我有
replicate(1000, t.test(rnorm(10)))
它的作用是从正态分布中抽取10号样本,对其执行一次t.test,并执行1000次.但对于我的任务,我只对p值感兴趣(问题是:零假设被拒绝了多少次).我如何仅获得p值,或者我可以添加已经说明零假设被拒绝多少次的内容(p值小于0.05的次数)
t.test返回一个类的对象,htest该对象是一个包含许多组件的列表,包括p.value(这是你想要的).
你有几个选择.
您可以将t.test结果保存在列表中,然后提取p.value组件
# simplify = FALSE to avoid coercion to array
ttestlist <- replicate(1000, t.test(rnorm(10)), simplify = FALSE)
ttest.pval <- sapply(ttestlist, '[[', 'p.value')
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或者您只能保存t.test对象的该组件
pvals <- replicate(1000, t.test(rnorm(10))$p.value)
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以下是我用来解决问题的步骤.注意我如何将其分解为最小的组件并逐步构建它:
#Let's look at the structure of one t.test to see where the p-value is stored
str(t.test(rnorm(10)))
#It is named "p.value, so let's see if we can extract it
t.test(rnorm(10))[["p.value"]]
#Now let's test if its less than your 0.05 value
ifelse(t.test(rnorm(10))[["p.value"]]< 0.05,1,0)
#That worked. Now let's replace the code above in your replicate function:
replicate(1000, ifelse(t.test(rnorm(10))[["p.value"]]< 0.05,1,0))
#That worked too, now we can just take the sum of that:
#Make it reproducible this time
set.seed(42)
sum(replicate(1000, ifelse(t.test(rnorm(10))[["p.value"]]< 0.05,1,0)))
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应该得到这个:
[1] 54
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