与p值矩阵的相关性 [英] Correlation with p values matrix

问题描述

我有以下数据:

x1 = sample(1:10, 100, replace=T)
x2 = sample(1:3, 100, replace=T)
x3 = sample(50:100, 100, replace=T)
y1 = sample(50:100, 100, replace=T)
y2 = sample(50:100, 100, replace=T)

mydf = data.frame(x1,x2,x3,y1,y2)
head(mydf)
  x1 x2 x3  y1 y2
1  2  2 96 100 73
2  5  2 77  93 52
3 10  1 86  54 80
4  3  2 98  59 94
5  2  2 85  94 85
6  9  2 56  79 99

我有以下数据:

我想进行相关并产生以下输出:

I want to do correlations and produce following output:

        x1                      x2                  x3
y1  r.value; p.value    r.value; p.value    r.value; p.value

y2  r.value; p.value    r.value; p.value    r.value; p.value

R值需要四舍五入为2位数字，p_value应当四舍五入为3位数字.

R value needs to be rounded to 2 digits and p_value to 3 digits.

这怎么办?感谢您的帮助.

How can this be done? Thanks for your help.

我尝试了以下操作:

library(Hmisc)
res = rcorr(as.matrix(mydf), type="pearson")
res

      x1    x2    x3    y1    y2
x1  1.00 -0.01 -0.16 -0.28 -0.21
x2 -0.01  1.00 -0.20 -0.10 -0.13
x3 -0.16 -0.20  1.00  0.14 -0.09
y1 -0.28 -0.10  0.14  1.00  0.12
y2 -0.21 -0.13 -0.09  0.12  1.00

n= 100 

P
   x1     x2     x3     y1     y2    
x1        0.9520 0.1089 0.0047 0.0364
x2 0.9520        0.0444 0.3463 0.1887
x3 0.1089 0.0444        0.1727 0.3948
y1 0.0047 0.3463 0.1727        0.2482
y2 0.0364 0.1887 0.3948 0.2482       

matrix(paste0(round(res[[1]][,1:3],2),';',round(res[[3]][1:2,],4)),ncol=3)
     [,1]           [,2]           [,3]          
[1,] "1;NA"         "-0.01;0.0444" "-0.16;NA"    
[2,] "-0.01;0.952"  "1;0.0047"     "-0.2;0.952"  
[3,] "-0.16;0.952"  "-0.2;0.3463"  "1;0.952"     
[4,] "-0.28;NA"     "-0.1;0.0364"  "0.14;NA"     
[5,] "-0.21;0.1089" "-0.13;0.1887" "-0.09;0.1089"

但是组合不正确.

推荐答案

尝试

r2 <- matrix(0, ncol=3, nrow=2, 
         dimnames=list( paste0('y',1:2), paste0('x',1:3)))
r2[] <- paste(round(res$r[4:5,1:3],2), round(res$P[4:5,1:3],4), sep="; ")

更新

您可以创建如下所示的函数

Update

You could create a function like below

 f1 <- function(df){
   df1 <- df[order(colnames(df))]
   indx <- sub('\\d+', '', colnames(df1))
   indx1 <- which(indx[-1]!= indx[-length(indx)])
   indx2 <- (indx1+1):ncol(df1) 
   r2 <- matrix(0, ncol=indx1, nrow=(ncol(df1)-indx1), 
           dimnames=list(colnames(df1)[indx2], colnames(df1)[1:indx1]))
   r1 <- rcorr(as.matrix(df1), type='pearson') 
   r2[] <- paste(round(r1$r[indx2,1:indx1],2), round(r1$P[indx2,1:indx1],4),
                       sep="; ")
   r2
  }


  f1(mydf) #using your dataset (`set.seed` is different)
  #        x1              x2             x3            
  #y1 "0.07; 0.4773"  "0.02; 0.84"   "0.21; 0.0385"
  #y2 "-0.08; 0.4363" "0.08; 0.4146" "0.02; 0.8599"

  Testing with unordered dataset

  f1(mydf1)
  #          x1              x2             x3              x4             
  #y1 "-0.08; 0.4086" "0.17; 0.0945" "-0.25; 0.0112" "-0.16; 0.1025"
  #y2 "0.07; 0.5174"  "-0.1; 0.3054" "0.03; 0.7478"  "-0.06; 0.5776"

Update2

如果您希望函数具有数字索引参数

Update2

If you want a function to have the numeric index argument

f2 <- function(df, v1, v2){
    r2 <- matrix(0, nrow=length(v2), ncol=length(v1),
          dimnames=list(colnames(df)[v2], colnames(df)[v1]))
    r1 <- rcorr(as.matrix(df), type='pearson')
   r2[] <- paste(round(r1$r[v2,v1],2), round(r1$P[v2,v1],4), sep="; ")
   r2
}

f2(mydf, 1:3, 4:5)

f2(mydf, c(1,3), c(2,4,5))

数据

 set.seed(29)
 x1 = sample(1:10, 100, replace=T)
 x2 = sample(1:3, 100, replace=T)
 x3 = sample(50:100, 100, replace=T)
 x4  <- sample(40:80, 100, replace=TRUE)
 y1 = sample(50:100, 100, replace=T)
 y2 = sample(50:100, 100, replace=T)

 mydfN = data.frame(x1,x2,x3,x4, y1,y2)

 set.seed(25)
 mydf1 <- mydfN[sample(colnames(mydfN))]

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