小编use*_*545的帖子

是否有一个R函数可以在文本文件中搜索s字符串？像unix grep这样的东西？

我想替代方法是逐行读取文件但是想知道是否可以通过这样的函数绕过它？

我正在尝试向ggplot两个直方图中的一个添加图例，这些图可能会重叠，因此希望它们稍微透明：

library(ggplot2)
set.seed(1)
plot.df <- data.frame(x=c(rnorm(1000,30,1),rnorm(10000,40,5)),
                      group=c(rep("a",1000),rep("b",10000)))

使用：

ggplot(plot.df,aes(x=x,fill=factor(group)))+ 
  geom_histogram(data=subset(plot.df,group=='a'),fill="red",alpha=0.5)+
  geom_histogram(data=subset(plot.df,group=='b'),fill="darkgray",alpha=0.5)+
  scale_colour_manual(name="group",values=c("red","darkgray"),labels=c("a","b"))+scale_fill_manual(name="group",values=c("red","darkgray"),labels=c("a","b"))

但我得到的是：

少了什么东西？

假设我有以下rle对象:

r = rle(c(rep("M",28),rep("N",4265),rep("M",16),rep("S",2),rep("N",400),rep("M",10)));

我想将其分解为以下字符串向量:

a = c("28M","4265N","16M2S","400N","10M");

含义我将"N"值和非"N"值及其对应的长度分隔为向量中的单独元素.

请注意,所有非Ns都是粘贴在一起,这就是结果为"16M2S"而不是"16M""2S"分开的原因.

最有效的方法是什么？

我有这个data.frame:

my.df = data.frame(mean = c(0.045729661,0.030416531,0.043202944,0.025600973,0.040526913,0.046167044,0.029352414,0.021477789,0.027580529,0.017614864,0.020324659,0.027547972,0.0268722,0.030804717,0.021502093,0.008342398,0.02295506,0.022386184,0.030849534,0.017291356,0.030957321,0.01871551,0.016945678,0.014143042,0.026686185,0.020877973,0.028612298,0.013227244,0.010710895,0.024460647,0.03704981,0.019832982,0.031858501,0.022194059,0.030575241,0.024632496,0.040815748,0.025595652,0.023839083,0.026474704,0.033000706,0.044125751,0.02714219,0.025724641,0.020767752,0.026480009,0.016794441,0.00709195), std.dev = c(0.007455271,0.006120299,0.008243454,0.005552582,0.006871527,0.008920899,0.007137174,0.00582671,0.007439398,0.005265133,0.006180637,0.008312494,0.006628951,0.005956211,0.008532386,0.00613411,0.005741645,0.005876588,0.006640122,0.005339993,0.008842722,0.006246828,0.005532832,0.005594483,0.007268493,0.006634795,0.008287031,0.00588119,0.004479003,0.006333063,0.00803285,0.006226441,0.009681048,0.006457784,0.006045368,0.006293256,0.008062195,0.00857954,0.008160441,0.006830088,0.008095485,0.006665062,0.007437581,0.008599525,0.008242957,0.006379928,0.007168385,0.004643819), parent.origin = c("paternal","paternal","paternal","paternal","paternal","paternal","maternal","maternal","maternal","maternal","maternal","maternal","paternal","paternal","paternal","paternal","paternal","paternal","maternal","maternal","maternal","maternal","maternal","maternal","maternal","maternal","maternal","maternal","maternal","maternal","paternal","paternal","paternal","paternal","paternal","paternal","maternal","maternal","maternal","maternal","maternal","maternal","paternal","paternal","paternal","paternal","paternal","paternal"), group = c("F1r:M","F1r:M","F1r:M","F1r:M","F1r:M","F1r:M","F1r:M","F1r:M","F1r:M","F1r:M","F1r:M","F1r:M","F1r:F","F1r:F","F1r:F","F1r:F","F1r:F","F1r:F","F1r:F","F1r:F","F1r:F","F1r:F","F1r:F","F1r:F","F1i:M","F1i:M","F1i:M","F1i:M","F1i:M","F1i:M","F1i:M","F1i:M","F1i:M","F1i:M","F1i:M","F1i:M","F1i:F","F1i:F","F1i:F","F1i:F","F1i:F","F1i:F","F1i:F","F1i:F","F1i:F","F1i:F","F1i:F","F1i:F"), replicate = c(1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6))

为此,我使用以下代码生成多面ggplot:

p1 = ggplot(data = my.df, aes(factor(replicate), color = factor(parent.origin)))
p1 = p1 + geom_boxplot(aes(fill = factor(parent.origin),lower = mean - std.dev, upper = mean + std.dev, middle = mean, ymin = mean - 3*std.dev, ymax = mean + 3*std.dev), position = position_dodge(width = 0), width = 0.5, alpha = 0.5, stat="identity") + facet_wrap(~group, ncol = 4)+scale_fill_manual(values = c("red","blue"),labels = c("maternal","paternal"),name = "parental …

我得到了这个 Rcpp实现到包的rmvnorm功能mvtnorm,我想知道我需要添加什么才能使用openmp,因此它可以利用多个内核.

我虽然应该这样做:

library(Rcpp)
library(RcppArmadillo)
library(inline)
settings <- getPlugin("RcppArmadillo")
settings$env$PKG_CXXFLAGS <- paste('-fopenmp', settings$env$PKG_CXXFLAGS)
settings$env$PKG_LIBS <- paste('-fopenmp -lgomp', settings$env$PKG_LIBS)

code <- '
#include <omp.h>
using namespace Rcpp;
int cores = 1;
cores = as<int>(cores_);
omp_set_num_threads(cores);
int n = as<int>(n_);
arma::vec mu = as<arma::vec>(mu_);
arma::mat sigma = as<arma::mat>(sigma_);
int ncols = sigma.n_cols;
#pragma omp parallel for schedule(static)
arma::mat Y = arma::randn(n, ncols);
return wrap(arma::repmat(mu, 1, n).t() + Y * arma::chol(sigma));
'

rmvnorm.rcpp <- cxxfunction(signature(n_="integer", mu_="numeric", sigma_="matrix", …

假设我想绘制我的数据:

my.df <- data.frame(mean = c(0.045729661,0.030416531,0.043202944,0.025600973,0.040526913,0.046167044,0.029352414,0.021477789,0.027580529,0.017614864,0.020324659,0.027547972,0.0268722,0.030804717,0.021502093,0.008342398,0.02295506,0.022386184,0.030849534,0.017291356,0.030957321,0.01871551,0.016945678,0.014143042,0.026686185,0.020877973,0.028612298,0.013227244,0.010710895,0.024460647,0.03704981,0.019832982,0.031858501,0.022194059,0.030575241,0.024632496,0.040815748,0.025595652,0.023839083,0.026474704,0.033000706,0.044125751,0.02714219,0.025724641,0.020767752,0.026480009,0.016794441,0.00709195), std.dev = c(0.007455271,0.006120299,0.008243454,0.005552582,0.006871527,0.008920899,0.007137174,0.00582671,0.007439398,0.005265133,0.006180637,0.008312494,0.006628951,0.005956211,0.008532386,0.00613411,0.005741645,0.005876588,0.006640122,0.005339993,0.008842722,0.006246828,0.005532832,0.005594483,0.007268493,0.006634795,0.008287031,0.00588119,0.004479003,0.006333063,0.00803285,0.006226441,0.009681048,0.006457784,0.006045368,0.006293256,0.008062195,0.00857954,0.008160441,0.006830088,0.008095485,0.006665062,0.007437581,0.008599525,0.008242957,0.006379928,0.007168385,0.004643819), parent.origin = c("paternal","paternal","paternal","paternal","paternal","paternal","maternal","maternal","maternal","maternal","maternal","maternal","paternal","paternal","paternal","paternal","paternal","paternal","maternal","maternal","maternal","maternal","maternal","maternal","maternal","maternal","maternal","maternal","maternal","maternal","paternal","paternal","paternal","paternal","paternal","paternal","maternal","maternal","maternal","maternal","maternal","maternal","paternal","paternal","paternal","paternal","paternal","paternal"), group = c("F1r:M","F1r:M","F1r:M","F1r:M","F1r:M","F1r:M","F1r:M","F1r:M","F1r:M","F1r:M","F1r:M","F1r:M","F1r:F","F1r:F","F1r:F","F1r:F","F1r:F","F1r:F","F1r:F","F1r:F","F1r:F","F1r:F","F1r:F","F1r:F","F1i:M","F1i:M","F1i:M","F1i:M","F1i:M","F1i:M","F1i:M","F1i:M","F1i:M","F1i:M","F1i:M","F1i:M","F1i:F","F1i:F","F1i:F","F1i:F","F1i:F","F1i:F","F1i:F","F1i:F","F1i:F","F1i:F","F1i:F","F1i:F"), replicate = c(1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6))

如下:

library(ggplot2)

p1 <- ggplot(data = my.df, aes(factor(replicate), color = factor(parent.origin)))
p1 <- p1 + geom_boxplot(aes(fill = factor(parent.origin),lower = mean - std.dev, upper = mean + std.dev, middle = mean, ymin = mean - 3*std.dev, ymax = mean + 3*std.dev), position = position_dodge(width = 0), width = 0.5, alpha = 0.5, stat="identity") + facet_wrap(~group, ncol = 4)+scale_fill_manual(values = c("red","blue"),labels = c("maternal","paternal"),name = …

我有这个例子data.frame:

df <- data.frame(id=c("a","a,b,c","d,e","d","h","e","i","b","c"), start=c(100,100,400,400,800,500,900,200,300), end=c(150,350,550,450,850,550,950,250,350), level = c(1,5,2,3,6,4,2,1,1))

> df
     id start end level
1     a   100 150     1
2 a,b,c   100 350     5
3   d,e   400 550     2
4     d   400 450     3
5     h   800 850     6
6     e   500 550     4
7     i   900 950     2
8     b   200 250     1
9     c   300 350     1

其中每一行是一个线性区间.如此示例所示,某些行是合并的间隔(第2行和第3行).

我想要做的是,每个合并的间隔要么消除所有单独的部分,df如果df$level合并的间隔大于其所有部分的部分,或者df$level合并的间隔小于其至少一个部分消除合并的间隔.

因此,对于此示例,输出应为:

> res.df
     id start end level
1 a,b,c   100 …

假设我生成了这个散点图：

plot(x = runif(20,-10,10), y = runif(20,-10,10), xlim = c(-10,10), ylim = c(-10,10))
abline(h = 0, col = "black")
abline(v = 0, col = "black")

所以abline's 将平面划分为四个笛卡尔象限。我想用不同的颜色为每个象限的背景着色。分别为 1-4 象限说蓝色、红色、绿色和黄色。

任何的想法？

假设我有一个数字向量,我想要舍入到"更漂亮"的数字,例如:

vec <- c(1.739362e-08,8.782537e-08,0.5339712)

我希望它是:

pretty.vec <- c(1.74e-08,8.78e-08,0.53)

我如何实现这一目标？使用round并没有真正帮助,因为它将前两个元素舍入为0:

> round(vec,2)
[1] 0.00 0.00 0.53

我正在尝试组合一系列R plotly绘图 - 垂直，但由于每个单独绘图的 y 轴标题是水平的，因此组合绘图变得混乱。

这是我的例子：

数据：

set.seed(1)
df <- data.frame(cluster=unlist(lapply(letters[1:10],function(i) rep(paste0("cluster:",i),200))),
                 group=rep(c(rep("A",100),rep("B",100)),10),val=rnorm(2000))
df$group <- factor(df$group,levels=c("A","B"))

绘制一个图列表density，每个df$cluster：

library(plotly)

plot.list <- lapply(unique(df$cluster),function(i){
  density.df <- do.call(rbind,lapply(c("A","B"),function(b){
    dens <- density(dplyr::filter(df,cluster == i,group == b)$val,adjust=1)
    return(data.frame(x=dens$x,y=dens$y,group=b,stringsAsFactors=F))
  }))
  density.df$group <- factor(density.df$group,levels=c("A","B"))
  
  if(i == unique(df$cluster)[1]){
    cluster.plot <- plot_ly(x=~density.df$x,y=~density.df$y,type='scatter',mode='lines',color=~density.df$group,line=list(width=3),showlegend=T) %>%
      layout(xaxis=list(title="Val",zeroline=F),yaxis=list(title=i,zeroline=F,showticklabels=F),legend=list(orientation="h",xanchor="center",x=0,y=1))
  } else{
    cluster.plot <- plot_ly(x=~density.df$x,y=~density.df$y,type='scatter',mode='lines',color=~density.df$group,line=list(width=3),showlegend=F) %>%
      layout(xaxis=list(title="Val",zeroline=F),yaxis=list(title=i,zeroline=F,showticklabels=F))
  }
  return(cluster.plot)
})

然后，使用以下方法组合它们：

subplot(plot.list,nrows=length(plot.list),shareX=T,shareY=T,titleX=T,titleY=T)

给出：

如何将每个图的 y 轴标题旋转plot.list为水平而不是垂直？