17

我想从以下位置添加变量dat2：

          concreteness familiarity typicality
amoeba            3.60        1.30       1.71
bacterium         3.82        3.48       2.13
leech             5.71        1.83       4.50

至dat1：

    ID  variable value
1    1    amoeba     0
2    2    amoeba     0
3    3    amoeba    NA
251  1 bacterium     0
252  2 bacterium     0
253  3 bacterium     0
501  1     leech     1
502  2     leech     1
503  3     leech     0

给出以下输出：

    X ID  variable value concreteness familiarity typicality
1   1  1    amoeba     0         3.60        1.30       1.71
2   2  2    amoeba     0         3.60        1.30       1.71
3   3  3    amoeba    NA         3.60        1.30       1.71
4 251  1 bacterium     0         3.82        3.48       2.13
5 252  2 bacterium     0         3.82        3.48       2.13
6 253  3 bacterium     0         3.82        3.48       2.13
7 501  1     leech     1         5.71        1.83       4.50
8 502  2     leech     1         5.71        1.83       4.50
9 503  3     leech     0         5.71        1.83       4.50

如您所见，来自的信息dat1必须在dat2.

这是我失败的尝试：

dat3 <- merge(dat1, dat2, by=intersect(dat1$variable(dat1), dat2$row.names(dat2)))

给出以下错误：

Error in as.vector(y) : attempt to apply non-function

请在此处找到复制示例：

数据1：

structure(list(ID = c(1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L), variable = structure(c(1L, 
1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L), .Label = c("amoeba", "bacterium", 
"leech", "centipede", "lizard", "tapeworm", "head lice", "maggot", 
"ant", "moth", "mosquito", "earthworm", "caterpillar", "scorpion", 
"snail", "spider", "grasshopper", "dust mite", "tarantula", "termite", 
"bat", "wasp", "silkworm"), class = "factor"), value = c(0L, 
0L, NA, 0L, 0L, 0L, 1L, 1L, 0L)), .Names = c("ID", "variable", 
"value"), row.names = c(1L, 2L, 3L, 251L, 252L, 253L, 501L, 502L, 
503L), class = "data.frame")

数据2：

structure(list(concreteness = c(3.6, 3.82, 5.71), familiarity = c(1.3, 
3.48, 1.83), typicality = c(1.71, 2.13, 4.5)), .Names = c("concreteness", 
"familiarity", "typicality"), row.names = c("amoeba", "bacterium", 
"leech"), class = "data.frame")
4

3 回答 3

15

您可以将连接变量添加到 dat2 然后使用合并：

dat2$variable <- rownames(dat2)
merge(dat1, dat2)
   variable ID value concreteness familiarity typicality
1    amoeba  1     0         3.60        1.30       1.71
2    amoeba  2     0         3.60        1.30       1.71
3    amoeba  3    NA         3.60        1.30       1.71
4 bacterium  1     0         3.82        3.48       2.13
5 bacterium  2     0         3.82        3.48       2.13
6 bacterium  3     0         3.82        3.48       2.13
7     leech  1     1         5.71        1.83       4.50
8     leech  2     1         5.71        1.83       4.50
9     leech  3     0         5.71        1.83       4.50
于 2012-12-31T14:12:12.710 回答
12

试试这个：

merge(dat1, dat2, by.x = 2, by.y = 0, all.x = TRUE)

这假设如果有任何dat1不匹配的行，那么dat2结果中的列应该被填充，NA如果有不匹配的值，dat2那么它们被忽略。例如：

dat2a <- dat2
rownames(2a)[3] <- "elephant"
# the above still works:
merge(dat1, dat2a, by.x = 2, by.y = 0, all.x = TRUE)

以上在 SQL 中称为左连接，可以在 sqldf 中像这样完成（忽略警告）：

library(sqldf)
sqldf("select * 
         from dat1 left join dat2 
         on dat1.variable = dat2.row_names", 
       row.names = TRUE)
于 2012-12-31T15:33:42.807 回答
7

@agstudy 的回答没有错，但是您可以通过创建一个匿名临时文件来实际修改 dat2 来做到这一点。添加 X 类似：

> merge(cbind(dat1, X=rownames(dat1)), cbind(dat2, variable=rownames(dat2)))
   variable ID value   X concreteness familiarity typicality
1    amoeba  1     0   1         3.60        1.30       1.71
2    amoeba  2     0   2         3.60        1.30       1.71
3    amoeba  3    NA   3         3.60        1.30       1.71
4 bacterium  1     0 251         3.82        3.48       2.13
5 bacterium  2     0 252         3.82        3.48       2.13
6 bacterium  3     0 253         3.82        3.48       2.13
7     leech  1     1 501         5.71        1.83       4.50
8     leech  2     1 502         5.71        1.83       4.50
9     leech  3     0 503         5.71        1.83       4.50
于 2012-12-31T14:26:10.140 回答