我了解了当只有两个重复项 block 时如何处理删除重复列的情况，但在我的真实数据中我有 3 个或更多重复项。我试图提出一些玩具示例数据集，其中有一组额外的重复列名称，我想将其折叠。是否有一种类似的简单方法可以使用 dplyr 和 tidyr 来解决这些困惑？
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
dplyr 应该如何处理:
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
更困难的情况:
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
dplyr 之后应该是什么样子:
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，输出数据帧都应该有两列，第一列和第二列。
感谢您的帮助!
最佳答案
这两种情况都只是简单的索引问题
拳头案例(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)
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
关于当存在多个重复列时，删除表中的重复列，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/29803379/