I am building charts that have two lines in the axis text. The first line contains the group name, the second line contains that group population. I build my axis labels as a single character string with the format "LINE1 n LINE2". Is it possible to assign different font faces and sizes to LINE1 and LINE2, even though they are contained within a single character string? I would like
我正在构建轴文本中有两行的图表。 第一行包含组名，第二行包含该组人口。 我使用格式“LINE1 n LINE2”将我的轴标签构建为单个字符串。 是否可以为LINE1和LINE2分配不同的字体大小和大小，即使它们包含在单个字符串中？ 我想让LINE1变大变粗，而LINE2变小变大。 以下是一些示例代码： Treatment <- rep(c('T','C'),each=2) Gender <- rep(c('Male','Female'),2) Response <- sample(1:100,4) test_df <- data.
I have a shapefile (which can be downloaded here http://www.car.gov.br/publico/municipios/downloads?sigla=MA (any would do)) where each polygon represent a rural private property, as the owner submited it. However, the polygons overlap. For the analysis I need to run, an area cannot be accounted twice (ie. be in two properties ). So, on the areas that do overlap, I wanted to assign this duplic
我有一个shapefile（可以在这里下载http://www.car.gov.br/publico/municipios/downloads?sigla=MA（任何会这样做）），其中每个多边形代表一个农村私有财产，因为所有者提交它。 但是，多边形重叠。 对于我需要运行的分析，一个区域不能被计入两次（即在两个属性中）。 因此，在重叠的区域，我想将这个重复区域分配给面积最小的多边形，并从另一个区域中擦除。 我看过其他问题，比如这个 。 但没有人真的这样做。 你可
I would like a chart that as a function of the day in the year, advances from 0 -> 100% in both x and y axes (where each axis is a separate metric). Depending on where the data is relative to the day in the year, I'd like to show whether that's good or bad. Quite simply, I can show it like this: So the above plot shows we're in a good sitaution because the "tip" (dark
我想要一张图表，作为一年中某一天的函数，在x和y轴（每个轴都是一个单独的度量标准）中，从0 - > 100％前进。 根据数据相对于一年中的某一天的情况，我想说明这是好还是不好。 很简单，我可以像这样展示它： 所以上面的图表显示我们处于良好的状态，因为“小费”（最黑暗的最大分数）超过了50％的标准（并且假设我们在一年中是50％）。 但我想在水平线和垂直线周围添加渐变线以显示更多细微差别。 下面是对这些区域的解
In the tidy model of data science (TM) implemented in modelr , resampled data are organized using list-columns: library(modelr) library(tidyverse) # create the k-folds df_heights_resampled = heights %>% crossv_kfold(k = 10, id = "Resample ID") It is possible to map a model to each of the training datasets in the list-column train and to compute a performance metric by map ping onto the l
在实施数据科学（TM）的整齐模型modelr ，重新采样数据使用列表列组织： library(modelr) library(tidyverse) # create the k-folds df_heights_resampled = heights %>% crossv_kfold(k = 10, id = "Resample ID") 它可以map模型，以列表中的每个列的训练数据集的train ，并通过计算性能度量map平榜上有名列test 。 如果需要使用多个模型来完成，则需要对每个模型重复此操作。 # create a list of formulas formula
I'd like to calculate the position of the sun given time, latitude, and longitude. I found this great question and answer here: Position of the sun given time of day, latitude and longitude. However, when I evaluate the function I get incorrect results. Given the quality of the answer, I'm almost certain there's something wrong on my end, but I'm asking this question as a recor
我想计算太阳给定时间，纬度和经度的位置。 我在这里发现了这个伟大的问题和答案：太阳在一天中的时间，纬度和经度的位置。 但是，当我评估函数时，我得到不正确的结果。 鉴于答案的质量，我几乎可以肯定，我的问题出现了问题，但我把这个问题作为试图解决问题的记录。 为便于说明，以下是重印后功能的代码： astronomersAlmanacTime <- function(x) { # Astronomer's almanach time is the number of # days since
I have a Run length encoded vector representing some value at every position on the genome, in order. As a toy example suppose I had just one chromosome of length 10, then I would have a vector looking like library(GenomicRanges) set.seed(1) toyData = Rle(sample(1:3,10,replace=TRUE)) I would like to coerce this into a GRanges object. The best I can come up with is gr = GRanges('toyChr',IRan
我有一个运行长度编码的向量，代表基因组上每个位置的某个值。 作为一个玩具的例子，假设我只有一个长度为10的染色体，那么我会看到一个向量 library(GenomicRanges) set.seed(1) toyData = Rle(sample(1:3,10,replace=TRUE)) 我想将它强制为一个GRanges对象。 我能拿出最好的是 gr = GRanges('toyChr',IRanges(cumsum(c(0,runLength(toyData)[-nrun(toyData)])), width=runLength(toyData))
I am supposed to write a function for Stirling Numbers of the Second Kind, given by the formula: For this, I have written the following function in R: stirling <- function(n, k) { sum = 0 for (i in 0:k) { sum = sum + (-1)^(k - i) * choose(k, i) * i^n } sum = sum / factorial(k) return(sum) } The next part of the question is to "create a plot for n = 20, k = 1,2,...,10&q
我应该为公式给出的第二类斯特林数的函数： 为此，我在R中编写了以下函数： stirling <- function(n, k) { sum = 0 for (i in 0:k) { sum = sum + (-1)^(k - i) * choose(k, i) * i^n } sum = sum / factorial(k) return(sum) } 问题的下一部分是“为n = 20，k = 1,2，...，10创建一个情节”。 我做了一些研究，我认为方法curve或plot可能会对我有所帮助。 然而，我猜这些方法是在y的形式为f(x) （即一个
Does anybody know where can I find an example of LSTM via MXNet (R package)? The basic task is prediction of x[t + 1] value by x[1 ... t] sequence. 如果你仍然在寻找它，那么有这个非常好的博客解释如何在mxnet中使用R绑定来使用RNN：https：//www.r-bloggers.com/recurrent-models-and-examples-with-mxnetr/。
有谁知道我在哪里可以通过MXNet（R软件包）找到LSTM的例子？ 基本任务是通过x [1 ... t]序列预测x [t + 1]值。 如果你仍然在寻找它，那么有这个非常好的博客解释如何在mxnet中使用R绑定来使用RNN：https：//www.r-bloggers.com/recurrent-models-and-examples-with-mxnetr/。
I have a graph and need the shortest distance between all nodes. Now I made the following function, shortestPath <- function(streets, length) { streets <- matrix(streets, byrow=TRUE, ncol=2) # from -> to g <- graph.data.frame(as.data.frame(streets)) # create graph, see plot(g) return <- shortest.paths(g, weights = length) # return routes lengths } Here str
我有一个图形，需要所有节点之间的最短距离。 现在我做了以下功能， shortestPath <- function(streets, length) { streets <- matrix(streets, byrow=TRUE, ncol=2) # from -> to g <- graph.data.frame(as.data.frame(streets)) # create graph, see plot(g) return <- shortest.paths(g, weights = length) # return routes lengths } 这里的streets是一个包含数据的向量，我们有一
This question already has an answer here: How can I load dependencies in an R package? 2 answers Since there's no answer yet... J_F writes Have you tried to put the three packages under "Depends" in your DESCRIPTON? I ran into this recently and Depends still seems to work where Imports does not.
这个问题在这里已经有了答案： 我如何加载R包中的依赖关系？ 2个答案 由于目前还没有答案...... J_F写道 您是否试图将三个软件包放在您的DESCRIPTON的“Depends”下？ 我最近遇到了这种情况， Depends似乎仍然适用于Imports没有的地方。