首页 \ 问答 \ Html 5 Drag n Drop Ajax(Html 5 Drag n Drop Ajax)

Html 5 Drag n Drop Ajax(Html 5 Drag n Drop Ajax)

 我试图处理Ajax Dom Elements Loaded上的dragstart事件，但不起作用。 以下是我如何处理Dragstart事件以获取要移动的元素的ID：  
  $(document).on(".dragItem", "dragstart", function (ev) {

        ev.originalEvent.dataTransfer.setData("text", ev.target.id);
    });
 
 对于页面上加载的任何元素，Dragstart由上面的代码处理。  
 但是，当我稍后在代码中使用JS创建Html div时，事件不会响应。  
<div class="dragItem" draggable='true'>My Drag Item </div>
 
 实际上我通过jquery Ajax接收JSON数据，然后使用HTML Created更新DOM。 

I m trying to handle dragstart event on Ajax Dom Elements Loaded but doesn't work. Here is how I handle the Dragstart Event to get the ID of the Element being Moved: 
  $(document).on(".dragItem", "dragstart", function (ev) {

        ev.originalEvent.dataTransfer.setData("text", ev.target.id);
    });
 
For any element loaded on the Page the Dragstart is handled by the code above. 
However, when I create an Html div using JS later in the Code, the Event doesn't respond. 
<div class="dragItem" draggable='true'>My Drag Item </div>
 
In fact I'm receiving JSON Data through jquery Ajax, then update The DOM with the HTML Created.

原文：https://stackoverflow.com/questions/34299381

更新时间：2021-06-22 22:06

最满意答案

 为了可视化概率回归模型的拟合优度，“标准”残差（例如，泊松或偏差）通常不那么具有信息性，因为它们主要捕获均值的建模而不是整个分布的建模。 有时使用的一种替代方案是（随机化的）分位数残差。 没有随机化，它们被定义为qnorm(pdist(y)) ，其中pdist()是拟合分布函数（这里是ZIP模型）， y是观察值， qnorm()是标准正态分布的分位数函数。 如果模型拟合，残差的分布应该是标准的，并且可以在QQ图中检查。 在离散分布的情况下（如此处），需要随机化来打破数据的离散性质。 有关更多详细信息，请参见Dunn＆Smyth（1996， Journal of Computational and Graphical Statistics ，5,236-244）。 在R中你可以使用R-Forge的countreg包（希望很快也会在CRAN上）。  
 另一种检查数据边际分布的替代方案是所谓的根图。 它在视觉上比较了计数0,1，......的观察和拟合频率。通常比显示随机分位数残差的QQ图更好地显示过量零和/或过度离散的问题。 有关详细信息，请参阅我们的论文Kleiber＆Zeileis（2016， The American Statistician ， 70 （3），296-303， doi：10.1080 / 00031305.2016.1173590 ）。  
 将这些应用于回归模型可以快速显示零膨胀泊松不能解释响应中的过度离散。 （计数高达100以上，基于泊松的分布几乎从不适合。）此外，零膨胀模型并不适合，因为对于assnage = 1和= 2，零很少，并且不需要零膨胀。 这导致零膨胀部分中的相应系数朝向-Inf具有非常大的标准误差（如在二元回归中的准分离中）。 因此，由两部分组成的障碍模型更适合，并且可能更容易解释。 最后，由于两个assnage组不同，我会将assnage编码为一个因素（我不清楚你是否已经这样做了）。  
 因此，为了分析您的数据，我使用您帖子中提供的yc并确保：  
yc$assnage <- factor(yc$assnage)
 
 为了第一次探索assnage的影响我绘制count是否为正（左：零障碍）和对count刻度上的正count （右：计数）。  
plot(factor(count > 0, levels = c(FALSE, TRUE), labels = c("=0", ">0")) ~ assnage,
  data = yc, ylab = "count", main = "Zero hurdle")
plot(count ~ assnage, data = yc, subset = count > 0,
  log = "y", main = "Count (positive)")
 
   
 然后，我使用R-Forge的countreg包装适合ZIP，ZINB和障碍NB模型。 这还包含zeroinfl()和hurdle()函数的更新版本。  
install.packages("countreg", repos = "http://R-Forge.R-project.org")
library("countreg")
zip <- zeroinfl(count ~ assnage * spawncob, offset = logoffset,
  data = yc, dist = "poisson")
zinb <- zeroinfl(count ~ assnage * spawncob, offset = logoffset,
  data = yc, dist = "negbin")
hnb <- hurdle(count ~ assnage * spawncob, offset = logoffset, data = yc,
  dist = "negbin")
 
 ZIP明显不合适，障碍NB略胜ZINB。  
BIC(zip, zinb, hnb)
##      df      BIC
## zip  20 7700.085
## zinb 21 3574.720
## hnb  21 3556.693
 
 如果检查summary(zinb)您还会看到零膨胀部分中的某些系数大约为10（对于虚拟变量），标准误差summary(zinb)到两个数量级。 这实质上意味着相应组中的零膨胀概率变为零，因为负二项分布已经具有足够的零响应概率权重（分组1和2）。  
 为了在HNB正确捕获响应时可视化ZIP模型不适合，我们现在可以使用根图。  
rootogram(zip, main = "ZIP", ylim = c(-5, 15), max = 50)
rootogram(hnb, main = "HNB", ylim = c(-5, 15), max = 50)
 
   
 ZIP的波浪状图案清楚地显示了模型中未适当捕获的数据中的过度离散。 相比之下，障碍相当合适。  
 作为最后的检查，我们还可以查看障碍模型中分位数残差的QQ图。 这些看起来相当正常，并且没有显示模型的可疑偏离。  
qqrplot(hnb, main = "HNB")
 
   
 由于残差是随机的，您可以重新运行代码几次以获得变化的印象。 qqrplot()也有一些参数让你在一个图中探索这种变化。 

For visualizing the goodness of fit of probabilistic regression models, "standard" residuals (e.g., Poisson or deviance) are often not so informative because they mostly capture the modeling of the mean but not of the entire distribution. One alternative that is somtimes used are (randomized) quantile residuals. Without randomization they are defined as qnorm(pdist(y)) where pdist() is the fitted distribution function (here a ZIP model), y are the observations, and qnorm() is the quantile function of the standard normal distribution. If the model fits, the distribution of the residuals should be standard normal and can be checked in a Q-Q plot. In case of discrete distributions (as here), randomization is needed to break up the discrete nature of the data. See Dunn & Smyth (1996, Journal of Computational and Graphical Statistics, 5, 236-244) for more details. In R you can use the countreg package from R-Forge (hopefully soon also on CRAN) for these. 
Another alternative which checks the marginal distribution of the data is the so-called rootogram. It visually compares the observed and fitted frequencies for the counts, 0, 1, ... It is often better at displaying the problems of excess zeros and/or overdispersion than Q-Q plots of randomized quantile residuals. See our paper Kleiber & Zeileis (2016, The American Statistician, 70(3), 296–303, doi:10.1080/00031305.2016.1173590) for more details. 
Applying these to your regression model quickly shows that a zero-inflated Poisson is not accounting for the overdispersion in the response. (With counts up to and beyond 100 a Poisson-based distribution almost never fits well.) Furthermore, a zero inflation model does not fit very well because for assnage = 1 and = 2 there are very few zeros and no zero inflation is needed. This leads to corresponding coefficients in the zero inflation part that go towards -Inf with very large standard errors (like in quasi-separation in binary regression). Therefore, a two-part hurdle model fits somewhat better and is likely easier to interpret. Finally, due to the two assnage groups being different I would code assnage as a factor (it is not clear to me whether you did that already). 
Thus, for analyzing your data I use yc as provided in your post and ensure: 
yc$assnage <- factor(yc$assnage)
 
For a first explorative look at the influence of assnage I plot whether or not the count is positive (left: zero hurdle) and the positive count on a log-scale (right: count). 
plot(factor(count > 0, levels = c(FALSE, TRUE), labels = c("=0", ">0")) ~ assnage,
  data = yc, ylab = "count", main = "Zero hurdle")
plot(count ~ assnage, data = yc, subset = count > 0,
  log = "y", main = "Count (positive)")
 
 
Then, I fit the ZIP, ZINB, and hurdle NB models using the countreg package from R-Forge. This also contains updated versions of the zeroinfl() and hurdle() functions. 
install.packages("countreg", repos = "http://R-Forge.R-project.org")
library("countreg")
zip <- zeroinfl(count ~ assnage * spawncob, offset = logoffset,
  data = yc, dist = "poisson")
zinb <- zeroinfl(count ~ assnage * spawncob, offset = logoffset,
  data = yc, dist = "negbin")
hnb <- hurdle(count ~ assnage * spawncob, offset = logoffset, data = yc,
  dist = "negbin")
 
The ZIP is clearly inappropriate and the hurdle NB is slightly better than the ZINB. 
BIC(zip, zinb, hnb)
##      df      BIC
## zip  20 7700.085
## zinb 21 3574.720
## hnb  21 3556.693
 
If you check summary(zinb) you will also see that some coefficients in the zero-inflation part are around 10 (for a dummy variable) with standard errors one or two orders of magnitude larger. This essentially means that the zero-inflation probability in the corresponding groups goes to zero because the negative binomial distribution already has more than enough probability weight for the zero responses (assnage groups 1 and 2). 
To visualize that the ZIP model does not fit while the HNB appropriately captures the response, we can now use the rootograms. 
rootogram(zip, main = "ZIP", ylim = c(-5, 15), max = 50)
rootogram(hnb, main = "HNB", ylim = c(-5, 15), max = 50)
 
 
The wave-like pattern for the ZIP clearly shows the overdispersion in the data that is not appropriately captured by the model. In contrast, the hurdle fits reasonably well. 
As a final check we can also look at the Q-Q plot of the quantile residuals in the hurdle model. These look fairly normal and show no suspicious departures from the model. 
qqrplot(hnb, main = "HNB")
 
 
As the residuals are randomized, you can re-run the code a couple of times to get an impression of the variation. qqrplot() has also some arguments that let's you explore this variation in a single plot.

Html 5 Drag n Drop Ajax(Html 5 Drag n Drop Ajax)

最满意答案

相关问答

R中的零膨胀准泊松模型（glmmadmb，pscl）(Zero-Inflated Quasi-Poisson Models in R (glmmadmb, pscl))[2024-01-01]

简单的自回归AR（1）残差，在R中(simple autoregression AR(1) residuals, in R)[2023-01-11]

R：列表中线性模型的残差(R: Residuals from Linear Models in List)[2023-03-11]

无法在R中获得AOV的残差(Unable to get Residuals for AOV in R)[2020-01-11]

R对数变换的残差绝对值(R absolute value of residuals with log transformation)[2022-06-01]

在python中R data.chisq $ residuals的等价物是什么？(What is the equivalent of R data.chisq$residuals in python?)[2024-02-04]

R包pscl中modelobject $残差和残差（modelobject）之间有什么区别？(Any difference between modelobject$residuals and residuals(modelobject) in R package pscl?)[2021-08-25]

检查残差和可视化零膨胀的裂缝r(examining residuals and visualizing zero-inflated poission r)[2023-10-29]

在R错误中绘制残差(plotting residuals in R error)[2022-02-05]

如何使用零膨胀泊松回归来修复错误(How to fix an error with zero-inflated Poisson regression)[2021-12-02]

相关文章

最新问答