小编Sul*_*esi的帖子

我有以下data.frame调用df.我的问题与y轴上的顺序有关.我希望y轴上的名称根据变量排序depth.

如果我这样做:

ggplot(df,aes(x=factor(name),y=depth)) + geom_bar(stat='identity') + coord_flip() + labs(y='depth',x='species')

我得到graph1下面没有订购.因此我按照这里的说明,name按照深度订购了我的因子水平:

df2=df[order(df$depth),]
df2$name=factor(df2$name,levels=df2$name)
ggplot(df2,aes(x=factor(name),y=depth)) + geom_bar(stat='identity') + coord_flip() + labs(y='depth',x='species')

我得到了Graph2以下.但对我来说,下一步是根据Mut变量对条形图进行不同的着色.

ggplot(df2,aes(x=factor(name),y=depth)) + geom_bar(stat='identity',data=subset(df2,df2$Mut==2),fill='red') + geom_bar(stat='identity',data=subset(df2,df2$Mut==1),fill='blue') + coord_flip() + labs(y='depth',x='species')

我得到Graph3的不再订购!

如何根据graph2中显示的顺序生成graph3

                    name depth Mut            x
25  A_rubrocinctus_GA070     8   2 -0.033318659
9      A_omanensis_GA051    10   2 -0.020387101
4  A_latifasciatus_GA083    12   1 -0.005645811
27      A_frenatus_GA068    12   1 -0.024190876
13       A_percula_GA017    15   1  0.034591721
14       A_percula_GA039    15   2 …

我正在使用@jbaums 在这里trees找到的包,并在这篇文章中解释过.

我的数据如下:

树由树干组成

    Trunk
[1] 13.60415

和分支机构

Tree
   TreeBranchLength TreeBranchID
1         10.004269            1
2          7.994269            2
3          9.028834           11
4         10.817401           12
5          8.551311          111
6         10.599798          112
7         11.073243          121
8         13.367392          122
9          9.625431         1111
10        10.793569         1112
11         9.896499        11121
12         8.687741        11122
13         7.791180         1211
14        12.506105         1212
15         6.768478         1221
16        10.441796         1222
17        10.751892         1121
18         9.458651         1122
19        10.768509        11221
20        10.150673        11222 …

我试图在下面的情节背景上有2个"阴影".这些阴影应分别代表橙色和蓝色点的密度.是否有意义？

以下是要改进的ggplot:

这是df我用来创建这个图的代码和数据(矩阵):

                         PC1           PC2 aa
A_akallopisos    0.043272525  0.0151023307  2
A_akindynos     -0.020707141 -0.0158198405  1
A_allardi       -0.020277664 -0.0221016281  2
A_barberi       -0.023165596  0.0389906701  2
A_bicinctus     -0.025354572 -0.0059122384  2
A_chrysogaster   0.012608835 -0.0339330213  2
A_chrysopterus  -0.022402365 -0.0092476009  1
A_clarkii       -0.014474658 -0.0127024469  1
A_ephippium     -0.016859412  0.0320034231  2
A_frenatus      -0.024190876  0.0238499714  2
A_latezonatus   -0.010718845 -0.0289904165  1
A_latifasciatus -0.005645811 -0.0183202248  2
A_mccullochi    -0.031664307 -0.0096059126  2
A_melanopus     -0.026915545  0.0308399009  2
A_nigripes       0.023420045  0.0293801537  2
A_ocellaris      0.052042539  0.0126144250  2
A_omanensis     -0.020387101  0.0010944998  2
A_pacificus      0.042406273 -0.0260308092  2 …

> matrix(c(c(0, 3.75882e-06, 3.71645e-05, 2.16088e-06, 1.357e-06, 1.19274e-06, NaN, 1.14748e-06, 9.3314e-07), c(3.75882e-06, 0, 3.94165e-05, 3.58464e-06, 3.60392e-06, 3.43881e-06, NaN, 3.39315e-06, 3.17616e-06), c(3.71645e-05, 3.94165e-05, 0, 3.78173e-05, 3.70121e-05, 3.68449e-05, NaN, 3.6798e-05, 3.65591e-05), c(2.16088e-06, 3.58464e-06, 3.78173e-05, 0, 2.00581e-06, 1.84085e-06, NaN, 1.79527e-06, 1.57976e-06), c(1.357e-06, 3.60392e-06, 3.70121e-05, 2.00581e-06, 0, 1.03709e-06, NaN, 9.91615e-07, 7.77135e-07), c(1.19274e-06, 3.43881e-06, 3.68449e-05, 1.84085e-06, 1.03709e-06, 0, NaN, 8.27333e-07, 6.12979e-07), c(NaN, NaN, NaN, NaN, NaN, NaN, 0, NaN, NaN), c(1.14748e-06, 3.39315e-06, 3.6798e-05, 1.79527e-06, 9.91615e-07, 8.27333e-07, NaN, 0, 5.67856e-07), c(9.3314e-07, 3.17616e-06, 3.65591e-05, 1.57976e-06, 7.77135e-07, 6.12979e-07, NaN, …

我有一个以下类型的data.frame

set.seed(12)
d = data.frame(a=sample(5,x=1:9), 
               b=sample(5,x=1:9),
               c=sample(5,x=1:9),
               d=sample(5,x=1:9),
               e=sample(5,x=1:9),
               f=sample(5,x=1:9))

d
#   a b c d e f
# 1 1 1 4 4 2 3
# 2 7 2 7 9 7 5
# 3 8 5 3 8 1 2
# 4 2 9 8 7 5 9
# 5 9 6 2 1 9 4

我想取前两列,将整数转换为字符并将同一行的两个元素粘贴在一起.然后重复每个连续的一对列的过程.

这是一个可以正确完成工作的脚本:

bar = function (twocols) {sapply(1:nrow(twocols), FUN=function(x) {paste(twocols[x,], collapse="")} )}

    count = 0
    out = matrix(0, ncol=ncol(d)/2, nrow=nrow(d))
    for (i in …