如何在R中的依赖数据上应用直方图？

Question

我想通过 R. 欢迎ggplot2 方法和任何其他方法！ 伪代码

• N11.1、...、N32.1 组的第二列和第三列直方图

代码

        N11.1 N22.1 N33.1 N44.1 N21.1 N31.1 N32.1
Sinus     1.0   0.0   0.0   0.0   0.0   0.0  12.0
Arr/AHB   1.0   0.0   0.0   0.1   0.0   0.0  20.9
        N11.1 N22.1 N33.1 N44.1 N21.1 N31.1 N32.1
Sinus     1.0   0.0   0.0   0.0   0.0   0.0   4.0
Arr/AHB   1.0   0.0   0.0   0.0   0.0   0.0  24.0

第一列有row.names。用数据编码

library("ggplot2")

data.female <- structure(list(N11.1 = structure(c(3L, 3L), .Label = c("", "0.0", 
"1.0", "N11"), class = "factor"), N22.1 = structure(c(2L, 2L), .Label = c("", 
"0.0", "2.0", "N22"), class = "factor"), N33.1 = structure(c(2L, 
2L), .Label = c("", "0.0", "N33"), class = "factor"), N44.1 = structure(2:3, .Label = c("", 
"0.0", "0.1", "0.2", "N44"), class = "factor"), N21.1 = structure(c(2L, 
2L), .Label = c("", "0.0", "N21"), class = "factor"), N31.1 = structure(c(2L, 
2L), .Label = c("", "0.0", "N31"), class = "factor"), N32.1 = structure(c(5L, 
7L), .Label = c("", "0.0", "10.8", "11.0", "12.0", "17.0", "20.9", 
"22.8", "24.0", "3.0", "4.0", "44.0", "N32"), class = "factor")), .Names = c("N11.1", 
"N22.1", "N33.1", "N44.1", "N21.1", "N31.1", "N32.1"), row.names = c("Sinus", 
"Arr/AHB"), class = "data.frame")

data.male <- structure(list(N11.1 = structure(c(3L, 3L), .Label = c("", "0.0", 
"1.0", "N11"), class = "factor"), N22.1 = structure(c(2L, 2L), .Label = c("", 
"0.0", "2.0", "N22"), class = "factor"), N33.1 = structure(c(2L, 
2L), .Label = c("", "0.0", "N33"), class = "factor"), N44.1 = structure(c(2L, 
2L), .Label = c("", "0.0", "0.1", "0.2", "N44"), class = "factor"), 
    N21.1 = structure(c(2L, 2L), .Label = c("", "0.0", "N21"), class = "factor"), 
    N31.1 = structure(c(2L, 2L), .Label = c("", "0.0", "N31"), class = "factor"), 
    N32.1 = structure(c(11L, 9L), .Label = c("", "0.0", "10.8", 
    "11.0", "12.0", "17.0", "20.9", "22.8", "24.0", "3.0", "4.0", 
    "44.0", "N32"), class = "factor")), .Names = c("N11.1", "N22.1", 
"N33.1", "N44.1", "N21.1", "N31.1", "N32.1"), row.names = c("Sinus", 
"Arr/AHB"), class = "data.frame")

尝试单个数据行

data.female.sinus <- data.female[1:1,1:7]
print(data.female.sinus)

g <- ggplot(data.female.sinus)
g + geom_bar()
#Warning messages:
#1: In min(x, na.rm = na.rm) :
#  no non-missing arguments to min; returning Inf
#2: In max(x, na.rm = na.rm) :
#  no non-missing arguments to max; returning -Inf
#3: In min(diff(sort(x))) : no non-missing arguments to min; returning Inf
#4: In is.na(x) : is.na() applied to non-(list or vector) of type 'NULL'
#5: Computation failed in `stat_count()`:
#arguments imply differing number of rows: 0, 1 
#null device 

预期输出：男性和女性比较的直方图，强调 Arr/AHB 是因变量

测试 hhh 的 answer

我不明白为什么你不能使用带有列名的给定数据，比如没有列名

Sinus <- c(1,0,0,0,0,0,12)
ArrAHB <- c(1,0,0,0.1,0,0,20.9)
# Things work with this data  

Sinus <- data.female[1, 1:7]
ArrAHB <- data.female[2, 1:7]
# Things do not work with this data which has column names

Labels <- c("N11.1","N22.2","N33.1","N44.1","N21.1","N31.1","N32.1")
ID <- c("Sinus","Arr/AHB")
data.female <- data.frame(Sinus,ArrAHB,row.names=Labels)
data.female <- t(data.female)

barchart(data.female,auto.key=list(space='right'))

R：3.3.1
操作系统：Debian 8.5

Answer 1

您的数据如下所示：

> data.female
        N11.1 N22.1 N33.1 N44.1 N21.1 N31.1 N32.1
Sinus     1.0   0.0   0.0   0.0   0.0   0.0  12.0
Arr/AHB   1.0   0.0   0.0   0.1   0.0   0.0  20.9
> data.male
        N11.1 N22.1 N33.1 N44.1 N21.1 N31.1 N32.1
Sinus     1.0   0.0   0.0   0.0   0.0   0.0   4.0
Arr/AHB   1.0   0.0   0.0   0.0   0.0   0.0  24.0

并且您想在多列上绘制每一行的直方图（如here），所以下面的演示。

1. Sinus 和 ArrAHB 组分开的每一行的直方图

您想为 Sinus 和 Arr/AHB 创建一个通用标识符，因此我们为此创建一个新的 ID 列。我们使用这种方法here 和 lattice pkg。

require(lattice)
Sinus<-c(1,0,0,0,0,0,12)
ArrAHB<-c(1,0,0,0.1,0,0,20.9)
Labels<-c("N11.1","N22.1","N33.1","N44.1","N21.1","N31.1","N32.1")
ID<-c("Sinus","Arr/AHB")
data.female<-data.frame(Sinus,ArrAHB,row.names=Labels)
data.female<-as.data.frame(t(data.female))
data.female$ID<-ID

barchart(N11.1+N22.1+N33.1+N44.1+N21.1+N31.1+N32.1 ~ ID,
         data=data.female,
         auto.key=list(space='right')
         )

相比之下，这是人类的图表：

1.2.您的因子数据必须转换为向量或更好：将原始文件直接读入向量，而不是因子！

您的输入数据格式错误为因子数据，这里很糟糕，这可能是误用 read.csv 的结果，例如缺少 hte 标志 na.strings="." 或某些格式错误的元素。更多：

"Sometimes when a data frame is read directly from a file, a column you’d thought would produce a numeric vector instead produces a factor. This is caused by a non-numeric value in the column, often a missing value encoded in a special way like . or -. To remedy the situation, coerce the vector from a factor to a character vector, and then from a character to a double vector. (Be sure to check for missing values after this process.) Of course, a much better plan is to discover what caused the problem in the first place and fix that; using the na.strings argument to read.csv() is often a good place to start.*

为了使用这种格式错误的数据，必须将因子元素转换为数值。类命令揭示了您在将原始数据读入 R 时的错误，因此

> class(data.female$N22.1)
[1] "factor"
> as.double(as.character(data.female$N22.1))
[1] 0 0

as.double(as.character(...)) 允许再次正确操作数据对象。所以代码

require(lattice)
data.female <- structure(list(N11.1 = structure(c(3L, 3L), .Label = c("", "0.0", "1.0", "N11"), class = "factor"),
                              N22.1 = structure(c(2L, 2L), .Label = c("", "0.0", "2.0", "N22"), class = "factor"),
                              N33.1 = structure(c(2L, 2L), .Label = c("", "0.0", "N33"), class = "factor"),
                              N44.1 = structure(2:3, .Label = c("", "0.0", "0.1", "0.2", "N44"), class = "factor"),
                              N21.1 = structure(c(2L, 2L), .Label = c("", "0.0", "N21"), class = "factor"),
                              N31.1 = structure(c(2L, 2L), .Label = c("", "0.0", "N31"), class = "factor"),
                              N32.1 = structure(c(5L, 7L), .Label = c("", "0.0", "10.8", "11.0", "12.0", "17.0", "20.9", "22.8", "24.0", "3.0", "4.0", "44.0", "N32"),
                                                class = "factor")), .Names = c("N11.1", "N22.1", "N33.1", "N44.1", "N21.1", "N31.1", "N32.1"),
                         row.names = c("Sinus", "Arr/AHB"), class = "data.frame")
data.female$ID<-c("Sinus","Arr/AHB")
data.female<-as.data.frame(data.female)

f<-function(x) as.double(as.character(x))   #factors converted to vectors

barchart(f(N11.1)+f(N22.1)+f(N33.1)+f(N44.1)+f(N21.1)+f(N31.1)+f(N32.1) ~ ID,
         data=data.female,
         auto.key=list(space='right')
         )

函数f 将因子转换为向量，唉，因子是具有类对象和属性值的特殊向量，更多here。

您需要自己操作图例的地方。

2。再次显示比例的条形图

数据输入变为可读格式（不是某些CSZ文件的输出）：N32.1中的值远大于其他列中的任何其他数据。

require(lattice)
Sinus<-c(1,0,0,0,0,0,12)
ArrAHB<-c(1,0,0,0.1,0,0,20.9)
Labels<-c("N11.1","N22.2","N33.1","N44.1","N21.1","N31.1","N32.1")
ID<-c("Sinus","Arr/AHB")
data.female<-data.frame(Sinus,ArrAHB,row.names=Labels)
data.female<-t(data.female)

barchart(data.female,auto.key=list(space='right'))

> data.female
       N11.1 N22.2 N33.1 N44.1 N21.1 N31.1 N32.1
Sinus      1     0     0   0.0     0     0  12.0
ArrAHB     1     0     0   0.1     0     0  20.9