blocks|key|4162332|text|据我所知，没有内置的函数可以将分布拟合到频率表中。请注意，从理论上讲，连续分布不适合表，因为数据是离散的。当然，对于足够大的N和足够精细的网格，这可以忽略不计。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4162333|如果您知道自己感兴趣的密度，则可以使用optim或任何其他优化器构建自己的模型拟合函数。我为伽马分布做了这个here+(对于那个特定的数据集来说，这是一个很糟糕的假设，但没关系)。|offset|length|style|CODE|4162334|下面重现的代码。|4162335|negll+<-+function(par,+x,+y)
{
++++shape+<-+par[1]
++++rate+<-+par[2]
++++mu+<-+dgamma(x,+shape,+rate)+*+sum(y)
++++-2+*+sum(dpois(y,+mu,+log=TRUE))
}


optim(c(1,+1),+negll,+x=seq_along(g$count),+y=g$count,+method="L-BFGS-B",+lower=c(.001,+.001))
$par
[1]+0.73034879+0.00698288

$value
[1]+62983.18

$counts
function+gradient+
++++++32+++++++32+

$convergence
[1]+0

$message
[1]+"CONVERGENCE:+REL_REDUCTION_OF_F+<=+FACTR*EPSMCH"|code-block|syntax|javascript|4162336|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/17225795/using-r-to-fit-data-from-a-csv-with-a-gamma-function/17227816#17227816^0|0|J|5|1I|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@$D|Y|E|Z|F|G]]|9|@$D|10|E|11|1|12]]|A|$]]|$1|H|3|I|5|6|7|13|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|14|8|@]|9|@]|A|$M|N]]|$1|O|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

There's no built-in function that I know of for fitting a distribution to a frequency table. Note that, in theory, a continuous distribution is inappropriate for a table, since the data is discrete. Of course, for large enough N and a fine enough grid, this can be ignored.

You can build your own model-fitting function using <code>optim</code> or any other optimizer, if you know the density that you're interested in. I did this <a href="https://stackoverflow.com/questions/17225795/using-r-to-fit-data-from-a-csv-with-a-gamma-function/17227816#17227816">here</a> for a gamma distribution (which was a bad assumption for that particular dataset, but never mind that).

Code reproduced below.

<pre><code>negll &lt;- function(par, x, y)
{
 shape &lt;- par[1]
 rate &lt;- par[2]
 mu &lt;- dgamma(x, shape, rate) * sum(y)
 -2 * sum(dpois(y, mu, log=TRUE))
}


optim(c(1, 1), negll, x=seq_along(g$count), y=g$count, method="L-BFGS-B", lower=c(.001, .001))
$par
[1] 0.73034879 0.00698288

$value
[1] 62983.18

$counts
function gradient 
 32 32 

$convergence
[1] 0

$message
[1] "CONVERGENCE: REL_REDUCTION_OF_F &lt;= FACTR*EPSMCH"
</code></pre>

blocks|key|2037220|text|对于泊松分布的拟合，你只需要样本的平均值。然后，平均值等于λ，这是泊松分布的唯一参数。示例：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2037221|set.seed(1111)
sample<-rpois(n=10000,l=10)
mean(sample)
[1]+10.0191|code-block|syntax|javascript|2037222|它几乎等于用于创建样本的lambda值(l=10)。较小的差异(0.0191)是由于泊松分布随机值生成器的随机性。随着n的增加，差异将变得越来越小。或者，您可以使用优化方法来拟合分布：|offset|length|style|CODE|2037223|library(fitdistrplus)
fitdist(sample,"pois")
set.seed(1111)

Fitting+of+the+distribution+'+pois+'+by+maximum+likelihood+
Parameters:
+++++++estimate+Std.+Error
lambda++10.0191+0.03165296|2037224|但这只是浪费时间。有关拟合频率数据的理论信息，请参阅我的答案here。|2037225|entityMap|0|LINK|mutability|MUTABLE|url|https://stats.stackexchange.com/questions/62337/negative-binomial-and-poisson/262322#262322^0|0|0|1N|1|0|0|U|4|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@$I|11|J|12|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|14|8|@]|9|@$I|15|J|16|1|17]]|A|$]]|$1|Q|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

For fitting a Poisson distribution, you only need the mean of your sample. Then the mean equals the lambda, which is the only parameter of the Poisson distribution. Example:

<pre><code>set.seed(1111)
sample&lt;-rpois(n=10000,l=10)
mean(sample)
[1] 10.0191
</code></pre>

which is almost equal to the lambda value put for creating the sample (l=10). The small difference (0.0191) is due to the randomness of the Poisson distribution random value generator. As you increase <code>n</code> the difference will get smaller.
Alternatively, you can fit the distribution using an optimization method:

<pre><code>library(fitdistrplus)
fitdist(sample,"pois")
set.seed(1111)

Fitting of the distribution ' pois ' by maximum likelihood 
Parameters:
 estimate Std. Error
lambda 10.0191 0.03165296
</code></pre>

but it's only a waste of time.
For theoritical information on fitting frequency data, you can see my answer <a href="https://stats.stackexchange.com/questions/62337/negative-binomial-and-poisson/262322#262322">here</a>.

blocks|key|702645|text|ForestFit包中的函数fixtmixturegrouped使用每个组的频率数据为其他分布模型执行此工作。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|702646|它可以拟合基于“伽玛”、“对数正态”、“倾斜正态”和“威布尔”的简单或混合分布模型。|702647|对于泊松分布，总体平均值是唯一需要的参数。对数据应用一个简单的汇总函数就足够了(正如ntzortzis所建议的那样)|702648|entityMap^0|0|9|E|I|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Q|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|R|8|@]|D|@]|E|$]]|$1|J|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|K|$]]

The function <code>fixtmixturegrouped</code> from the package <code>ForestFit</code> does the job for other distribution models using frequency-by-group data.

It can fit simple or mixture distribution models based on "gamma", "log-normal", "skew-normal", and "weibull".

For a Poisson distribution, the population mean is the only parameter that is needed. Applying a simple summary function on your data would suffice (as suggested by ntzortzis)

Is there a function that could be used to fit a frequency distribution in R? I'm aware of <code>fitdistr</code> but as far as I can tell it only works for data vectors (random samples). Also, I know that converting between the two formats is trivial but frequencies are so large that memory is a concern. 

For example, <code>fitdistr</code> may be used the following way:

<pre><code>x&lt;-rpois(100, lambda=10)
fitdistr(x,"poisson")
</code></pre>

Is there a function that would do the same fitting on a frequency table? Something along the lines:

<pre><code>freqt &lt;- as.data.frame(table(x))
fitfreqtable(freqt$x, weights=freqt$Freq, "poisson")
</code></pre>

Thanks!

How to fit frequency distributions 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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有没有一个函数可以用来拟合R中的频率分布？我知道fitdistr，但据我所知，它只适用于数据向量(随机样本)。此外，我知道两种格式之间的转换是微不足道的，但频率太大了，内存是一个问题。例如，fitdistr可以按以下方式使用：x<-rpois(100, lambda=10)fitdistr(x,"poisson")有没...

问如何在R中拟合频率分布？
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