如何使用SapplyforPearson系数遍历列

[i]表示我必须在列上迭代皮尔逊系数，以及如何将其转换为附加到变量的数据框？

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')
})


我认为您应该尝试：

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


它将提取各列与数据集的列years的比较的皮尔逊系数。

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


您将获得以下输出：

> 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