可以在没有聚合函数的情况下使用 dcast 吗？ [复制]

可能的重复：
这个 R 重塑应该很简单，但是 https://stackoverflow.com/questions/12829995/this-r-reshaping-should-be-simple-but

dcast from reshape2无需公式即可工作，不存在重复项。以这些示例数据为例：

df <- structure(list(id = c("A", "B", "C", "A", "B", "C"), cat = c("SS", 
"SS", "SS", "SV", "SV", "SV"), val = c(220L, 222L, 223L, 224L, 
225L, 2206L)), .Names = c("id", "cat", "val"), class = "data.frame", row.names = c(NA, 
-6L))

我想dcast这些数据只是将值制成表格，而不对value.var包括默认的length.

在这种情况下，效果很好。

> dcast(df, id~cat, value.var="val")
  id  SS   SV
1  A 220  224
2  B 222  225
3  C 223 2206

但当存在重复变量时，fun默认为length。有办法避免吗？

df2 <- structure(list(id = c("A", "B", "C", "A", "B", "C", "C"), cat = c("SS", 
"SS", "SS", "SV", "SV", "SV", "SV"), val = c(220L, 222L, 223L, 
224L, 225L, 220L, 1L)), .Names = c("id", "cat", "val"), class = "data.frame", row.names = c(NA, 
-7L))

> dcast(df2, id~cat, value.var="val")
Aggregation function missing: defaulting to length
  id SS SV
1  A  1  1
2  B  1  1
3  C  1  2

理想情况下，我正在寻找的是添加一个fun = NA，就像不要尝试聚合value.var。 dcasting df2 时我想要的结果：

 id  SS  SV
1  A 220 224
2  B 222 225
3  C 223 220
4. C NA  1

我认为没有办法直接做到这一点，但我们可以添加一个额外的列，这将帮助我们

df2 <- structure(list(id = c("A", "B", "C", "A", "B", "C", "C"), cat = c("SS", 
"SS", "SS", "SV", "SV", "SV", "SV"), val = c(220L, 222L, 223L, 
224L, 225L, 220L, 1L)), .Names = c("id", "cat", "val"), class = "data.frame", row.names = c(NA, 
-7L))

library(reshape2)
library(plyr)
# Add a variable for how many times the id*cat combination has occured
tmp <- ddply(df2, .(id, cat), transform, newid = paste(id, seq_along(cat)))
# Aggregate using this newid and toss in the id so we don't lose it
out <- dcast(tmp, id + newid ~ cat, value.var = "val")
# Remove newid if we want
out <- out[,-which(colnames(out) == "newid")]
> out
#  id  SS  SV
#1  A 220 224
#2  B 222 225
#3  C 223 220
#4  C  NA   1