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法律研究如何在 R 数据框中对所有列执行 Wilcoxon 检验？
在 R 数据框中对所有列执行 Wilcoxon 检验，意味着我们要对单个样本使用该检验，单个样本的 Wilcoxon 检验用于检验样本的中位数，中位数是否等于某个值。如果我们不提供任何值，则零为参考值。可以通过借助 apply 函数和 wilcox.test 来对所有列执行 Wilcoxon 检验，如下面的示例所示。
考虑以下数据框 −
示例
x1<-rnorm(20,5,0.31) x2<-rnorm(20,2,0.025) x3<-rpois(20,4) x4<-rpois(20,2) x5<-rpois(20,5) x6<-rpois(20,1) x7<-round(rnorm(20,3,1.1),2) x8<-round(rnorm(20,10,2.25),2) df<-data.frame(x1,x2,x3,x4,x5,x6,x7,x8) df
输出
x1 x2 x3 x4 x5 x6 x7 x8 1 4.667097 2.032878 5 1 8 0 3.82 5.68 2 4.556913 1.952845 7 3 5 0 4.62 12.57 3 5.274511 1.947305 3 0 2 1 3.27 7.03 4 4.621090 1.960653 4 3 4 0 3.37 9.71 5 4.808041 1.955832 4 3 4 0 2.96 7.58 6 4.509070 2.084535 2 4 10 1 3.04 6.42 7 5.230658 1.981629 2 2 5 0 2.26 7.25 8 4.724433 1.986739 3 2 5 0 3.76 10.46 9 5.123489 1.959177 3 1 1 0 3.97 10.55 10 5.179769 1.970168 3 3 4 0 3.91 7.68 11 5.133287 2.006720 5 1 7 0 1.89 10.85 12 4.677813 2.007699 3 0 6 2 1.81 9.01 13 4.662342 2.064619 7 2 5 3 2.72 8.42 14 5.375585 1.994618 2 1 3 0 4.09 9.83 15 5.574414 1.997730 2 4 3 0 3.14 9.58 16 5.279330 1.985777 8 4 10 2 2.95 9.60 17 5.258145 2.019408 2 3 6 1 3.42 10.68 18 5.051640 2.017030 6 3 4 2 3.91 11.66 19 5.064925 2.007080 4 1 6 0 3.06 8.74 20 4.957406 1.964513 9 2 5 0 4.59 11.11
对 df 的所有列执行 Wilcoxon 检验 −
apply(df,2,wilcox.test) $x1
Wilcoxon signed rank exact test data: newX[, i] V = 210, p-value = 1.907e-06 alternative hypothesis: true location is not equal to 0
$x2
Wilcoxon signed rank exact test data: newX[, i] V = 210, p-value = 1.907e-06 alternative hypothesis: true location is not equal to 0
$x3
Wilcoxon signed rank test with continuity correction data: newX[, i] V = 210, p-value = 8.966e-05 alternative hypothesis: true location is not equal to 0
$x4
Wilcoxon signed rank test with continuity correction data: newX[, i] V = 171, p-value = 0.0001896 alternative hypothesis: true location is not equal to 0
$x5
Wilcoxon signed rank test with continuity correction data: newX[, i] V = 210, p-value = 9.095e-05 alternative hypothesis: true location is not equal to 0
$x6
Wilcoxon signed rank test with continuity correction data: newX[, i] V = 28, p-value = 0.0206 alternative hypothesis: true location is not equal to 0
$x7
Wilcoxon signed rank test with continuity correction data: newX[, i] V = 210, p-value = 9.556e-05 alternative hypothesis: true location is not equal to 0
$x8
Wilcoxon signed rank exact test data: newX[, i] V = 210, p-value = 1.907e-06 alternative hypothesis: true location is not equal to 0
警告信息 −
1: In wilcox.test.default(newX[, i], ...) : cannot compute exact p-value with ties 2: In wilcox.test.default(newX[, i], ...) : cannot compute exact p-value with ties 3: In wilcox.test.default(newX[, i], ...) : cannot compute exact p-value with zeroes 4: In wilcox.test.default(newX[, i], ...) : cannot compute exact p-value with ties 5: In wilcox.test.default(newX[, i], ...) : cannot compute exact p-value with ties 6: In wilcox.test.default(newX[, i], ...) : cannot compute exact p-value with zeroes 7: In wilcox.test.default(newX[, i], ...) : cannot compute exact p-value with ties
更新于：06-2-2021
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