来自
dcast
的 reshape2
没有没有重复的公式。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,只是将值列表,而不对value.var
应用任何功能,包括默认length
。
在这种情况下,它可以正常工作。
> dcast(df, id~cat, value.var="val")
id SS SV
1 A 220 224
2 B 222 225
3 C 223 2206
但是当存在重复变量时,fun
默认为length
。有没有办法避免它?
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fun = NA
,因为不要尝试聚合value.var
。当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
答案 0 :(得分:19)
我认为没有办法直接进行,但我们可以添加一个额外的列来帮助我们
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
答案 1 :(得分:9)
当Dason回答我的时候,我想出了同样的解决方案。
我意识到dcast
根本不知道如何处理重复项。我弄清楚如何欺骗它的方式是添加另一个独特的标识符,这样它就不会被重复项混淆。
在这个例子中:
df <- ddply(df2, .(cat), function(x){ x$id2 = 1:nrow(x); x})
> dcast(df, id+id2~cat, value.var="val")[,-2]
id SS SV
1 A 220 224
2 B 222 225
3 C 223 220
4 C NA 1