如何使用 Sapply 迭代列以获得 Pearson 系数

Question

[i] 指示我必须在列上迭代皮尔逊系数的位置以及如何将其转换为附加到变量的数据帧？

代码示例：

*INSTEAD OF DOING THIS*
F.ReedBunting.pear<- cor.test(W_farmland_mean$Years,W_farmland_mean$ReedBunting,method='pearson')
F.Whitethroat.pear<- cor.test(W_farmland_mean$Years,W_farmland_mean$Whitethroat,method='pearson')
F.Rook.pear<- cor.test(W_farmland_mean$Years,W_farmland_mean$Rook,method='pearson')
.
.
.
*HOW CAN IT BE DONE QUICKLY WITH THIS*
workspaceone <- sapply(W_farmland_mean, function(x){
    cor.test(W_farmland_mean$Years, W_farmland_mean[, 1[i]], method = 'pearson')
})

Answer 1

我认为你应该试试：

result_cor <- apply(W_farmland_mean,2,function(x){cor.test(W_farmland_mean$Years,x, method = 'pearson')$estimate})

它将提取每列与数据集第 years 列比较的 Pearson 系数。

例子使用 mtcars 数据集：

df <- mtcars[c(1:10),]

> df
                   mpg cyl  disp  hp drat    wt  qsec vs am gear carb
Mazda RX4         21.0   6 160.0 110 3.90 2.620 16.46  0  1    4    4
Mazda RX4 Wag     21.0   6 160.0 110 3.90 2.875 17.02  0  1    4    4
Datsun 710        22.8   4 108.0  93 3.85 2.320 18.61  1  1    4    1
Hornet 4 Drive    21.4   6 258.0 110 3.08 3.215 19.44  1  0    3    1
Hornet Sportabout 18.7   8 360.0 175 3.15 3.440 17.02  0  0    3    2
Valiant           18.1   6 225.0 105 2.76 3.460 20.22  1  0    3    1
Duster 360        14.3   8 360.0 245 3.21 3.570 15.84  0  0    3    4
Merc 240D         24.4   4 146.7  62 3.69 3.190 20.00  1  0    4    2
Merc 230          22.8   4 140.8  95 3.92 3.150 22.90  1  0    4    2
Merc 280          19.2   6 167.6 123 3.92 3.440 18.30  1  0    4    4

如果我们应用函数：

result_cor = apply(df,2, function(x){cor.test(x,df$mpg,method ='pearson')$estimate})

你得到以下输出：

> result_cor
       mpg        cyl       disp         hp       drat         wt       qsec 
 1.0000000 -0.8614165 -0.7739868 -0.8937223  0.5413585 -0.5991894  0.5494131 
        vs         am       gear       carb 
 0.4796102  0.2919683  0.6646449 -0.3711956

如何使用 Sapply 迭代列以获得 Pearson 系数

How to iterate over columns with Sapply for Pearson coefficient

loops

r

sapply