5

我有以下宽数据框（mydf.wide）：

DAY JAN F1  FEB F2  MAR F3  APR F4  MAY F5  JUN F6  JUL F7  AUG F8  SEP F9  OCT F10 NOV F11 DEC F12
1   169 0   296 0   1095    0   599 0   1361    0   1746    0   2411    0   2516    0   1614    0   908 0   488 0   209 0
2   193 0   554 0   1085    0   1820    0   1723    0   2787    0   2548    0   1402    0   1633    0   897 0   411 0   250 0
3   246 0   533 0   1111    0   1817    0   2238    0   2747    0   1575    0   1912    0   705 0   813 0   156 0   164 0
4   222 0   547 0   1125    0   1789    0   2181    0   2309    0   1569    0   1798    0   1463    0   878 0   241 0   230 0

我想制作以下“半长”：

DAY variable_month value_month value_F
1 JAN 169 0

我试过：

library(reshape2)
mydf.long <- melt(mydf.wide, id.vars=c("YEAR","DAY"), measure.vars=c("JAN","FEB","MAR","APR","MAY","JUN","JUL","AUG","SEP","OCT","NOV","DEC"))

但这跳过了 F 变量，我不知道如何处理两个变量......

4

2 回答 2

6

这是reshape(...)在基础 R 中是更好选择的情况之一。

months    <- c(2,4,6,8,10,12,14,16,18,20,22,24)   # column numbers of months
F         <- c(3,5,7,9,11,13,15,17,19,21,23,25)   # column numbers of Fn
mydf.long <- reshape(mydf.wide,idvar=1,
             times=colnames(mydf.wide)[months],
             varying=list(months,F),
             v.names=c("value_month","value_F"),
             direction="long")
colnames(mydf.long)[2] <- "variable_month"
head(mydf.long)
#       DAY variable_month value_month value_F
# 1.JAN   1            JAN         169       0
# 2.JAN   2            JAN         193       0
# 3.JAN   3            JAN         246       0
# 4.JAN   4            JAN         222       0
# 1.FEB   1            FEB         296       0
# 2.FEB   2            FEB         554       0

您也可以通过 2 次调用来执行此操作melt(...)

library(reshape2)
months    <- c(2,4,6,8,10,12,14,16,18,20,22,24)   # column numbers of months
F         <- c(3,5,7,9,11,13,15,17,19,21,23,25)   # column numbers of Fn
z.1 <- melt(mydf.wide,id=1,measure=months,
            variable.name="variable_month",value.name="value_month")
z.2 <- melt(mydf.wide,id=1,measure=F,value.name="value_F")
mydf.long <- cbind(z.1,value_F=z.2$value_F)
head(mydf.long)
#   DAY variable_month value_month z.2$value_F
# 1   1            JAN         169           0
# 2   2            JAN         193           0
# 3   3            JAN         246           0
# 4   4            JAN         222           0
# 5   1            FEB         296           0
# 6   2            FEB         554           0
于 2014-02-27T18:28:30.170 回答
2

melt()并且dcast()可以从reshape2和data.table包中获得。最新版本data.table允许melt同时使用多个列。该patterns()参数可用于通过正则表达式指定两组列：

library(data.table)   # CRAN version 1.10.4 used
regex_month <- toupper(paste(month.abb, collapse = "|"))
mydf.long <- melt(setDT(mydf.wide), measure.vars = patterns(regex_month, "F\\d"),
                  value.name = c("MONTH", "F"))
# rename factor levels
mydf.long[, variable := forcats::lvls_revalue(variable, toupper(month.abb))][]

    DAY variable MONTH F
 1:   1      JAN   169 0
 2:   2      JAN   193 0
 3:   3      JAN   246 0
 4:   4      JAN   222 0
 5:   1      FEB   296 0
...
44:   4      NOV   241 0
45:   1      DEC   209 0
46:   2      DEC   250 0
47:   3      DEC   164 0
48:   4      DEC   230 0
    DAY variable MONTH F

请注意，"F\\d"在 中用作正则表达式patterns()。一个简单"F"的函数会捕获FEB以及F1,F2等产生意想不到的结果。

还要注意mydf.wide需要被强制到一个data.table对象。否则，reshape2::melt()将在无法识别的 data.frame 对象上调度patterns()。

数据

library(data.table)
mydf.wide <- fread(
"DAY JAN F1  FEB F2  MAR F3  APR F4  MAY F5  JUN F6  JUL F7  AUG F8  SEP F9  OCT F10 NOV F11 DEC F12
  1   169 0   296 0   1095    0   599 0   1361    0   1746    0   2411    0   2516    0   1614    0   908 0   488 0   209 0
  2   193 0   554 0   1085    0   1820    0   1723    0   2787    0   2548    0   1402    0   1633    0   897 0   411 0   250 0
  3   246 0   533 0   1111    0   1817    0   2238    0   2747    0   1575    0   1912    0   705 0   813 0   156 0   164 0
  4   222 0   547 0   1125    0   1789    0   2181    0   2309    0   1569    0   1798    0   1463    0   878 0   241 0   230 0",
data.table = FALSE)
于 2017-07-04T05:53:45.767 回答