, <code>spline</code>, and <code>smooth.spline</code> all produce gibberish on a dataset like this with any set of parameters I have tried, perhaps due to their tendency to map to every point, which does not work for noisy data.

The <code>loess</code>, <code>lowess</code>, and <code>approx</code> functions all produce usable results, although just barely for <code>approx</code>. This is the code for each using lightly optimized parameters:

<pre><code>loessFit &lt;- loess(y~x, dat, span = 0.6)
loessFit &lt;- data.frame(x=loessFit$x,y=loessFit$fitted)
loessFit &lt;- loessFit[order(loessFit$x),]

approxFit &lt;- approx(dat,n = 15)

lowessFit &lt;-data.frame(lowess(dat,f = .6,iter=1))
</code></pre>

And the results:

<pre><code>plot(dat,col='gray')
curve(sigmoid,0,200,add=TRUE,col='blue',)
lines(lowessFit,col='red')
lines(loessFit,col='green')
lines(approxFit,col='purple')
legend(150,.6,
 legend=c("Sigmoid","Loess","Lowess",'Approx'),
 lty=c(1,1),
 lwd=c(2.5,2.5),col=c("blue","green","red","purple"))
</code></pre>

<a href="https://i.stack.imgur.com/wDBiO.png" rel="noreferrer"><img src="https://i.stack.imgur.com/wDBiO.png" alt="Fits"></a>

As you can see, <code>lowess</code> produces a near perfect fit to the original generating curve. <code>Loess</code> is close, but experiences a strange deviation at both tails.

Although your dataset will be very different, I have found that other datasets perform similarly, with both <code>loess</code> and <code>lowess</code> capable of producing good results. The differences become more significant when you look at benchmarks:

<pre><code>&gt; microbenchmark::microbenchmark(loess(y~x, dat, span = 0.6),approx(dat,n = 20),lowess(dat,f = .6,iter=1),times=20)
Unit: milliseconds
 expr min lq mean median uq max neval cld
 loess(y ~ x, dat, span = 0.6) 153.034810 154.450750 156.794257 156.004357 159.23183 163.117746 20 c
 approx(dat, n = 20) 1.297685 1.346773 1.689133 1.441823 1.86018 4.281735 20 a 
 lowess(dat, f = 0.6, iter = 1) 9.637583 10.085613 11.270911 11.350722 12.33046 12.495343 20 b 
</code></pre>

<code>Loess</code> is extremely slow, taking 100x as long as <code>approx</code>. <code>Lowess</code> produces better results than <code>approx</code>, while still running fairly quickly (15x faster than loess).

<code>Loess</code> also becomes increasingly bogged down as the number of points increases, becoming unusable around 50,000. 

EDIT: Additional research shows that <code>loess</code> gives better fits for certain datasets. If you are dealing with a small dataset or performance is not a consideration, try both functions and compare the results.

blocks|key|1171268|text|在ggplot2中，您可以通过多种方式进行平滑处理，例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1171269|library(ggplot2)
ggplot(mtcars,+aes(wt,+mpg))+%2B+geom_point()+%2B
++geom_smooth(method+=+"gam",+formula+=+y+~+poly(x,+2))+
ggplot(mtcars,+aes(wt,+mpg))+%2B+geom_point()+%2B
++geom_smooth(method+=+"loess",+span+=+0.3,+se+=+FALSE)+|code-block|syntax|javascript|1171270|​|1171271|📷|atomic|offset|length|1171272|1171273|1171274|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/bfa79428438e6159a91e9f9891582d05.jpeg|imageAlt|1|https://ask.qcloudimg.com/http-save/yehe-900000/728239a24dec922d9e607aff01803f53.jpeg^0|0|0|0|0|1|0|0|0|1|1|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|K|7|13|8|@]|9|@$L|14|M|15|1|16]]|A|$]]|$1|N|3|J|5|K|7|17|8|@]|9|@$L|18|M|19|1|1A]]|A|$]]|$1|O|3|H|5|6|7|1B|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]]|Q|$R|$5|S|T|U|A|$V|W|X|-4]]|Y|$5|S|T|U|A|$V|Z|X|-4]]]]

In ggplot2 you can do smooths in a number of ways, for example:

<pre><code>library(ggplot2)
ggplot(mtcars, aes(wt, mpg)) + geom_point() +
 geom_smooth(method = "gam", formula = y ~ poly(x, 2)) 
ggplot(mtcars, aes(wt, mpg)) + geom_point() +
 geom_smooth(method = "loess", span = 0.3, se = FALSE) 
</code></pre>

<a href="https://i.stack.imgur.com/HlGSgm.png" rel="noreferrer"><img src="https://i.stack.imgur.com/HlGSgm.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/zT33Gm.png" rel="noreferrer"><img src="https://i.stack.imgur.com/zT33Gm.png" alt="enter image description here"></a>

blocks|key|1171368|text|我没有看到显示的这个方法，所以如果其他人希望这样做，我发现ggplot文档建议了一种使用gam方法的技术，该方法在处理小数据集时产生与loess相似的结果。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1171369|library(ggplot2)
x+<-+1:10
y+<-+c(2,4,6,8,7,8,14,16,18,20)

df+<-+data.frame(x,y)
r+<-+ggplot(df,+aes(x+=+x,+y+=+y))+%2B+geom_smooth(method+=+"gam",+formula+=+y+~+s(x,+bs+=+"cs"))%2Bgeom_point()
r|code-block|syntax|javascript|1171370|First+with+the+loess+method+and+auto+formula+Second+with+the+gam+method+with+suggested+formula|1171371|entityMap|0|LINK|mutability|MUTABLE|url|https://i.stack.imgur.com/ArzPW.png|1|https://i.stack.imgur.com/zx1dy.png^0|18|3|1V|5|0|0|0|18|0|19|1D|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@]|D|@$9|13|A|14|1|15]|$9|16|A|17|1|18]]|E|$]]|$1|M|3|-4|5|6|7|19|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]|U|$5|P|Q|R|E|$S|V]]]]

I didn't see this method shown, so if someone else is looking to do this I found that ggplot documentation suggested a technique for using the <code>gam</code> method that produced similar results to <code>loess</code> when working with small data sets. 

<pre><code>library(ggplot2)
x &lt;- 1:10
y &lt;- c(2,4,6,8,7,8,14,16,18,20)

df &lt;- data.frame(x,y)
r &lt;- ggplot(df, aes(x = x, y = y)) + geom_smooth(method = "gam", formula = y ~ s(x, bs = "cs"))+geom_point()
r
</code></pre>

<a href="https://i.stack.imgur.com/ArzPW.png" rel="nofollow noreferrer">First with the loess method and auto formula</a>
<a href="https://i.stack.imgur.com/zx1dy.png" rel="nofollow noreferrer">Second with the gam method with suggested formula</a>

I'm trying to draw a smooth curve in <code>R</code>. I have the following simple toy data:

<pre><code>&gt; x
 [1] 1 2 3 4 5 6 7 8 9 10
&gt; y
 [1] 2 4 6 8 7 12 14 16 18 20
</code></pre>

Now when I plot it with a standard command it looks bumpy and edgy, of course:

<pre><code>&gt; plot(x,y, type='l', lwd=2, col='red')
</code></pre>

How can I make the curve smooth so that the 3 edges are rounded using estimated values? I know there are many methods to fit a smooth curve but I'm not sure which one would be most appropriate for this type of curve and how you would write it in <code>R</code>.

How to fit a smooth curve to my data in R?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在尝试在R中绘制一条平滑的曲线。我有以下简单的玩具数据：> x [1]  1  2  3  4  5  6  7  8  9 10> y [1]  2  4  6  8  7 12 14 16 18 20现在，当我使用一个标准命令绘制它时，它看起来当然是凹凸不平的：> plot(x,y, type='l', lwd=2, col='red')如何使曲线平滑，以便使用估计值对3条边进行舍入？我知

问如何将平滑曲线拟合到R中的数据？
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