小编use*_*545的帖子

我有一份运行由slurm管理的linux机器的工作.既然工作运行了几个小时,我意识到我低估了它完成所需的时间,因此--time我指定的参数值不够.有没有办法通过slurm为现有的正在运行的工作增加时间？

我想这是一个非常微不足道的问题,但是,我正在寻找(我猜测的)sacct命令,它将显示slurm作业ID使用的CPU时间和内存.

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 = …

我有一个data.frame(或矩阵或任何其他表格数据结构对象):

df = data.frame(field1 = c(1,1,1),field2 = c(2,2,2),field3 = c(3,3,3))

我想复制其部分列 - 在下面的向量中给出:

fields = c("field1","field2")

到已有1列或更多列的新data.table:

dt = data.table(fieldX = c("x","x","x"))

我正在寻找比以下更有效(和优雅)的东西:

for(f in 1:length(fields))
{
dt[,fields[f]] = df[,fields[f]]
}

下面是使用ggplot2生成boxplot的代码我正在尝试修改以适应我的问题:

library(ggplot2)
set.seed(1)
# create fictitious data
a <- rnorm(10)
b <- rnorm(12)
c <- rnorm(7)
d <- rnorm(15)

# data groups
group <- factor(rep(1:4, c(10, 12, 7, 15)))

# dataframe
mydata <- data.frame(c(a,b,c,d), group)
names(mydata) <- c("value", "group")

# function for computing mean, DS, max and min values
min.mean.sd.max <- function(x) {
  r <- c(min(x), mean(x) - sd(x), mean(x), mean(x) + sd(x), max(x))
  names(r) <- c("ymin", "lower", "middle", "upper", "ymax")
  r
}

# ggplot code
p1 <- ggplot(aes(y = …

我想expression在一个ggplot2方面标签上有一个R.

假设我正在策划tips data.frame:

library(reshape2)
> head(tips)
  total_bill  tip    sex smoker day   time size
1      16.99 1.01 Female     No Sun Dinner    2
2      10.34 1.66   Male     No Sun Dinner    3
3      21.01 3.50   Male     No Sun Dinner    3
4      23.68 3.31   Male     No Sun Dinner    2
5      24.59 3.61 Female     No Sun Dinner    4
6      25.29 4.71   Male     No Sun Dinner    4

如下:

library(ggplot2)
sp <- ggplot(tips, aes(x=total_bill, y=tip/total_bill)) + 
         geom_point(shape=1) + 
         facet_wrap(~sex, ncol = …

我正在按照这个例子说明如何用带有R's的树状图创建聚簇热图plotly.这是一个例子:

library(ggplot2)
library(ggdendro)
library(plotly)

#dendogram data
x <- as.matrix(scale(mtcars))
dd.col <- as.dendrogram(hclust(dist(x)))
dd.row <- as.dendrogram(hclust(dist(t(x))))
dx <- dendro_data(dd.row)
dy <- dendro_data(dd.col)

# helper function for creating dendograms
ggdend <- function(df) {
  ggplot() +
    geom_segment(data = df, aes(x=x, y=y, xend=xend, yend=yend)) +
    labs(x = "", y = "") + theme_minimal() +
    theme(axis.text = element_blank(), axis.ticks = element_blank(),
          panel.grid = element_blank())
}

# x/y dendograms
px <- ggdend(dx$segments)
py <- ggdend(dy$segments) + coord_flip()

# heatmap
col.ord …

我想创建一个矩阵数组,我首先创建一个具有NA值的k矩阵数组,然后循环k并通过数组更新每个第k个矩阵.

有什么建议？

我有一组不同ID的间隔.例如:

df <- data.frame(id=c(rep("a",4),rep("b",2),rep("c",3)), start=c(100,250,400,600,150,610,275,600,700), end=c(200,300,550,650,275,640,325,675,725))

每个id的间隔不重叠,但不同id的间隔可以重叠.这是一张图片:

plot(range(df[,c(2,3)]),c(1,nrow(df)),type="n",xlab="",ylab="",yaxt="n")
for ( ii in 1:nrow(df) ) lines(c(df[ii,2],df[ii,3]),rep(nrow(df)-ii+1,2),col=as.numeric(df$id[ii]),lwd=2)
legend("bottomleft",lwd=2,col=seq_along(levels(df$id)),legend=levels(df$id))

我正在寻找的是两个函数:1.一个将这些区间联合起来的函数.对于上面的示例,它将返回此data.frame:

union.df <- data.frame(id=rep("a,b,c",4), start=c(100,400,600,700), end=c(325,550,675,725))

2. 一个与这些间隔相交的函数,只有在所有id与该范围重叠时才保持范围.对于上面的示例,它将返回此data.frame:

intersection.df <- data.frame(id="a,b,c", start=610, end=640)

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))

我有这个代码用于绘制ggplot2 geom_boxplot:

p1 = ggplot(data = my.df, aes(factor(replicate), color = factor(parent.origin)))
p1 = p1 + geom_boxplot(aes(fill = factor(parent.origin), width = 0.3, lower = mean - std.dev, upper = mean + std.dev, middle = mean, ymin = mean - 3*std.dev, ymax = mean + 3*std.dev), stat="identity") + facet_wrap(~group, ncol = 4)+scale_color_manual(values = c("red","blue"),labels = c("maternal","paternal"),name = "parental allele")+scale_fill_manual(values = c("red","blue"))

我的问题是:1.我想为每个盒子添加一条中心线(比如黑色)2.我想让盒子变窄,这样任何给定复制品中的蓝色和红色盒子都不会相互重叠3我想让胡须线虚线或点缀

任何的想法？