循环以合并数据文件并绑定这些

Question

我有一个像https://stackoverflow.com/questions/5965698/r-merge-unequal-dataframes-and-replace-missing-rows-with-0这样的问题

以下是这个问题的数据：

df1 = data.frame(x=c('a', 'b', 'c', 'd', 'e'))
df2 = data.frame(x=c('a', 'b', 'c'),y = c(0,1,0)) 
df3 = data.frame(x=c('a', 'b', 'c', 'd'),y = c(1,1,1,0))
df4 = data.frame(x=c('b', 'a', 'e'),y = c(0,1,0))
zz <- merge(df1, df2, all = TRUE)
zz[is.na(zz)] <- 0

在本例中，我将df1与df2合并。现在，我想创建一个循环，以便将df3和df4合并，并创建更多与df1合并的数据格式。问题是，列表中的结果必须是cbind，才能生成最终的数据。

有谁可以帮我？

谢谢!

编辑！这是我创建的循环。变量术语与10个变量的列表进行对比。变量是interestedGO中列表的名称。选择第一个interestedGO，计算结果为变量结果。这个结果必须与x合并，因为它是一个循环，所以所有10个结果都必须是cbind，才能创建最终的dataframe。

for (i in 1:length(goterms)){
  goilmn<-as.data.frame(interestedGO[i])
  resultILMN<-match(goilmn[,1], rownames(xx2),nomatch=0)
  resultILMN[resultILMN] <- 1
  result<-cbind(goilmn,resultILMN)
  colnames(result) <- c('x','result')

  zz<-merge(x, result, all=TRUE)
  resultloop<-zz[is.na(zz)]<-0
  standard[i]<-cbind(resultloop)
}

goterms:
[1] "GO:0009611" "GO:0007596" "GO:0050817" "GO:0061082" "GO:0007599"
[6] "GO:0050776" "GO:0006910" "GO:0034383" "GO:0019932" "GO:0002720"

interestedGO:
$`GO:0009611` 
[1] "ILMN_1651346" "ILMN_1651354" "ILMN_1651599" "ILMN_1651950" "ILMN_1652287"
[6] "ILMN_1652445" "ILMN_1652693" "ILMN_1652825" "ILMN_1653324" "ILMN_1653395"

$`GO:0007596`
[1] "ILMN_1651599" "ILMN_1652693" "ILMN_1652825" "ILMN_1653324" "ILMN_1655595"
[6] "ILMN_1656057" "ILMN_1659077" "ILMN_1659923" "ILMN_1659947" "ILMN_1662619"
[11] "ILMN_1664565" "ILMN_1665132" "ILMN_1665859" "ILMN_1666175" "ILMN_1668052"
[16] "ILMN_1670229" "ILMN_1670305" "ILMN_1670490" "ILMN_1670708"
"ILMN_1671766"                              

$`GO:0050817`
[1] "ILMN_1651599" "ILMN_1652693" "ILMN_1652825" "ILMN_1653324" "ILMN_1655595"
[6] "ILMN_1656057" "ILMN_1659077" "ILMN_1659923" "ILMN_1659947" "ILMN_1662619"
[11] "ILMN_1664565" "ILMN_1665132" "ILMN_1665859" "ILMN_1666175" "ILMN_1668052"
[16] "ILMN_1670229" "ILMN_1670305" "ILMN_1670490" "ILMN_1670708" "ILMN_1671766"
[21] "ILMN_1671928" "ILMN_1675083" "ILMN_1678049" "ILMN_1678728" 
"ILMN_1680805"                            

$`GO:0061082`
[1] "ILMN_1661695" "ILMN_1665132" "ILMN_1716446" "ILMN_1737314" "ILMN_1772387"
[6] "ILMN_1784863" "ILMN_1796094" "ILMN_1800317" "ILMN_1800512" "ILMN_1807074"

X是所有ILMN代码的引用。这里是x变量的头。x1:100，

 [1] ILMN_1343291 ILMN_1343295 ILMN_1651228 ILMN_1651229 ILMN_1651238
 [6] ILMN_1651254 ILMN_1651259 ILMN_1651260 ILMN_1651262 ILMN_1651278
 [11] ILMN_1651282 ILMN_1651285 ILMN_1651286 ILMN_1651303 ILMN_1651310
 [16] ILMN_1651315 ILMN_1651330 ILMN_1651336 ILMN_1651343 ILMN_1651346
 [21] ILMN_1651347 ILMN_1651354 ILMN_1651358 ILMN_1651370 ILMN_1651373
 [26] ILMN_1651385 ILMN_1651396 ILMN_1651415 ILMN_1651428 ILMN_1651430
 [31] ILMN_1651433 ILMN_1651437 ILMN_1651438 ILMN_1651456 ILMN_1651457

Answer 1

我不确定我是否正确地理解了你想要的，但像这样？

> zz <- Reduce(function(a,b)merge(a,b,all=TRUE, by="x"), list(df1, df2, df3, df4))
> zz[is.na(zz)] <- 0
> zz
  x y.x y.y y
1 a   0   1 1
2 b   1   1 0
3 c   0   1 0
4 d   0   0 0
5 e   0   0 0

您可以通过使用Reduce来避免循环，但请注意，它并不一定会导致性能的提高。

如果您想要单独的数据文件，那么Map (仅仅是mapply的包装器)是有用的：

> zz <- Map(function(b)merge(df1,b,all=TRUE, by="x"), list(df2, df3, df4))
> zz
[[1]]
  x  y
1 a  0
2 b  1
3 c  0
4 d NA
5 e NA

[[2]]
  x  y
1 a  1
2 b  1
3 c  1
4 d  0
5 e NA

[[3]]
  x  y
1 a  1
2 b  0
3 c NA
4 d NA
5 e  0

并通过do.call绑定它们。

> zz <- do.call("cbind", zz)
> zz[is.na(zz)] <- 0
> zz
  x y x y x y
1 a 0 a 1 a 1
2 b 1 b 1 b 0
3 c 0 c 1 c 0
4 d 0 d 0 d 0
5 e 0 e 0 e 0