重新排列数据：从水年转换为日历年
rearrange data: convert from water year to calendar year
我有一个 table，流量计的数据排列如下：
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"water years"排列，每年5月开始，次年4月结束（第一栏可见）。
我想将其转换为包含三列的数据框：Calendar.Year -- Month -- Flow.Measurement
我已经使用 tidyr 中的 "separate" 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 "gather" 来完成其余的转换，但我仍然不知道如何创建 Calendar.Year 列使用 year1 用于列 May 到 Dec 和 year2 1 月 至 4 月。
任何帮助将不胜感激。
好的，这个怎么样。它是 reshape 和 base R 的混搭。
你一发布我就用了你的数据集。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，1 月至 4 月是第 2 年。可能有点复杂，但它完成了工作。
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$year2
fixDF<-fixDF[,3:5]
colnames(fixDF)<-c("Month","Flow.Measurement", "Calendar.Year")
好的，我刚刚意识到您在 structure() 中提供的月份可能使用不同的语言。我将只使用我创建的数据，它使用英文版的 Months。这样人们就可以看到相应的英文解决方案了。
library(tidyr) # for separate function
library(reshape2) # for melt function
library(dplyr) # for pipe operator and arrange function
# Reproducible Data
weather = structure(list(Water.Year = structure(1:3, .Label = c("1953-1954",
"1954-1955", "1955-1956"), class = "factor"),
May = c(55.55, 23.49, 9.87),
Jun = c(43.62, 81.35, 51.59),
Jul = c(30.46, 46.71, 55.36),
Aug = c(26.17, 29.33, 63.03),
Sep = c(26.76, 67.83, 154.08),
Oct = c(41.74, 133.3, 98.15),
Nov = c(19.92, 37.62, 104.06),
Dec = c(41.25, 30.16, 32.85),
Jan = c(28.77, 21.07, 22.89),
Feb = c(20.96, 19.38, 17.3), Mar = c(12.47, 13.87, 15.68),
Apr = c(10.51, 10.63, 10.88)), .Names = c("Water.Year", "May",
"Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec", "Jan", "Feb",
"Mar", "Apr"), class = "data.frame", row.names = c(NA, -3L))
编码从这里开始：
df = separate(weather, Water.Year, c("year1","year2"))
# Split into two datasets
columns1 = c("year1", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Dec")
df1 = subset(df, select = c(year1, May:Dec))
df2 = subset(df, select = c(year2, Jan:Apr))
longdf1 = melt(df1, variable.name = "Month", id.vars = "year1",
value.name = "Flow.Measurement")
names(longdf1)[1] = "Calendar.Year"
longdf2 = melt(df2, variable.name = "Month", id.vars = "year2",
value.name = "Flow.Measurement")
names(longdf2)[1] = "Calendar.Year"
# Combine the two datasets
final_df = rbind(longdf1, longdf2)
# Releveling the Month
final_df$Month = factor(final_df$Month, levels = month.abb)
final_df = arrange(final_df, Calendar.Year, Month)
最终数据帧：
> 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
另一个想法（使用带有英文月份的@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 列分成 Year1 和 Year2，并使用 gather() 将数据重塑为长格式。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
我决定使用我得到的所有答案中的一些片段。
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)
我有一个 table，流量计的数据排列如下：
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"water years"排列，每年5月开始，次年4月结束（第一栏可见）。 我想将其转换为包含三列的数据框：Calendar.Year -- Month -- Flow.Measurement
我已经使用 tidyr 中的 "separate" 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 "gather" 来完成其余的转换，但我仍然不知道如何创建 Calendar.Year 列使用 year1 用于列 May 到 Dec 和 year2 1 月 至 4 月。
任何帮助将不胜感激。
好的，这个怎么样。它是 reshape 和 base R 的混搭。
你一发布我就用了你的数据集。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，1 月至 4 月是第 2 年。可能有点复杂，但它完成了工作。
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")
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。我将只使用我创建的数据，它使用英文版的 Months。这样人们就可以看到相应的英文解决方案了。
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)
最终数据帧：
> 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
另一个想法（使用带有英文月份的@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 列分成 Year1 和 Year2，并使用 gather() 将数据重塑为长格式。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），并用 if_else() 分配相应的年份。最后，我们通过 Year 和 Month
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
我决定使用我得到的所有答案中的一些片段。 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)