【发布时间】:2019-12-09 16:15:46
【问题描述】:
我一直在使用一个函数来计算 t 检验,使用每个条件下的均值、标准差和 n。
ttest <- function(m1,m2,s1,s2,n1,n2,m0=0,equal.variance=FALSE)
{
if( equal.variance==FALSE )
{
se <- sqrt( (s1^2/n1) + (s2^2/n2) )
df <- ( (s1^2/n1 + s2^2/n2)^2 )/( (s1^2/n1)^2/(n1-1) + (s2^2/n2)^2/(n2-1) )
} else
{
se <- sqrt( (1/n1 + 1/n2) * ((n1-1)*s1^2 + (n2-1)*s2^2)/(n1+n2-2) )
df <- n1+n2-2
}
t <- (m1-m2-m0)/se
sdpooled_n <- ((n1-1)*(s1^2)) + ((n2-1)*(s2^2))
sdpooled_df <- ((n1 + n2) - 2) # 115
sdpooled <- sqrt(sdpooled_n/sdpooled_df)
cohend <- (m1-m2)/sdpooled
dat <- c(m1-m2, se, t, df,2*pt(-abs(t),df),cohend)
names(dat) <- c("Mean difference", "Std Error", "t","df", "p-value","Cohen's d")
return(dat)
}
它很好用,但是 R 使用科学记数法返回结果很烦人。例如这段代码:
ttest(1.44,1.02,0.80,0.75,133,123,m0=0,equal.variance=FALSE)
返回:
Mean difference Std Error t df p-value Cohen's d
4.200000e-01 9.687725e-02 4.335383e+00 2.539507e+02 2.099250e-05 5.409679e-01
如果它返回这个就好了:
Mean difference Std Error t df p-value Cohen's d
0.42 0.10 4.34 253.95 0.000 0.54
我尝试过使用 round() 函数(见下文),它运行良好,只是它只显示 p 值的两位小数。重要的是我报告了准确的 p 值,除非 p<.001>
这是我改进后的代码版本:
ttest <- function(m1,m2,s1,s2,n1,n2,m0=0,equal.variance=FALSE)
{
if( equal.variance==FALSE )
{
se <- sqrt( (s1^2/n1) + (s2^2/n2) )
df <- ( (s1^2/n1 + s2^2/n2)^2 )/( (s1^2/n1)^2/(n1-1) + (s2^2/n2)^2/(n2-1) )
} else
{
se <- sqrt( (1/n1 + 1/n2) * ((n1-1)*s1^2 + (n2-1)*s2^2)/(n1+n2-2) )
df <- n1+n2-2
}
mdif <- m1-m2
t <- (mdif-m0)/se
sdpooled_n <- ((n1-1)*(s1^2)) + ((n2-1)*(s2^2))
sdpooled_df <- ((n1 + n2) - 2) # 115
sdpooled <- sqrt(sdpooled_n/sdpooled_df)
cohend <- (m1-m2)/sdpooled
p <- 2*pt(-abs(t),df)
mdif <- round(mdif,2)
se <- round(se,2)
t <- round(t,2)
df <- round(df,2)
p <- round(p,3)
cohend <- round(cohend,2)
dat <- c(mdif, se, t, df,p,cohend)
names(dat) <- c("Mean difference", "Std Error", "t","df", "p-value","Cohen's d")
return(dat)
}
但如您所见,它不会将 p 值返回到小数点后 3 位:
Mean difference Std Error t df p-value Cohen's d
0.42 0.10 4.34 253.95 0.00 0.54
【问题讨论】:
标签: r function output rounding