将年月字符串转换为带有间隔的三个月垃圾箱

本文介绍了将年月字符串转换为带有间隔的三个月垃圾箱-如何分配连续的升序值？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我已使用下面的代码将year.month字符串 bin到三个月的bin中。问题是我希望每个垃圾箱都有一个数字，该数字对应于垃圾箱按时间顺序出现的位置（即第一个垃圾箱= 1，第二垃圾箱= 2，依此类推）。现在，第一个月的垃圾箱已分配给数字4，但我不确定为什么。任何帮助将不胜感激！

I have used the code below to "bin" a year.month string into three month bins. The problem is that I want each of the bins to have a number that corresponds where the bin occurs chronologically (i.e. first bin =1, second bin=2, etc.). Right now, the first month bin is assigned to the number 4, and I am not sure why. Any help would be highly appreciated!

> head(Master.feed.parts.gn$yr.mo, n=20) [1] "2007.10" "2007.10" "2007.10" "2007.11" "2007.11" "2007.11" "2007.11" "2007.12" "2008.01" [10] "2008.01" "2008.01" "2008.01" "2008.01" "2008.02" "2008.03" "2008.03" "2008.03" "2008.04" [19] "2008.04" "2008.04" > > yearmonth_to_integer <- function(xx) { + yy_mm <- as.integer(unlist(strsplit(xx, '.', fixed=T))) + return( (yy_mm[1] - 2006) + (yy_mm[2] %/% 3) ) + } > > Cluster.GN <- sapply(Master.feed.parts.gn$yr.mo, yearmonth_to_integer) > Cluster.GN 2007.10 2007.10 2007.10 2007.11 2007.11 2007.11 2007.11 2007.12 2008.01 2008.01 2008.01 4 4 4 4 4 4 4 5 2 2 2 2008.01 2008.01 2008.02 2008.03 2008.03 2008.03 2008.04 2008.04 2008.04 2008.04 2008.05 2 2 2 3 3 3 3 3 3 3 3 2008.05 2008.05 2008.06 2008.10 2008.11 2008.11 2008.12 <NA> 2009.05 2009.05 2009.05 3 3 4 5 5 5 6 NA 4 4 4 2009.06 2009.07 2009.07 2009.07 2009.09 2009.10 2009.11 2010.01 2010.02 2010.02 2010.02 5 5 5 5 6 6 6 4 4 4 4

更新：

我被要求提供示例输入（年）和所需的输出（Cluster.GN）。我有一个年月字符串，每个月的观察值数量不同，有些月份没有任何观察值。我想做的是对具有数据的连续三个月中的每个月进行bin，为每个三个月 bin分配一个数字，如下所示。

I was asked to provide sample input (year) and the desired output (Cluster.GN).I have a year-month string that has varying numbers of observations for each month, and some months don't have any observations. What I want to do is bin each of the three consecutive months that have data, assigning each three month "bin" a number as shown below.

yr.mo Cluster.GN 1 2007.10 1 2 2007.10 1 3 2007.10 1 4 2007.10 1 5 2007.10 1 6 2007.11 1 7 2007.11 1 8 2007.11 1 9 2007.11 1 10 2007.12 1 11 2007.12 1 12 2007.12 1 13 2007.12 1 14 2008.10 2 15 2008.10 2 16 2008.10 2 17 2008.10 2 18 2008.12 2 19 2008.12 2 20 2008.12 2 21 2008.12 2 22 2008.12 2

推荐答案

1）将字符串转换为动物园的 yearqtr 类，然后转换为整数：

1) Convert the strings to zoo's "yearqtr" class and then to integers:

s <- c("2007.10", "2007.10", "2007.10", "2007.11", "2007.11", "2007.11", "2007.11", "2007.12", "2008.01", "2008.01", "2008.01", "2008.01", "2008.01", "2008.02", "2008.03", "2008.03", "2008.03", "2008.04", "2008.04", "2008.04") library(zoo) yq <- as.yearqtr(s, "%Y.%m") as.numeric(factor(yq)) ## [1] 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 3 3 3

最后一行可能是： 4 *（yq-yq [1]）+ 1

请注意，问题2007.12与2007.10和2007.11位于不同的季度;但是，它们都在同一季度，因此我们假设您没有打算这样做。

Note that in the question 2007.12 is classified as in a different quarter than 2007.10 and 2007.11; however, they are all in the same quarter and we assume you did not intend this.

2） ：

f <- factor(s) nlev <- nlevels(f) levels(f) <- gl(nlev, 3, nlev) f ## [1] 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 3 3 3 ## Levels: 1 2 3

如果缺少月份，那么给出的答案将不同于（1）因此，一切都取决于您要寻找的东西。

IF there are missing months then this will give a different answer than (1) so it all depends on what you are looking for.