我有一个表格，其中包含来自流量计的数据：
Water.Year May Jun Jul Aug Sep Oct Nov Dec Jan Feb Mar Apr
1 1953-1954 55.55 43.62 30.46 26.17 26.76 41.74 19.92 41.25 28.77 20.96 12.47 10.51
2 1954-1955 23.49 81.35 46.71 29.33 67.83 133.30 37.62 30.16 21.07 19.38 13.87 10.63
3 1955-1956 9.87 51.59 55.36 63.03 154.08 98.15 104.06 32.85 22.89 17.30 15.68 10.88
> data <- structure(list(Water.Year = structure(1:6, .Label = c("1953-1954", "1954-1955", "1955-1956", "1956-1957", "1957-1958", "1958-1959", "1959-1960", "1960-1961", "1961-1962", "1962-1963", "1963-1964", "1964-1965", "1965-1966", "1966-1967", "1967-1968", "1968-1969", "1969-1970", "1970-1971", "1971-1972", "1972-1973", "1973-1974", "1974-1975", "1975-1976", "1976-1977", "1977-1978", "1978-1979", "1979-1980", "1980-1981", "1981-1982", "1982-1983", "1983-1984", "1984-1985", "1985-1986", "1986-1987", "1987-1988", "1988-1989", "1989-1990", "1990-1991", "1991-1992", "1992-1993", "1993-1994", "1994-1995", "1995-1996", "1996-1997", "1997-1998", "1998-1999", "1999-2000", "2000-2001"), class = "factor"), May = c(55.55, 23.49, 9.87, 18.03, 17.46, 11.37), Jun = c(43.62, 81.35, 51.59, 28.61, 15.14, 29.48), Jul = c(30.46, 46.71, 55.36, 24.36, 20.09, 19.48), Ago = c(26.17, 29.33, 63.03, 22.01, 16.97, 16.86), Set = c(26.76, 67.83, 154.08, 28.51, 27.24, 21.01), Oct = c(41.74, 133.3, 98.15, 53.72, 35.78, 19.78), Nov = c(19.92, 37.62, 104.06, 115.78, 20.35, 18.69), Dic = c(41.25, 30.16, 32.85, 32.04, 22, 18.86), Ene = c(28.77, 21.07, 22.89, 25.44, 13.27, 14.89), Feb = c(20.96, 19.38, 17.3, 14.53, 10.37, 10.4), Mar = c(12.47, 13.87, 15.68, 10.78, 8.77, 8.79), Abr = c(10.51, 10.63, 10.88, 9.33, 7.69, 8.99)), .Names = c("Water.Year", "May", "Jun", "Jul", "Ago", "Set", "Oct", "Nov", "Dic", "Ene", "Feb", "Mar", "Abr"), row.names = c(NA, 6L), class = "data.frame")
按照“水年”安排，每年从5月开始到明年4月结束（这可以在第1栏看到）。 我想将其转换为包含三列的数据框： Calendar.Year - 月 - Flow.Measurement
我已经将 Water.Year 列分解为使用“分离”来自tidyr的两列：
> df = separate(data, Water.Year, c("year1","year2"))
year1 year2 May Jun Jul Aug Sep Oct Nov Dec Jan Feb Mar Apr
1 1953 1954 55.55 43.62 30.46 26.17 26.76 41.74 19.92 41.25 28.77 20.96 12.47 10.51
2 1954 1955 23.49 81.35 46.71 29.33 67.83 133.30 37.62 30.16 21.07 19.38 13.87 10.63
现在我打算使用tidyr的“聚集”来完成转换的其余部分，但我仍然坚持如何使用 year1 <创建 Calendar.Year 列/ em>列可以到 Dec 和 year2 Jan 到 Apr 。
任何帮助将不胜感激。
答案 0 :(得分：3)
另一个想法（使用带有英语月份的@useR数据）
library(dplyr)
library(tidyr)
df %>%
separate(Water.Year, c("Year1", "Year2")) %>%
gather(Month, Value, -(Year1:Year2)) %>%
group_by(Year1, Year2) %>%
mutate(Year = if_else(match(Month, month.abb) >= 5, Year1, Year2),
Month = factor(Month, levels = month.abb)) %>%
ungroup() %>%
select(Year, Month, Value) %>%
arrange(Year, Month)
我们将Water.Year列分为Year1和Year2，然后使用gather()将数据重新整形为长格式。然后，对于每个组，我们使用match()和month.abb来检查月份是否大于或等于5（5月），并将相应的年份指定为if_else()。最后，我们删除了arrange()和Year
Month
## A tibble: 36 × 3
# Year Month Value
# <chr> <fctr> <dbl>
#1 1953 May 55.55
#2 1953 Jun 43.62
#3 1953 Jul 30.46
#4 1953 Aug 26.17
#5 1953 Sep 26.76
#6 1953 Oct 41.74
#7 1953 Nov 19.92
#8 1953 Dec 41.25
#9 1954 Jan 28.77
#10 1954 Feb 20.96
## ... with 26 more rows
答案 1 :(得分：1)
好的，这个怎么样。它是重塑和基础R之间的混搭。
发布后我使用了您的数据集。谢谢你提供它。
data <- structure(list(Water.Year = structure(1:6, .Label = c("1953-1954", "1954-1955", "1955-1956", "1956-1957", "1957-1958", "1958-1959", "1959-1960", "1960-1961", "1961-1962", "1962-1963", "1963-1964", "1964-1965", "1965-1966", "1966-1967", "1967-1968", "1968-1969", "1969-1970", "1970-1971", "1971-1972", "1972-1973", "1973-1974", "1974-1975", "1975-1976", "1976-1977", "1977-1978", "1978-1979", "1979-1980", "1980-1981", "1981-1982", "1982-1983", "1983-1984", "1984-1985", "1985-1986", "1986-1987", "1987-1988", "1988-1989", "1989-1990", "1990-1991", "1991-1992", "1992-1993", "1993-1994", "1994-1995", "1995-1996", "1996-1997", "1997-1998", "1998-1999", "1999-2000", "2000-2001"), class = "factor"), May = c(55.55, 23.49, 9.87, 18.03, 17.46, 11.37), Jun = c(43.62, 81.35, 51.59, 28.61, 15.14, 29.48), Jul = c(30.46, 46.71, 55.36, 24.36, 20.09, 19.48), Ago = c(26.17, 29.33, 63.03, 22.01, 16.97, 16.86), Set = c(26.76, 67.83, 154.08, 28.51, 27.24, 21.01), Oct = c(41.74, 133.3, 98.15, 53.72, 35.78, 19.78), Nov = c(19.92, 37.62, 104.06, 115.78, 20.35, 18.69), Dic = c(41.25, 30.16, 32.85, 32.04, 22, 18.86), Ene = c(28.77, 21.07, 22.89, 25.44, 13.27, 14.89), Feb = c(20.96, 19.38, 17.3, 14.53, 10.37, 10.4), Mar = c(12.47, 13.87, 15.68, 10.78, 8.77, 8.79), Abr = c(10.51, 10.63, 10.88, 9.33, 7.69, 8.99)), .Names = c("Water.Year", "May", "Jun", "Jul", "Ago", "Set", "Oct", "Nov", "Dic", "Ene", "Feb", "Mar", "Abr"), row.names = c(NA, 6L), class = "data.frame")
我决定使用您之前的年份信息，并在此基础上添加日历年。因为我们知道5月到12月是1年级，1月到4月是2年级。也许有点复杂但是它完成了工作。
df = separate(data, Water.Year, c("year1","year2"))
library(reshape2)
fixDF<-melt(df)
fixDF$CalendarYear<-rep(NA,nrow(fixDF))
fixDF$CalendarYear[min(which(fixDF$variable=="May")):max(which(fixDF$variable=="Dic"))]<-df$year1
fixDF$CalendarYear[min(which(fixDF$variable=="Ene")):max(which(fixDF$variable=="Abr"))]<-df$year2
fixDF<-fixDF[,3:5]
colnames(fixDF)<-c("Month","Flow.Measurement", "Calendar.Year")
答案 2 :(得分：1)
好的，我刚刚意识到您在structure()中提供的月份可能使用不同的语言。我将坚持使用我创建的数据，它使用英文版的Months。这样人们就可以用英语看到相应的解决方案。
library(tidyr) # for separate function
library(reshape2) # for melt function
library(dplyr) # for pipe operator and arrange function
# Reproducible Data
weather = structure(list(Water.Year = structure(1:3, .Label = c("1953-1954",
"1954-1955", "1955-1956"), class = "factor"),
May = c(55.55, 23.49, 9.87),
Jun = c(43.62, 81.35, 51.59),
Jul = c(30.46, 46.71, 55.36),
Aug = c(26.17, 29.33, 63.03),
Sep = c(26.76, 67.83, 154.08),
Oct = c(41.74, 133.3, 98.15),
Nov = c(19.92, 37.62, 104.06),
Dec = c(41.25, 30.16, 32.85),
Jan = c(28.77, 21.07, 22.89),
Feb = c(20.96, 19.38, 17.3), Mar = c(12.47, 13.87, 15.68),
Apr = c(10.51, 10.63, 10.88)), .Names = c("Water.Year", "May",
"Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec", "Jan", "Feb",
"Mar", "Apr"), class = "data.frame", row.names = c(NA, -3L))
编码从这里开始：
df = separate(weather, Water.Year, c("year1","year2"))
# Split into two datasets
columns1 = c("year1", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Dec")
df1 = subset(df, select = c(year1, May:Dec))
df2 = subset(df, select = c(year2, Jan:Apr))
longdf1 = melt(df1, variable.name = "Month", id.vars = "year1",
value.name = "Flow.Measurement")
names(longdf1)[1] = "Calendar.Year"
longdf2 = melt(df2, variable.name = "Month", id.vars = "year2",
value.name = "Flow.Measurement")
names(longdf2)[1] = "Calendar.Year"
# Combine the two datasets
final_df = rbind(longdf1, longdf2)
# Releveling the Month
final_df$Month = factor(final_df$Month, levels = month.abb)
final_df = arrange(final_df, Calendar.Year, Month)
最终数据框：
> final_df
Calendar.Year Month Flow.Measurement
1 1953 May 55.55
2 1953 Jun 43.62
3 1953 Jul 30.46
4 1953 Aug 26.17
5 1953 Sep 26.76
6 1953 Oct 41.74
7 1953 Nov 19.92
8 1953 Dec 41.25
9 1954 Jan 28.77
10 1954 Feb 20.96
11 1954 Mar 12.47
12 1954 Apr 10.51
13 1954 May 23.49
14 1954 Jun 81.35
15 1954 Jul 46.71
16 1954 Aug 29.33
17 1954 Sep 67.83
18 1954 Oct 133.30
19 1954 Nov 37.62
20 1954 Dec 30.16
21 1955 Jan 21.07
22 1955 Feb 19.38
23 1955 Mar 13.87
24 1955 Apr 10.63
25 1955 May 9.87
26 1955 Jun 51.59
27 1955 Jul 55.36
28 1955 Aug 63.03
29 1955 Sep 154.08
30 1955 Oct 98.15
31 1955 Nov 104.06
32 1955 Dec 32.85
33 1956 Jan 22.89
34 1956 Feb 17.30
35 1956 Mar 15.68
36 1956 Apr 10.88
答案 3 :(得分：0)
我决定使用我得到的所有答案的一些部分。 这是我写的代码：
library(dplyr)
library(tidyr)
#separate the year column into two years
df_years <- df %>%
separate(Water.Year, c("Year1", "Year2"))
#create two different dataframes for each section of the year
df1 <- subset(df_years, select = c(Year1, May:Dec))
df2 <- subset(df_years, select = c(Year2, Jan:Apr))
#rename both years' columns using the same name
colnames(df2)[1] <- "Year"
colnames(df1)[1] <- "Year"
#join both dataframes
cleandata <- full_join(df1, df2, by = "Year")
#sort months chronologically
cleandata <- cleandata[, c("Year", "Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec")]
#convert to tidy data set
cleandata <- gather(cleandata, "Month", "Flow", 2:13)
#sort by year and month
cleandata <- arrange(cleandata, Year, Month)