我有宽格式的数据,它有两组不同的值列:包含质量(Mass1,Mass2等)的那些和包含相应日期的那些(Mass1_date,Mass2_date等)
library(tidyr)
library(dplyr)
library(lubridate)
df <- structure(list(Year = 2004, Nest_no = 21, Mass1 = 2325, Mass1_date = structure(1081987200, class = c("POSIXct",
"POSIXt"), tzone = "UTC"), Mass2 = 2000, Mass2_date = structure(1082851200, class = c("POSIXct",
"POSIXt"), tzone = "UTC"), Mass3 = 1750, Mass3_date = structure(1083715200, class = c("POSIXct",
"POSIXt"), tzone = "UTC")), class = c("tbl_df", "tbl", "data.frame"
), row.names = c(NA, -1L), .Names = c("Year", "Nest_no", "Mass1",
"Mass1_date", "Mass2", "Mass2_date", "Mass3", "Mass3_date"))
df
## Source: local data frame [1 x 8]
##
## Year Nest_no Mass1 Mass1_date Mass2 Mass2_date Mass3 Mass3_date
## (dbl) (dbl) (dbl) (time) (dbl) (time) (dbl) (time)
## 1 2004 21 2325 2004-04-15 2000 2004-04-25 1750 2004-05-05
我想将数据“整理”为长格式,其中两组值列gather
ed melt
ed}到两个不同的值列,一列包含'质量列'的值,另一列包含'日期列'的值:
## Source: local data frame [3 x 5]
##
## Year Nest_no capture date weight
## (dbl) (dbl) (dbl) (date) (dbl)
## 1 2004 21 1 2004-04-15 2325
## 2 2004 21 2 2004-04-25 2000
## 3 2004 21 3 2004-05-05 1750
起初,我以为我可以使用tidyr
并分两步完成。
gather(df, capture, date, contains("Date")) %>%
gather(capture2, weight, contains("Mass"))
## Source: local data frame [9 x 6]
##
## Year Nest_no capture date capture2 weight
## (dbl) (dbl) (chr) (time) (chr) (dbl)
## 1 2004 21 Mass1_date 2004-04-15 Mass1 2325
## 2 2004 21 Mass2_date 2004-04-25 Mass1 2325
## 3 2004 21 Mass3_date 2004-05-05 Mass1 2325
## 4 2004 21 Mass1_date 2004-04-15 Mass2 2000
## 5 2004 21 Mass2_date 2004-04-25 Mass2 2000
## 6 2004 21 Mass3_date 2004-05-05 Mass2 2000
## 7 2004 21 Mass1_date 2004-04-15 Mass3 1750
## 8 2004 21 Mass2_date 2004-04-25 Mass3 1750
## 9 2004 21 Mass3_date 2004-05-05 Mass3 1750
但是,它没有按预期工作。经过几次尝试,我走了过来 有了这个解决方案:
df <- gather(df, capture2, weight, contains("Mass"), convert = T) %>%
mutate(capture = extract_numeric(capture2))
## Warning: attributes are not identical across measure variables; they will
## be dropped
df$capture2 <- ifelse(grepl("date", df$capture2), "date", "weight")
df <- spread(df, capture2, weight) %>%
mutate(date = as.Date(as.POSIXct(date, origin = "1970-01-01")))
df
## Source: local data frame [3 x 5]
##
## Year Nest_no capture date weight
## (dbl) (dbl) (dbl) (date) (dbl)
## 1 2004 21 1 2004-04-15 2325
## 2 2004 21 2 2004-04-25 2000
## 3 2004 21 3 2004-05-05 1750
我想知道是否有更好的方式来实现这个目标?
谢谢Philippe
答案 0 :(得分:4)
我们可以使用melt
中的data.table
轻松完成此操作。 measure
可以使用多个patterns
列名称并转换广泛的&#39;长期&#39;格式。
library(data.table)
melt(as.data.table(df), measure=patterns('\\d$', 'date$'),
variable.name='capture', value.name= c('weight', 'date'))
# Year Nest_no capture weight date
#1: 2004 21 1 2325 2004-04-15
#2: 2004 21 2 2000 2004-04-25
#3: 2004 21 3 1750 2004-05-05