重新排列数据：从水年转换为日历年答案

【问题标题】：rearrange data: convert from water year to calendar year重新排列数据：从水年转换为日历年
【发布时间】：2017-03-05 17:56:46
【问题描述】：

我有一张表，其中包含来自流规的数据，排列如下：

  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“水年”排列，每年5月开始，次年4月结束（第一栏可见）。我想将其转换为具有三列的数据框：Calendar.Year -- Month -- Flow.Measurement

我已经使用 tidyr 的“分离”将 Water.Year 列分解为两列：

> 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 中的“收集”来完成其余的转换，但我不知道如何使用 year1Calendar.Year 列/em> 表示 May 到 Dec 列，year2 表示 Jan 到 Apr。

任何帮助将不胜感激。

【问题讨论】：

请使用dput(data) 发布可重现的问题。如果不手动输入数据，就无法复制您提供的数据。
library(reshape2); melt(df)

标签： r date time-series tidyr

【解决方案1】：

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，并使用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（五月），并使用if_else() 分配相应的年份。最后，我们通过Year 和Month 删除不必要的列和arrange()

## 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

【讨论】：

使用dplyr 的绝佳解决方案！但是你的最终数据的排序有点偏离，因为你的“1953 Dec”没有跟在“1954 Jan”之后。看看我的解决方案。
@user 哦，对了！添加了factor() 部分。谢谢！

【解决方案2】：

好吧，这个怎么样。这是 reshape 和 base R 之间的混搭。

您发布后我使用了您的数据集。感谢提供。

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")

我决定使用您之前获得的年份信息，并在此基础上添加日历年。因为我们知道 5 月至 12 月是第 1 年，而 1 月至 4 月是第 2 年。可能有点复杂，但它可以完成工作。

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")

【讨论】：

【解决方案3】：

好的，我刚刚意识到您在structure() 中提供的月份可能是不同的语言。我将坚持使用我创建的数据，它使用英文版的 Months。这样大家就可以看到对应的英文解法了。

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

【讨论】：

【解决方案4】：

我决定使用我得到的所有答案中的一部分。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)

【讨论】：