blocks|key|1227256|text|我是这样做的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1227257|foo+<-+rnorm(100,+mean=1,+sd=2)
hist(foo,+prob=TRUE)
curve(dnorm(x,+mean=mean(foo),+sd=sd(foo)),+add=TRUE)|code-block|syntax|javascript|1227258|一个额外的练习是使用ggplot2包来做这件事。|1227259|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|$]]

Here's the way I do it:

<pre><code>foo &lt;- rnorm(100, mean=1, sd=2)
hist(foo, prob=TRUE)
curve(dnorm(x, mean=mean(foo), sd=sd(foo)), add=TRUE)
</code></pre>

A bonus exercise is to do this with ggplot2 package ...

blocks|key|1227318|text|使用ggplot2很容易做到这一点|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1227319|library(ggplot2)
dataset+<-+data.frame(X+=+c(rep(65,+times=5),+rep(25,+times=5),+
++++++++++++++++++++++++++++rep(35,+times=10),+rep(45,+times=4)))
ggplot(dataset,+aes(x+=+X))+%2B+
++geom_histogram(aes(y+=+..density..))+%2B+
++geom_density()|code-block|syntax|javascript|1227320|或者模仿德克解决方案的结果|1227321|ggplot(dataset,+aes(x+=+X))+%2B+
++geom_histogram(aes(y+=+..density..),+binwidth+=+5)+%2B+
++geom_density()|1227322|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|$]]

Such thing is easy with ggplot2

<pre><code>library(ggplot2)
dataset &lt;- data.frame(X = c(rep(65, times=5), rep(25, times=5), 
 rep(35, times=10), rep(45, times=4)))
ggplot(dataset, aes(x = X)) + 
 geom_histogram(aes(y = ..density..)) + 
 geom_density()
</code></pre>

or to mimic the result from Dirk's solution

<pre><code>ggplot(dataset, aes(x = X)) + 
 geom_histogram(aes(y = ..density..), binwidth = 5) + 
 geom_density()
</code></pre>

blocks|key|1227447|text|Dirk解释了如何在直方图上绘制密度函数。但有时，您可能希望采用更强的假设，即偏态正态分布，并绘制该分布，而不是密度。您可以估计分布的参数，并使用sn+package将其绘制出来|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1227448|>+sn.mle(y=c(rep(65,+times=5),+rep(25,+times=5),+rep(35,+times=10),+rep(45,+times=4)))
$call
sn.mle(y+=+c(rep(65,+times+=+5),+rep(25,+times+=+5),+rep(35,+
++++times+=+10),+rep(45,+times+=+4)))

$cp
++++mean+++++s.d.+skewness+
41.46228+12.47892++0.99527+|code-block|syntax|javascript|1227449|​|1227450|📷|atomic|1227451|1227452|这可能在偏斜-正常的数据上效果更好：|1227453|1227454|1227455|1227456|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/1497697/103225|1|http://cran.r-project.org/web/packages/sn/index.html|2|IMAGE|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/dcb462ef0c7fcef4683258e2b851ba4b.png|imageAlt|3|https://ask.qcloudimg.com/http-save/yehe-900000/e2bdf10f499d4f4fed087c6e0c4517f1.png^0|0|4|0|21|A|1|0|0|0|0|1|2|0|0|0|0|0|1|3|0|0^^$0|@$1|2|3|4|5|6|7|1B|8|@]|9|@$A|1C|B|1D|1|1E]|$A|1F|B|1G|1|1H]]|C|$]]|$1|D|3|E|5|F|7|1I|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|1J|8|@]|9|@]|C|$]]|$1|K|3|L|5|M|7|1K|8|@]|9|@$A|1L|B|1M|1|1N]]|C|$]]|$1|N|3|J|5|6|7|1O|8|@]|9|@]|C|$]]|$1|O|3|P|5|6|7|1P|8|@]|9|@]|C|$]]|$1|Q|3|J|5|6|7|1Q|8|@]|9|@]|C|$]]|$1|R|3|L|5|M|7|1R|8|@]|9|@$A|1S|B|1T|1|1U]]|C|$]]|$1|S|3|J|5|6|7|1V|8|@]|9|@]|C|$]]|$1|T|3|-4|5|6|7|1W|8|@]|9|@]|C|$]]]|U|$V|$5|W|X|Y|C|$Z|10]]|11|$5|W|X|Y|C|$Z|12]]|13|$5|14|X|15|C|$16|17|18|-4]]|19|$5|14|X|15|C|$16|1A|18|-4]]]]

<a href="https://stackoverflow.com/a/1497697/103225">Dirk</a> has explained how to plot the density function over the histogram. But sometimes you might want to go with the stronger assumption of a skewed normal distribution and plot that instead of density. You can estimate the parameters of the distribution and plot it using the <a href="http://cran.r-project.org/web/packages/sn/index.html" rel="nofollow noreferrer">sn package</a>:

<pre><code>&gt; sn.mle(y=c(rep(65, times=5), rep(25, times=5), rep(35, times=10), rep(45, times=4)))
$call
sn.mle(y = c(rep(65, times = 5), rep(25, times = 5), rep(35, 
 times = 10), rep(45, times = 4)))

$cp
 mean s.d. skewness 
41.46228 12.47892 0.99527 
</code></pre>

<img src="https://i.stack.imgur.com/xQJYD.png" alt="Skew-normal distributed data plot">

This probably works better on data that is more skew-normal:

<img src="https://i.stack.imgur.com/HgX0f.png" alt="Another skew-normal plot">

blocks|key|1227564|text|我也有同样的问题，但德克的解决方案似乎不起作用。我每次都会收到这样的警告信息|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1227565|"prob"+is+not+a+graphical+parameter|code-block|syntax|javascript|1227566|我通读了?hist，找到了关于freq:+a+logical+vector+set+TRUE+by+default.的信息|offset|length|style|CODE|1227567|对我有效的代码是|1227568|hist(x,freq=FALSE)
lines(density(x),na.rm=TRUE)|1227569|entityMap^0|0|0|4|5|F|17|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]|$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|R|$]]

I had the same problem but Dirk's solution didn't seem to work. 
I was getting this warning messege every time

<pre><code>"prob" is not a graphical parameter
</code></pre>

I read through <code>?hist</code> and found about <code>freq: a logical vector set TRUE by default.</code>

the code that worked for me is

<pre><code>hist(x,freq=FALSE)
lines(density(x),na.rm=TRUE)
</code></pre>

blocks|key|1442581|text|这是内核密度估计，请点击this+link查看概念及其参数的一个很好的插图。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1442582|曲线的形状主要取决于两个元素:+1)核(通常是Epanechnikov+or+Gaussian)，它通过输入和加权所有数据来估计x坐标中的每个值在y坐标中的点；它是对称的，通常是集成到一个中的正函数；2)带宽越大，曲线越平滑，曲线越小，摆动越大。|1442583|对于不同的需求，需要使用不同的包，您可以参考这个文档：Density+estimation+in+R。对于多变量，你可以求助于multivariate+kernel+density+estimation。|1442584|entityMap|0|LINK|mutability|MUTABLE|url|https://mathisonian.github.io/kde/|1|https://stats.stackexchange.com/q/215835/103153|2|https://vita.had.co.nz/papers/density-estimation.pdf|3|https://bookdown.org/egarpor/NP-UC3M/kde-ii-mult.html^0|C|9|0|0|N|O|1|0|R|N|2|1R|12|3|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@$A|W|B|X|1|Y]]|C|$]]|$1|D|3|E|5|6|7|Z|8|@]|9|@$A|10|B|11|1|12]]|C|$]]|$1|F|3|G|5|6|7|13|8|@]|9|@$A|14|B|15|1|16]|$A|17|B|18|1|19]]|C|$]]|$1|H|3|-4|5|6|7|1A|8|@]|9|@]|C|$]]]|I|$J|$5|K|L|M|C|$N|O]]|P|$5|K|L|M|C|$N|Q]]|R|$5|K|L|M|C|$N|S]]|T|$5|K|L|M|C|$N|U]]]]

It's the kernel density estimation, and please hit <a href="https://mathisonian.github.io/kde/" rel="nofollow noreferrer">this link</a> to check a great illustration for the concept and its parameters.
The shape of the curve depends mostly on two elements: 1) the kernel(usually <a href="https://stats.stackexchange.com/q/215835/103153">Epanechnikov or Gaussian</a>) that estimates a point in the y coordinate for every value in the x coordinate by inputting and weighing all data; and it is symmetric and usually a positive function that integrates into one; 2) the bandwidth, the larger the smoother the curve, and the smaller the more wiggled the curve.
For different requirements, different packages should be applied, and you can refer to this document: <a href="https://vita.had.co.nz/papers/density-estimation.pdf" rel="nofollow noreferrer">Density estimation in R</a>. And for multivariate variables, you can turn to the <a href="https://bookdown.org/egarpor/NP-UC3M/kde-ii-mult.html" rel="nofollow noreferrer">multivariate kernel density estimation</a>.

Is there a function in R that fits a curve to a histogram?

Let's say you had the following histogram

<pre><code>hist(c(rep(65, times=5), rep(25, times=5), rep(35, times=10), rep(45, times=4)))
</code></pre>

It looks normal, but it's skewed. I want to fit a normal curve that is skewed to wrap around this histogram.

This question is rather basic, but I can't seem to find the answer for R on the internet.

Fitting a density curve to a histogram 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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R中是否有一个函数可以将曲线拟合为直方图？假设你有下面的直方图hist(c(rep(65, times=5), rep(25, times=5), rep(35, times=10), rep(45, times=4)))它看起来很正常，但它是倾斜的。我想要拟合一条正常的曲线，它是倾斜的，围绕着这个直方图。这个问题很基...

问将密度曲线拟合到R中的直方图
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