Question

我有一个相当大的data.frame，包含12374行（基因）和785列（单元格）。我想根据rowMeans将行分组到20个bin中。在每个区域内，我想对该区域内所有基因的分散度量（方差/平均值）进行z归一化，以便识别异常基因，即使与具有相似平均表达的基因相比，其表达值也是高度可变的。然后，我想提取超过z分数阈值1.7的基因，以从每个箱中鉴定显着可变的基因。

我的数据看起来像这样：

> head(temp[,1:5])
                         Cell1                Cell2                 Cell3              Cell4                 Cell5
0610007P14RIK            0.1439444            0.0000000             0.000000            0.8759335            0.0000000
0610009B22RIK            0.0000000            0.6776718             0.000000            0.0000000            0.0000000
0610009O20RIK            0.1439444            0.0000000             0.000000            0.2735741            0.0000000
0610010B08RIK            1.4769893            1.1369215             1.124842            0.8759335            1.9544187
0610010F05RIK            0.7944809            0.0000000             0.000000            0.7016789            0.9144108
0610010K14RIK            0.1439444            0.0000000             1.124842            0.7016789            0.0000000

我尝试使用dplyr来执行此操作，但遇到与（我认为是）垃圾箱数量相关的错误。这是我的尝试：

library(dplyr)
library(genefilter)
n_bins = 20
temp = data
temp$dispersion = rowMeans(temp)/rowVars(temp)
outscore = temp %>% mutate(bin=ntile(dispersion,n_bins)) %>% 
  group_by(bin) %>% mutate(zscore=scale(dispersion),outlier=abs(zscore)>1.7)

返回的错误为rror: dims [product 619] do not match the length of object [618]

Answer 1

修订：以下是基础R的解决方案，只需dplyr的一点帮助：

library(dplyr)

# I called the data set 'mydata'
colnames(mydata)[1]<-"ID"
a<-which(colnames(mydata)== "ID")

##from: http://www.inside-r.org/packages/cran/metaMA/docs/rowVars

rowVars<-function (x,na.rm = TRUE) 
  {
    sqr = function(x) x * x
    n = rowSums(!is.na(x))
    n[n <= 1] = NA
    return(rowSums(sqr(x - rowMeans(x,na.rm = na.rm)), na.rm = na.rm)/(n - 1))
  }

mydata$dispersion<-rowMeans(mydata[,-a])/rowVars(mydata[,-a])
nbins = 2 # for you, use 20, or however many you want.
mydata$bin<-ntile(mydata$dispersion, nbins)


b<-which(colnames(mydata)== "bin")
temp<-NULL
mydata$Z<-0

for(i in unique(mydata$bin)){
    temp<-mydata[mydata$bin == i, -c(a,b)]$dispersion
    temp<-(temp-mean(temp))/sd(temp)
    mydata[mydata$bin == i, -c(a,b)]$Z<-temp
  }

mydata$outlier<-ifelse(abs(mydata$Z) > 1.7, 1, 0)
mydata.small<-mydata[,c(1,7:10)] ##for display purposes
mydata.small

           ID dispersion bin          Z outlier
0610007P14RIK   1.406851   1 -0.9370254       0
0610009B22RIK   1.475641   1 -0.1158566       0
0610009O20RIK   5.502857   2  0.1333542       0
0610010B08RIK   7.553503   2  0.9266318       0
0610010F05RIK   2.418036   2 -1.0599860       0
0610010K14RIK   1.573546   1  1.0528820       0

Bin行和每个bin计算分散和返回异常值

1 个答案: