我找不到代码中的错误和/或逻辑中的缺陷。我有一个矩阵X
,0和1以及一个连续值的y
向量,我想在R
中进行2个样本t检验,其中行{{1}表示X
的不同组。
例如:
y
因此,我希望使用此代码进行6次t检验,其中x = matrix(rbinom(60,1,.5),ncol=10)
y = abs(rnorm(ncol(x)))
apply(x,1,function(x,y=y)t.test(y[x==1],y[x==0]))
的每一行对应两组X
。但是,当我运行代码时出现此错误:
y
有人可以解释错误并修改我的代码以获得我想要的内容。
答案 0 :(得分:2)
问题来自于在函数参数中重用变量名。这应该有效:
apply(x,1,function(x.f,y.f=y)t.test(y.f[x.f==1],y.f[x.f==0]))
答案 1 :(得分:1)
怎么样?
apply(x,1,function(x,z)t.test(y[x==1],y[x==0]),y)
如果你想在函数中使用第二个参数,你也应该将它传递给apply
答案 2 :(得分:1)
以下作品:
> apply(x,1,function(a)t.test(y[a==1],y[a==0]))
[[1]]
您应该为data.frames和向量中的数据提供更好的名称,以便x和y等可以用作常规变量。此外,无需向函数发送y,因为它对所有测试都是相同的。
输出:
Welch Two Sample t-test
data: y[a == 1] and y[a == 0]
t = 0.43835, df = 5.377, p-value = 0.6782
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.6356057 0.9036413
sample estimates:
mean of x mean of y
0.5807408 0.4467230
[[2]]
Welch Two Sample t-test
data: y[a == 1] and y[a == 0]
t = -0.80208, df = 5.5382, p-value = 0.4555
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-1.0985419 0.5644195
sample estimates:
mean of x mean of y
0.4337110 0.7007722
[[3]]
Welch Two Sample t-test
data: y[a == 1] and y[a == 0]
t = 0.58194, df = 7.3884, p-value = 0.5779
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.5584942 0.9283034
sample estimates:
mean of x mean of y
0.6329878 0.4480832
[[4]]
Welch Two Sample t-test
data: y[a == 1] and y[a == 0]
t = 1.1148, df = 4.8236, p-value = 0.3174
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.4919082 1.2308641
sample estimates:
mean of x mean of y
0.7622223 0.3927443
[[5]]
Welch Two Sample t-test
data: y[a == 1] and y[a == 0]
t = 0.23436, df = 5.5539, p-value = 0.8231
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.7818960 0.9439901
sample estimates:
mean of x mean of y
0.5729543 0.4919073
[[6]]
Welch Two Sample t-test
data: y[a == 1] and y[a == 0]
t = -1.015, df = 7.9168, p-value = 0.3401
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-1.0152988 0.3954558
sample estimates:
mean of x mean of y
0.3855747 0.6954962
仅适用于p值:
> apply(x,1,function(a)t.test(y[a==1],y[a==0])$p.value)
[1] 0.6781895 0.4555338 0.5779255 0.3173567 0.8231019 0.3400979