Question

假设我有这个data.table（实际数据是25061 x 5862）：

require(data.table)
df
  # gene     P1     P2     P3     P4     P5
 # 1: gene1  0.111  0.319  0.151     NA -0.397
 # 2: gene10  1.627  2.252  1.462 -1.339 -0.644
 # 3: gene2 -1.766 -0.056 -0.369  1.910  0.981
 # 4: gene3 -1.346  1.283  0.322 -0.465  0.403
 # 5: gene4 -0.783     NA -0.005  1.761  0.066
 # 6: gene5  0.386 -0.309 -0.886 -0.072  0.161
 # 7: gene6  0.547 -0.144 -0.725 -0.133  1.059
 # 8: gene7  0.785 -1.827  0.986  1.555 -0.798
 # 9: gene8 -0.186     NA  0.401  0.900 -1.075
# 10: gene9 -0.177  1.497 -1.370 -1.628 -1.044

我想知道如何利用data.table结构，我可以有效地计算每对基因值，有多少对没有NA的夫妇。例如，对于对gene1，gene2，我想结果是4。

使用基数R，我这样做：

calc_nonNA <- !is.na(df[, -1, with=F])
Effectifs <- calc_nonNA %*% t(calc_nonNA)
# or, as suggested by @DavidArenburg and @Khashaa, more efficiently:
Effectifs <- tcrossprod(calc_nonNA)

但是，如果df很大，则需要数小时......

我想要的输出，提供的示例如下：

       gene1 gene10 gene2 gene3 gene4 gene5 gene6 gene7 gene8 gene9
gene1      4      4     4     4     3     4     4     4     3     4
gene10     4      5     5     5     4     5     5     5     4     5
gene2      4      5     5     5     4     5     5     5     4     5
gene3      4      5     5     5     4     5     5     5     4     5
gene4      3      4     4     4     4     4     4     4     4     4
gene5      4      5     5     5     4     5     5     5     4     5
gene6      4      5     5     5     4     5     5     5     4     5
gene7      4      5     5     5     4     5     5     5     4     5
gene8      3      4     4     4     4     4     4     4     4     4
gene9      4      5     5     5     4     5     5     5     4     5

数据

df <- structure(list(gene = c("gene1", "gene10", "gene2", "gene3", 
"gene4", "gene5", "gene6", "gene7", "gene8", "gene9"), P1 = c(0.111, 
1.627, -1.766, -1.346, -0.783, 0.386, 0.547, 0.785, -0.186, -0.177
), P2 = c(0.319, 2.252, -0.056, 1.283, NA, -0.309, -0.144, -1.827, 
NA, 1.497), P3 = c(0.151, 1.462, -0.369, 0.322, -0.005, -0.886, 
-0.725, 0.986, 0.401, -1.37), P4 = c(NA, -1.339, 1.91, -0.465, 
1.761, -0.072, -0.133, 1.555, 0.9, -1.628), P5 = c(-0.397, -0.644, 
0.981, 0.403, 0.066, 0.161, 1.059, -0.798, -1.075, -1.044)), .Names = c("gene", 
"P1", "P2", "P3", "P4", "P5"), class = c("data.table", "data.frame"
), row.names = c(NA, -10L), .internal.selfref = <pointer: 0x022524a0>)

Answer 1

使用 dplyr ，将数据扩展为long，然后加入自身并进行汇总。不确定它是否比你的解决方案更有效率，有些人会对任何人进行基准测试吗？

library(dplyr)
library(tidyr)

# reshaping from wide to long
x <- df %>% gather(key = P, value = value, -c(1)) %>% 
  mutate(value=(!is.na(value)))

# result
left_join(x,x,by="P") %>% 
  group_by(gene.x,gene.y) %>% 
  summarise(N=sum(value.x & value.y)) %>% 
  spread(gene.y,N)

修改遗憾的是，这个 dplyr 解决方案失败了更大的数据集2600x600，无法加入自身 - internal vecseq reached physical limit，大约2 ^ 31行......

顺便说一下，这里是t vs tcrossprod的基准：

library(ggplot2) library(microbenchmark) op <- microbenchmark( BASE_t={ calc_nonNA <- !is.na(df[, -1, with=F]) calc_nonNA %*% t(calc_nonNA) }, BASE_tcrossprod={ calc_nonNA <- !is.na(df[, -1, with=F]) tcrossprod(calc_nonNA) }, times=10 ) qplot(y=time, data=op, colour=expr) + scale_y_log10()

Answer 2

我用25061x5862的随机数据尝试了这个，它很快就嚼掉了50gb的ram（包括交换空间），因此，比使用tcrossprod更省内存效率但是如果你有一个淫秽内存量然后可能（但可能不是）这可能会更快。

#generate cross columns for all matches
crossDT<-data.table(gene=rep(df1[,unique(gene)],length(df1[,unique(gene)])),gene2=rep(df1[,unique(gene)],each=length(df1[,unique(gene)])))
#create datatable with row for each combo
df2<-merge(df1,crossDT,by="gene")
setkey(df2,gene2)
setkey(df1,gene)
#make datatable with a set of P columns for each gene
df3<-df1[df2]
#find middle column and then make name vectors
pivotcol<-match("i.gene",names(df3))
names1<-names(df3)[2:(pivotcol-1)]
names2<-names(df3)[(pivotcol+1):ncol(df3)]
names3<-paste0("new",names1)
#make third set of P columns where the new value is False if either of the previous sets of P columns is NA
df3[,(names3):=lapply(1:length(names1),function(x) !any(is.na(c(get(names1[x]),get(names2[x]))))),by=c("gene","i.gene")]
#delete first sets of P columns
df3[,c(names1,names2):=NULL]
#sum up true columns
df3[,Sum:=rowSums(.SD),.SDcols=names3]
#delete set of P columns
df3[,(names3):=NULL]
#cast results to desired shape
dcast.data.table(df3,gene~i.gene,value.var='Sum')

使用data.table进行计算：变量中有多少2x2非NA值？

2 个答案: