2

当只有两个重复块时，我看到了如何处理删除重复列的情况，但在我的真实数据中，我有 3 个或更多。我试图想出一些玩具示例数据集，其中有一组额外重复的列名，我想折叠它们。dplyr有没有一种类似的直接方法可以用and解开这些混乱tidyr？

更简单的情况：

structure(list(x = c("a", "a", NA, "a", "a", NA, "a"), y = c(1, 
5, NA, 15, 19, NA, 27), z = c(2, 6, NA, 16, 20, NA, 28), x.1 = c("b", 
"b", "b", "b", "b", "b", "b"), y.1 = c(3, 7, 11, 17, 21, 23, 
29), z.1 = c(4, 8, 12, 18, 22, 24, 30), x.2 = c(NA, NA, "a", 
NA, NA, "a", NA), y.2 = c(NA, NA, 13, NA, NA, 25, NA), z.2 = c(NA, 
NA, 14, NA, NA, 26, NA)), .Names = c("x", "y", "z", "x.1", "y.1", 
"z.1", "x.2", "y.2", "z.2"), row.names = c(NA, -7L), class = "data.frame")

这看起来像在 R 中：

     x  y  z x.1 y.1 z.1  x.2 y.2 z.2
1    a  1  2   b   3   4 <NA>  NA  NA
2    a  5  6   b   7   8 <NA>  NA  NA
3 <NA> NA NA   b  11  12    a  13  14
4    a 15 16   b  17  18 <NA>  NA  NA
5    a 19 20   b  21  22 <NA>  NA  NA
6 <NA> NA NA   b  23  24    a  25  26
7    a 27 28   b  29  30 <NA>  NA  NA

它应该如何照顾dplyr：

  x  y  z x.1 y.1 z.1
1 a  1  2   b   3   4
2 a  5  6   b   7   8
3 a 13 14   b  11  12
4 a 15 16   b  17  18
5 a 19 20   b  21  22
6 a 25 26   b  23  24
7 a 27 28   b  29  30

更难的情况：

structure(list(x = c("a", "b", NA, "a", "a", NA, "a"), y = c(1, 
7, 9, 15, 19, NA, 27), z = c(2, 8, 10, 16, 20, NA, 28), x.1 = c("b", 
NA, "b", "b", "b", "b", "b"), y.1 = c(3, NA, 11, 17, 21, 23, 
29), z.1 = c(4, NA, 12, 18, 22, 24, 30), x.2 = c(NA, "a", "a", 
NA, NA, "a", NA), y.2 = c(NA, 5, 13, NA, NA, 25, NA), z.2 = c(NA, 
6, 14, NA, NA, 26, NA)), .Names = c("x", "y", "z", "x.1", "y.1", 
"z.1", "x.2", "y.2", "z.2"), row.names = c(NA, -7L), class = "data.frame")

这看起来像在 R 中：

     x  y  z  x.1 y.1 z.1  x.2 y.2 z.2
1    a  1  2    b   3   4 <NA>  NA  NA
2    b  7  8 <NA>  NA  NA    a   5   6
3 <NA>  9 10    b  11  12    a  13  14
4    a 15 16    b  17  18 <NA>  NA  NA
5    a 19 20    b  21  22 <NA>  NA  NA
6 <NA> NA NA    b  23  24    a  25  26
7    a 27 28    b  29  30 <NA>  NA  NA

之后应该是什么样子dplyr：

  x  y  z x.1 y.1 z.1
1 a  1  2   b   3   4
2 a  5  6   b   7   8
3 a 13 14   b  11  12
4 a 15 16   b  17  18
5 a 19 20   b  21  22
6 a 25 26   b  23  24
7 a 27 28   b  29  30

在这两种情况下，输出数据框都应该有两列，第一列和第二列。

谢谢你的帮助！

4

2 回答 2

2

这两种情况都只是简单的索引问题

拳套（最简单的）

indx <- is.na(df$x)
df[indx, 1:3] <- df[indx, 7:9]
df[1:6]
#   x  y  z x.1 y.1 z.1
# 1 a  1  2   b   3   4
# 2 a  5  6   b   7   8
# 3 a 13 14   b  11  12
# 4 a 15 16   b  17  18
# 5 a 19 20   b  21  22
# 6 a 25 26   b  23  24
# 7 a 27 28   b  29  30

第二种情况（较难的一种）

indx <- 1:3
indx2 <- as.logical(rowSums(is.na(df2[indx + 3])))
indx3 <- as.logical(rowSums(is.na(df2[indx])))

df2[indx2, indx + 3] <- df2[indx2, indx]
df2[indx3, indx] <- df2[indx3, indx + 6]
df2[1:6]
#   x  y  z x.1 y.1 z.1
# 1 a  1  2   b   3   4
# 2 b  7  8   b   7   8
# 3 a 13 14   b  11  12
# 4 a 15 16   b  17  18
# 5 a 19 20   b  21  22
# 6 a 25 26   b  23  24
# 7 a 27 28   b  29  30
于 2015-04-22T17:44:38.373 回答
0

这将作为静态校正工作，但根据重复的数量，您可以将其转换为函数以使其更具动态性。

library(stringr)
# Method One (Works when you have true duplicates from some join methods)
for(i in 1:length(df))
{
  Cols = which(colnames(df)==colnames(df)[i])
  if(length(Cols)>1){
    df[Cols[1]] = NULL 
  }
}


# Method Two 
for(i in 1:length(df))
{

Val = which(strsplit(colnames(df)[i], "")[[1]]==".")
if(length(Val) >= 1 ){

  Cols = which(colnames(df)==paste(substr(colnames(df)[i],1,Val-1),".2",sep=''))

    df[Cols[1]] = NULL 
   }

}
于 2015-04-22T17:14:14.970 回答