如何对R数据帧中的所有列执行shapiro测试？
shapiro检验用于检验变量的正态性，该检验的零假设是变量是正态分布的。如果我们在R数据框中有数字列，则可以检查所有变量的正态性。可以在apply函数的帮助下完成此操作，shapiro.test如以下示例所示。
示例
考虑以下数据帧-
set.seed(321) x1<−rnorm(20,2,0.34) x2<−rpois(20,5) x3<−rpois(20,2) x4<−rpois(20,5) x5<−rpois(20,6) x6<−runif(20,1,5) x7<−rexp(20,0.62) x8<−rpois(20,10) df<−data.frame(x1,x2,x3,x4,x5,x6,x7,x8) df输出结果
x1 x2 x3 x4 x5 x6 x7 x8 1 2.579667 7 0 2 4 4.712527 2.69354358 9 2 1.757907 4 0 3 3 1.519762 2.63275896 9 3 1.905485 5 2 5 4 3.087971 1.83827735 5 4 1.959319 7 0 10 14 3.564951 1.19092513 10 5 1.957853 7 3 5 5 4.576069 0.61126332 10 6 2.091182 4 0 4 10 3.316821 2.56506184 8 7 2.247126 3 4 5 7 1.636518 1.88751338 9 8 2.079266 8 4 7 7 3.018356 0.11237261 8 9 2.115299 3 2 7 4 4.516734 0.17862062 13 10 1.812349 3 0 6 5 3.009659 0.57255735 8 11 2.118218 5 2 6 4 1.025079 0.09536165 10 12 2.504761 4 1 3 4 1.936312 3.11482640 14 13 2.064031 1 0 5 7 2.388424 2.96859719 13 14 2.830708 2 4 9 6 3.779138 0.61244047 6 15 1.607831 6 5 7 7 2.740338 1.15703781 12 16 1.726412 6 3 5 7 4.690268 2.78394417 10 17 2.155064 3 2 8 11 4.043131 0.12627601 7 18 2.142913 3 4 8 4 1.481830 0.14825531 8 19 2.196379 4 2 3 6 1.490243 4.61761476 5 20 2.151761 6 1 5 2 1.914817 0.26060923 11
在df的所有列上应用shapiro测试-
示例
apply(df,2,shapiro.test)输出结果
$x1 Shapiro-Wilk normality test data: newX[, i] W = 0.94053, p-value = 0.2453 $x2 Shapiro-Wilk normality test data: newX[, i] W = 0.95223, p-value = 0.4022 $x3 Shapiro-Wilk normality test data: newX[, i] W = 0.88855, p-value = 0.02529 $x4 Shapiro-Wilk normality test data: newX[, i] W = 0.96244, p-value = 0.5938 $x5 Shapiro-Wilk normality test data: newX[, i] W = 0.87904, p-value = 0.017 $x6 Shapiro-Wilk normality test data: newX[, i] W = 0.93067, p-value = 0.1591 $x7 Shapiro-Wilk normality test data: newX[, i] W = 0.88531, p-value = 0.02208 $x8 Shapiro-Wilk normality test data: newX[, i] W = 0.96271, p-value = 0.5992