所以我使用r,包Bioconductor(oligo(,(limma(来分析一些微阵列数据。
我在配对分析中遇到问题。
所以这是我的表型数据
ph@dataph@data
index filename group
WT1 WT WT1 WT
WT2 WT WT2 WT
WT3 WT WT3 WT
WT4 WT WT4 WT
LT1 LT LT1 LT
LT2 LT LT2 LT
LT3 LT LT3 LT
LT4 LT LT4 LT
TG1 TG TG1 TG
TG2 TG TG2 TG
TG3 TG TG3 TG
TG4 TG TG4 TG
所以为了分析,我做了这个代码:
colnames(ph@data)[2]="source"
ph@data
groups = ph@data$source
f = factor(groups,levels = c("WT","LT","TG"))
design = model.matrix(~ 0 + f)
colnames(design)=c("WT","LT","TG")
design
data.fit = lmFit(normData,design)
> design
WT LT TG
1 1 0 0
2 1 0 0
3 1 0 0
4 1 0 0
5 0 1 0
6 0 1 0
7 0 1 0
8 0 1 0
9 0 0 1
10 0 0 1
11 0 0 1
12 0 0 1
attr(,"assign")
[1] 1 1 1
attr(,"contrasts")
attr(,"contrasts")$f
[1] "contr.treatment"
然后我起诉这个以比较我的小组:
contrast.matrix = makeContrasts(LT-WT,TG-WT,LT-TG,levels=design)
data.fit.con = contrasts.fit(data.fit,contrast.matrix)
data.fit.eb = eBayes(data.fit.con)
所以在所有这些之后,我想比较我的小组:
ph@data[ ,2] = c("control","control","control","control","littermate","littermate","littermate","littermate","transgenic","transgenic","transgenic","transgenic")
colnames(ph@data)[2]="Treatment"
ph@data[ ,3] = c("B1","B2","B3","B4","B1","B2","B3","B4","B1","B2","B3","B4")
colnames(ph@data)[3]="BioRep"
ph@data```
groupsB = ph@data$BioRep
groupsT = ph@data$Treatment
fb = factor(groupsB,levels=c("B1","B2","B3","B4"))
ft = factor(groupsT,levels=c("control","littermate","transgenic"))```
paired.design = (~ fb + ft)
所以现在我的表型数据看起来像这样:
index Treatment BioRep
WT1 WT control B1
WT2 WT control B2
WT3 WT control B3
WT4 WT control B4
LT1 LT littermate B1
LT2 LT littermate B2
LT3 LT littermate B3
LT4 LT littermate B4
TG1 TG transgenic B1
TG2 TG transgenic B2
TG3 TG transgenic B3
TG4 TG transgenic B4```
B1是生物复制品,然后我有野生型,同窝和转基因的组
比较我的样品,我正在尝试这个
colnames(paired.design)=c("Intercept","B4vsB1","B3vsB1","B2vsB1","B4vsB2","B3vsB2","B4vsB3","littermatevscontrol","transgenicvscontrol")
但后来我得到了这个错误:Error in `colnames<-`(`*tmp*`, value = c("Intercept", "WTvsLT", "WTvsTG", :
attempt to set 'colnames' on an object with less than two dimensions
我做错了什么,这是比较我的数据的正确方法吗?
data.fit = lmFit(data.rma,paired.design)
data.fit$coefficients
我制作了 ft 和 fb,因为我没有您的数据:
fb <- structure(c(1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L), .Label = c("B1",
"B2", "B3", "B4"), class = "factor")
ft <- structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L), .Label = c("control", "littermate", "transgenic"), class = "factor")
这一行有一个错别字:
paired.design = (~ fb + ft)
dim(paired.design)
NULL
它应该是:
paired.design = model.matrix(~ fb + ft)
head(paired.design)
(Intercept) fbB2 fbB3 fbB4 ftlittermate fttransgenic
1 1 0 0 0 0 0
2 1 1 0 0 0 0
如果你看一下你的model.matrix,它有6列,这不是你试图分配的。因此,当您指定~ fb + ft
时,选择因子的第一个水平作为参考,并且您会发现与此相关的其他水平的影响。例如,对于 fb,B1 是参考,您可以估计 B2,B3,B4 的影响。
要进行所需的成对比较,对于所有内容与引用,您只需要指定列本身,例如,对于 B4 与 B1,它只是 fbB4,因为 B4 是相对于 B1 估计的。对于其他人,您可以像以前一样指定"-":
contrast.matrix = makeContrasts(fbB4,fbB3,fbB2,fbB4-fbB2,
fbB3-fbB2,fbB4-fbB3,ftlittermate,fttransgenic,levels=paired.design)
现在我们可以根据需要分配 col 名称:
colnames(contrast.matrix) <-c("B4vsB1","B3vsB1","B2vsB1","B4vsB2","B3vsB2","B4vsB3",
"littermatevscontrol","transgenicvscontrol")
我为 normData 模拟了一些数据以拟合模型和对比:
set.seed(100)
normData = matrix(rpois(100*12,100),100,12)
data.fit = lmFit(normData,paired.design)
data.fit.con = contrasts.fit(data.fit,contrast.matrix)
请注意,有一个警告:
Warning message:
In contrasts.fit(data.fit, contrast.matrix) :
row names of contrasts don't match col names of coefficients
因为"(拦截("被重命名为拦截。
您可以查看 data.fit.con 下的系数,以查看您所做的比较的倍数变化
Contrasts
B4vsB1 B3vsB1 B2vsB1 B4vsB2 B3vsB2 B4vsB3
[1,] 14.333333 11.000000 5.0000000 9.333333 6.000000 3.333333
[2,] -3.333333 2.000000 7.0000000 -10.333333 -5.000000 -5.333333
[3,] 3.666667 -2.666667 6.3333333 -2.666667 -9.000000 6.333333
[4,] -22.666667 -1.666667 -10.3333333 -12.333333 8.666667 -21.000000
[5,] -8.333333 -3.666667 8.3333333 -16.666667 -12.000000 -4.666667
[6,] 15.333333 -5.666667 -0.3333333 15.666667 -5.333333 21.000000
Contrasts
littermatevscontrol transgenicvscontrol
[1,] 1.25 -7.50
[2,] 3.50 10.50
[3,] -3.75 2.75
[4,] 7.75 14.00
[5,] 16.75 -2.50
[6,] 2.00 9.50
您可以与原始拟合系数进行交叉检查,以查看是否进行了正确的对比度:
head(data.fit$coefficients)
(Intercept) fbB2 fbB3 fbB4 ftlittermate fttransgenic
[1,] 95.41667 5.0000000 11.000000 14.333333 1.25 -7.50
[2,] 94.33333 7.0000000 2.000000 -3.333333 3.50 10.50
[3,] 97.66667 6.3333333 -2.666667 3.666667 -3.75 2.75
[4,] 93.41667 -10.3333333 -1.666667 -22.666667 7.75 14.00
[5,] 98.91667 8.3333333 -3.666667 -8.333333 16.75 -2.50
[6,] 97.16667 -0.3333333 -5.666667 15.333333 2.00 9.50