如何将曲线拟合到直方图

本文介绍了如何将曲线拟合到直方图的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 29岁程序员，3月因学历无情被辞！ 我探讨了与此主题相关的类似问题，但在直方图上绘制漂亮的曲线时遇到了一些麻烦。我知道有些人可能会认为这是重复的，但是我目前没有找到任何东西可以解决我的问题。 尽管此处的数据不可见，但此处是我正在使用的一些变量，以便您可以在下面的代码中看到它们的含义。 差异< -subset（Score_Differences ，select =差，drop = T）m =平均值（差异） std = sqrt（var（差异）） 这是我制作的第一条曲线（代码似乎最常见且最容易产生，但是曲线本身不太适合）。 hist（差异（密度，15 =断点= 15，概率= TRUE，xlab =得分差异，ylim = c（0，.1），main =得分差异的正态曲线） curve（dnorm（x，m，std），col = Red，lwd = 2，add = TRUE） 我真的很喜欢，但是不喜欢曲线进入负区域。 hist（差异，概率= TRUE）线（密度（差异），col =红色，lwd = 2）线（密度（差异，调整= 2），lwd = 2，col =蓝色） 这与第一个直方图相同，但具有频率。 h = hist（差异，密度= 15，中断= 15，xlab =得分差异，主要=分数差异的正态曲线） xfit = seq（min（差异），max（差异）） yfit = dnorm（xfit，m，std） yfit = yfit * diff（h $ mids [1：2]）* length（Differences）行（xfit，yfit，col = Red，lwd = 2） 另一种尝试，但没有运气。可能是因为我使用的是 qnorm ，但数据显然不正常。曲线再次变为负方向。 sample_x = seq（qnorm（.001，m，std），qnorm（。 999，m，std），length.out = l） binwidth = 3 breaks = seq（floor（min（Differences）），ceiling（max（Differences）），binwidth） hist（差异，中断）行（sample_x，l * dnorm（sample_x，m，std）* binwidth，col = Red） 唯一在视觉上看起来不错的曲线是第二，但是曲线落在负方向上。 我的问题是是否有标准方法在曲线上放置直方图？ 这个数据当然是不正常的。我在这里介绍的程序中的3个来自类似的帖子，但是显然我遇到了一些麻烦。我觉得所有拟合曲线的方法都取决于您使用的数据。 更新解决方案 感谢李哲远和其他人！我将把它留给我自己，也希望其他人也能参考。 hist（差异，概率= TRUE）行（密度（差异，剪切= 0），col =红色，lwd = 2）行（密度（差异，调整= 2，剪切= 0），lwd = 2，col =蓝色 ） 解决方案好的，因此您正在为密度超出自然范围这一事实而苦苦挣扎。好吧，只需将 cut设置为0 。您可能想阅读 注意，通过 cut = 0 ，密度估计严格在 range（x）范围内进行。超出此范围，密度为0。 I've explored similar questions asked about this topic but I am having some trouble producing a nice curve on my histogram. I understand that some people may see this as a duplicate but I haven't found anything currently to help solve my problem.Although the data isn't visible here, here is some variables I am using just so you can see what they represent in the code below.Differences <- subset(Score_Differences, select = Difference, drop = T)m = mean(Differences)std = sqrt(var(Differences))Here is the very first curve I produce (the code seems most common and easy to produce but the curve itself doesn't fit that well).hist(Differences, density = 15, breaks = 15, probability = TRUE, xlab = "Score Differences", ylim = c(0,.1), main = "Normal Curve for Score Differences")curve(dnorm(x,m,std),col = "Red", lwd = 2, add = TRUE)I really like this but don't like the curve going into the negative region.hist(Differences, probability = TRUE)lines(density(Differences), col = "Red", lwd = 2)lines(density(Differences, adjust = 2), lwd = 2, col = "Blue")This is the same histogram as the first, but with frequencies. Still doesn't look that nice.h = hist(Differences, density = 15, breaks = 15, xlab = "Score Differences", main = "Normal Curve for Score Differences")xfit = seq(min(Differences),max(Differences))yfit = dnorm(xfit,m,std)yfit = yfit*diff(h$mids[1:2])*length(Differences)lines(xfit, yfit, col = "Red", lwd = 2)Another attempt but no luck. Maybe because I am using qnorm, when the data obviously isn't normal. The curve goes into the negative direction again.sample_x = seq(qnorm(.001, m, std), qnorm(.999, m, std), length.out = l)binwidth = 3breaks = seq(floor(min(Differences)), ceiling(max(Differences)), binwidth)hist(Differences, breaks)lines(sample_x, l*dnorm(sample_x, m, std)*binwidth, col = "Red")The only curve that visually looks nice is the 2nd, but the curve falls into the negative direction.My question is "Is there a "standard way" to place a curve on a histogram?" This data certainly isn't normal. 3 of the procedures I presented here are from similar posts but I am having some troubles obviously. I feel like all methods of fitting a curve will depend on the data you're working with.Update with solutionThanks to Zheyuan Li and others! I will leave this up for my own reference and hopefully others as well.hist(Differences, probability = TRUE)lines(density(Differences, cut = 0), col = "Red", lwd = 2)lines(density(Differences, adjust = 2, cut = 0), lwd = 2, col = "Blue") 解决方案 OK, so you are just struggling with the fact that density goes beyond "natural range". Well, just set cut = 0. You possibly want to read plot.density extends "xlim" beyond the range of my data. Why and how to fix it? for why. In that answer, I was using from and to. But now I am using cut.## consider a mixture, that does not follow any parametric distribution family## note, by construction, this is a strictly positive random variableset.seed(0)x <- rbeta(1000, 3, 5) + rexp(1000, 0.5)## (kernel) density estimation offers a flexible nonparametric approachd <- density(x, cut = 0)## you can plot histogram and density on the density scalehist(x, prob = TRUE, breaks = 50)lines(d, col = 2)Note, by cut = 0, density estimation is done strictly within range(x). Outside this range, density is 0. 这篇关于如何将曲线拟合到直方图的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！