Question

我有这个数据集，并尝试转换为日期，以便我可以绘制它：

dput(y)
structure(c(23L, 17L, 41L, 1L, 47L, 35L, 29L, 7L, 63L, 58L, 53L, 
12L, 24L, 18L, 42L, 2L, 48L, 36L, 30L, 8L, 64L, 59L, 54L, 13L, 
25L, 19L, 43L, 3L, 49L, 37L, 31L, 9L, 65L, 60L, 55L, 14L, 26L, 
20L, 44L, 4L, 50L, 38L, 32L, 10L, 66L, 61L, 56L, 15L, 27L, 21L, 
45L, 5L, 51L, 39L, 33L, 11L, 67L, 62L, 57L, 16L, 28L, 22L, 46L, 
6L, 52L, 40L, 34L), .Label = c("APR-07", "APR-08", "APR-09", 
"APR-10", "APR-11", "APR-12", "AUG-07", "AUG-08", "AUG-09", "AUG-10", 
"AUG-11", "DEC-07", "DEC-08", "DEC-09", "DEC-10", "DEC-11", "FEB-07", 
"FEB-08", "FEB-09", "FEB-10", "FEB-11", "FEB-12", "JAN-07", "JAN-08", 
"JAN-09", "JAN-10", "JAN-11", "JAN-12", "JUL-07", "JUL-08", "JUL-09", 
"JUL-10", "JUL-11", "JUL-12", "JUN-07", "JUN-08", "JUN-09", "JUN-10", 
"JUN-11", "JUN-12", "MAR-07", "MAR-08", "MAR-09", "MAR-10", "MAR-11", 
"MAR-12", "MAY-07", "MAY-08", "MAY-09", "MAY-10", "May-11", "MAY-12", 
"NOV-07", "NOV-08", "NOV-09", "NOV-10", "NOV-11", "OCT-07", "OCT-08", 
"OCT-09", "OCT-10", "OCT-11", "SEP-07", "SEP-08", "SEP-09", "SEP-10", 
"SEP-11"), class = "factor")

我试过

 y1 <- strptime(y, format = "%b-%y")

y1获得所有NA，任何想法？

Answer 1

您需要一天时间将其转换为日期：

##Make each date the first of the month
strptime(paste("1", as.character(y)), format="%d %b-%y")

Answer 2

如果你将以regalr为基础在月份范围内处理时间序列，你可能还想看看动物园包，它作为'yearmon'类，不需要添加“月初值” 。它有许多相关的绘图和计算方法。广泛使用的'xts'包基于'动物园'结构。

 require(zoo)
 zym <- as.yearmon(as.character(y), "%b-%y")
 str(zym)
#Class 'yearmon'  num [1:67] 2007 2007 2007 2007 2007 ...
 head(zym)
#[1] "Jan 2007" "Feb 2007" "Mar 2007" "Apr 2007" "May 2007" "Jun 2007"
 format(head(zym), "%b %y")
#[1] "Jan 07" "Feb 07" "Mar 07" "Apr 07" "May 07" "Jun 07"

在R中将因子转换为日期

2 个答案: