Question

我有相当大的代谢物数据数据集。有些集合具有未标记的重复（没有列表示重复）。下面是一个小例子。

a<-structure(list(ABBRC = structure(c(1L, 2L, 2L, 3L, 4L, 4L, 4L
        ), .Label = c("X1", "X2", "X3", "X4"), class = "factor"), X = 1:7, 
            Y = 1:7, Year = c(2009L, 2009L, 2009L, 2009L, 2009L, 2009L, 
            2009L)), .Names = c("ABBRC", "X", "Y", "Year"), class = "data.frame", row.names = c(NA, 
        -7L))
        b<-structure(list(ABBRC = structure(c(1L, 2L, 3L, 4L, 4L, 4L, 4L
        ), .Label = c("X1", "X2", "X3", "X4"), class = "factor"), Z = c(1L, 
        2L, 4L, 5L, 6L, 7L, 8L), A = c(1L, 2L, 4L, 5L, 6L, 7L, 8L), Year = c(2009L, 
        2009L, 2009L, 2009L, 2009L, 2009L, 2009L)), .Names = c("ABBRC", 
        "Z", "A", "Year"), class = "data.frame", row.names = c(NA, -7L
        ))
    merge(a,b)
ABBRC Year X Y Z A
1     X1 2009 1 1 1 1
2     X2 2009 2 2 2 2
3     X2 2009 3 3 2 2
4     X3 2009 4 4 4 4
5     X4 2009 5 5 5 5
6     X4 2009 5 5 6 6
7     X4 2009 5 5 7 7
8     X4 2009 5 5 8 8
9     X4 2009 6 6 5 5
10    X4 2009 6 6 6 6
11    X4 2009 6 6 7 7
12    X4 2009 6 6 8 8
13    X4 2009 7 7 5 5
14    X4 2009 7 7 6 6
15    X4 2009 7 7 7 7
16    X4 2009 7 7 8 8

合并时，输出重复行的组合。这是预期的行为，但这不是我想要的。我希望将数据合并，就好像它们是重复一样（它们是）。是否有一个函数来进行这种合并，或者更容易标记重复然后合并？如果标签更容易，那么这样做的好方法是什么？

期望输出

structure(list(ABBRC = structure(c(1L, 2L, 2L, 3L, 4L, 4L, 4L, 
4L), .Label = c("X1", "X2", "X3", "X4"), class = "factor"), X = c(1L, 
2L, 3L, 4L, 5L, 6L, 7L, NA), Y = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 
NA), Z = c(1L, 2L, NA, 4L, 5L, 6L, 7L, 8L), A = c(1L, 2L, NA, 
4L, 5L, 6L, 7L, 8L), Year = c(2009L, 2009L, 2009L, 2009L, 2009L, 
2009L, 2009L, 2009L)), .Names = c("ABBRC", "X", "Y", "Z", "A", 
"Year"), class = "data.frame", row.names = c(NA, -8L))
ABBRC  X  Y  Z  A Year
1    X1  1  1  1  1 2009
2    X2  2  2  2  2 2009
3    X2  3  3 NA NA 2009
4    X3  4  4  4  4 2009
5    X4  5  5  5  5 2009
6    X4  6  6  6  6 2009
7    X4  7  7  7  7 2009
8    X4 NA NA  8  8 2009

Answer 1

删除我的第一次痛苦尝试后，这是另一种方法，但不如您自己的plyr方法。它涉及首先生成一个虚拟time变量。

a$time <- as.numeric(ave(as.character(a$ABBRC), a$ABBRC, a$Year, FUN=seq_along))
b$time <- as.numeric(ave(as.character(b$ABBRC), b$ABBRC, b$Year, FUN=seq_along))
library(reshape2)
ab.long <- rbind(melt(a, id.vars=c("ABBRC", "Year", "time")),
                 melt(b, id.vars=c("ABBRC", "Year", "time")))
dcast(ab.long, ABBRC + Year + time ~ variable)
#   ABBRC Year time  X  Y  Z  A
# 1    X1 2009    1  1  1  1  1
# 2    X2 2009    1  2  2  2  2
# 3    X2 2009    2  3  3 NA NA
# 4    X3 2009    1  4  4  4  4
# 5    X4 2009    1  5  5  5  5
# 6    X4 2009    2  6  6  6  6
# 7    X4 2009    3  7  7  7  7
# 8    X4 2009    4 NA NA  8  8

Answer 2

不确定回答你自己的问题是否很酷，但我想出了如何通过创建一个索引变量来做到这一点。感谢Hadley对plyr / seq_along（）的一些建议。

require(plyr)
a<-ddply(a, .(ABBRC), transform, rep=seq_along(ABBRC))
b<-ddply(b, .(ABBRC), transform, rep=seq_along(ABBRC))
merge(a,b, all=T)

  ABBRC Year rep  X  Y  Z  A
1    X1 2009   1  1  1  1  1
2    X2 2009   1  2  2  2  2
3    X2 2009   2  3  3 NA NA
4    X3 2009   1  4  4  4  4
5    X4 2009   1  5  5  5  5
6    X4 2009   2  6  6  6  6
7    X4 2009   3  7  7  7  7
8    X4 2009   4 NA NA  8  8

Answer 3

有几种方法可以解决这个问题。一种方法是在合并之前识别重复项

merge(a, b[!duplicatesFromA, ])
#    ABBRC Year X Y Z A
#  1    X4 2009 5 5 8 8
#  2    X4 2009 6 6 8 8
#  3    X4 2009 7 7 8 8

当然，有几种方法可以找到重复的内容这是一个使用嵌套的apply循环的colSums。

duplicatesFromA <- 
    colSums(apply(b, 1, function(row.b) {
        apply(a, 1, function(row.a) {
            all(row.b==row.a)
        })
    })) > 0

合并具有重复行的数据帧

3 个答案: