R-分割数据到水文小区
R-分割数据到水文小区
提问于 2015-08-30 04:52:46
我希望根据水文年的定义将我的数据集划分为年份季度。维基百科说，“由于气象和地理因素，水年的定义各不相同”。在美国，水文年是从一年的10月1日到次年的9月30日之间的一个时期。我使用了波兰水文年的定义(11月1日开始，10月31日结束)。
样本数据集看起来是折叠的：
sampleData <- structure(list(date = structure(c(15946, 15947, 15875, 15910, 15869, 15888, 15823, 16059, 16068, 16067), class = "Date"),`example value` = c(-0.325806595888448, 0.116001346459147, 1.68884381116696, -0.480527505762716, -0.50307381813168,-1.12032214801472, -0.659699514672226, -0.547101497279717, 0.729148872679021,-0.769760735764215)), .Names = c("date", "example value"), row.names = c(NA, -10L), class = "data.frame")由于某些原因，函数“剪切”在我的代码中抱怨“打断”和“标签”的长度不同(但它们没有)。如果我省略了剪切函数中的“标签”选项(如下所示)，则效果非常好。标签有什么问题？
ToHydroQuarters <-function(df)
{
result <- df
yearStart <- as.numeric(format(min(df$date),'%Y'))-1
#Hydrological year in Poland starts at November 1st
DateStart <- as.Date(paste(yearStart,"-11-01",sep=""))
breaks <- seq(from=DateStart, to=max(df$date)+90, by="quarter")
breakYear <- format(breaks,'%Y')
#Please, do not create labels in such way.
#Please note that for November and December we have next hydrological year - since it started at 1st November. So, we need to check month to decide which year we have (?) or use cut function again as mentioned here: http://stackoverflow.com/questions/22073881/hydrological-year-time-series
labels <- c(paste("Winter",breakYear[1]),
paste("Spring",breakYear[2]),
paste("Summer",breakYear[3]),
paste("Autumn",breakYear[4]),
paste("Autumn",breakYear[5]))
######Here is problem - once I add labels parameter, function complains about different lengths
result$hydroYear <- cut(df$date, breaks)
result
}回答 1
Stack Overflow用户
回答已采纳
发布于 2015-08-30 06:06:24
首先，我认为在函数中使用标签作为“硬编码”变量是不明智的，因为没有某种可重复的例子是不可能检查的，但是我可以看到您试图实现的目标。
您声称您的中断和标签应该是正确的长度，但是函数本身并不总是工作的(这没有标签，即使标签确实存在，cut函数也没有处理日期的最后一部分)。
例如：
library(lubridate)
x <- ymd(c("09-01-01", "09-01-02", "11-09-03"))
df <- data.frame(date=as.Date(seq(from=min(x), to=max(x), by="day")))
a <- ToHydroQuarters(df)
tail(a)返回：
date hydroYear
971 2011-08-29 <NA>
972 2011-08-30 <NA>
973 2011-08-31 <NA>
974 2011-09-01 <NA>
975 2011-09-02 <NA>
976 2011-09-03 <NA>做像breaks <- seq(from=DateStart, to=max(df$date)+90, by="quarter")这样的事情确实解决了这个问题，因为它迫使中断实际上存在。这可能解决您的标签问题，您在您的功能，但它没有使功能“通用”。
就个人而言，在编码方面，我认为更好的做法是分别转换月份和年份部分，因为这将更容易理解。例如，您可以使用library(lubridate)轻松提取月份，并像通常那样指定中断和标签。我在想这个函数应该是这样的：
thq <- function(date) {
mnth <- cut(month(date), breaks=c(1,4,7, 10, 12),
right=FALSE, include.lowest=TRUE,
labels=c("Spring", "Summer", "Autumn", "Winter"))
return(paste(mnth, ifelse(mnth == "Winter", year(date)+1, year(date))))
}所以用一些假数据..。
library(lubridate)
x <- ymd(c("09-01-01", "09-01-02", "11-09-03"))
df <- data.frame(date=as.Date(seq(from=min(x), to=max(x), by="month")))
thq <- function(date) {
mnth <- cut(month(date), breaks=c(1,4,7, 10, 12),
right=FALSE, include.lowest=TRUE,
labels=c("Spring", "Summer", "Autumn", "Winter"))
return(paste(mnth, ifelse(mnth == "Winter", year(date)+1, year(date))))
}
df$newdate <- thq(df$date)它的输出如下：
date newdate
1 2009-01-01 Spring 2009
2 2009-02-01 Spring 2009
3 2009-03-01 Spring 2009
4 2009-04-01 Summer 2009
5 2009-05-01 Summer 2009
6 2009-06-01 Summer 2009
7 2009-07-01 Autumn 2009
8 2009-08-01 Autumn 2009
9 2009-09-01 Autumn 2009
10 2009-10-01 Winter 2010
11 2009-11-01 Winter 2010
12 2009-12-01 Winter 2010
13 2010-01-01 Spring 2010
14 2010-02-01 Spring 2010
15 2010-03-01 Spring 2010
16 2010-04-01 Summer 2010
17 2010-05-01 Summer 2010
18 2010-06-01 Summer 2010
19 2010-07-01 Autumn 2010
20 2010-08-01 Autumn 2010
21 2010-09-01 Autumn 2010
22 2010-10-01 Winter 2011
23 2010-11-01 Winter 2011
24 2010-12-01 Winter 2011
25 2011-01-01 Spring 2011
26 2011-02-01 Spring 2011
27 2011-03-01 Spring 2011
28 2011-04-01 Summer 2011
29 2011-05-01 Summer 2011
30 2011-06-01 Summer 2011
31 2011-07-01 Autumn 2011
32 2011-08-01 Autumn 2011
33 2011-09-01 Autumn 2011你可以用模操作符换个月如果它的顺序很奇怪的话.
thq <- function(date) {
mnth <- cut(((month(df$date)+1) %% 12), breaks=c(0, 3, 6, 9, 12),
right=FALSE, include.lowest=TRUE,
labels=c("Nov_Jan", "Feb_Apr", "May_Jul", "Aug_Oct")
)
# you will need to alter the return statement yourself, because
# I feel there is enough information for you to do it, rather than
# me changing it every time you change the question.
return(paste(mnth, ifelse(mnth == "Winter", year(date)+1, year(date))))
}
library(lubridate)
x <- ymd(c("09-01-01", "09-01-02", "11-09-03"))
df <- data.frame(date=as.Date(seq(from=min(x), to=max(x), by="day")))
df$new <- thq(df$date)
head(df)产出：
> head(df)
date new
1 2009-01-01 Nov_Jan 2009
2 2009-01-02 Nov_Jan 2009
3 2009-01-03 Nov_Jan 2009
4 2009-01-04 Nov_Jan 2009
5 2009-01-05 Nov_Jan 2009
6 2009-01-06 Nov_Jan 2009页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/32293285
复制相关文章
相似问题
腾讯云开发者
