我一直试图将年份作为以下示例数据集的新列。
dta<-structure(list(time = structure(c(1L, 3L, 5L, 7L, 9L, 11L, 13L,
15L, 17L, 19L, 21L, 22L, 2L, 4L, 6L, 8L, 10L, 12L, 14L, 16L,
18L, 20L), .Label = c("1/01/2000", "1/01/2001", "1/02/2000",
"1/02/2001", "1/03/2000", "1/03/2001", "1/04/2000", "1/04/2001",
"1/05/2000", "1/05/2001", "1/06/2000", "1/06/2001", "1/07/2000",
"1/07/2001", "1/08/2000", "1/08/2001", "1/09/2000", "1/09/2001",
"1/10/2000", "1/10/2001", "1/11/2000", "1/12/2000"), class = "factor"),
site = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "A", class = "factor")), .Names = c("time",
"site"), class = "data.frame", row.names = c(NA, -22L))
可以使用以下dput访问数据集。
library(zoo)
dta$y_m = as.yearmon(as.Date(dta$time,"%m/%d/%Y"))
1] "Jan 2000" "Jan 2000" "Jan 2000" "Jan 2000" "Jan 2000" "Jan 2000" "Jan 2000"
[8] "Jan 2000" "Jan 2000" "Jan 2000" "Jan 2000" "Jan 2000" "Jan 2001" "Jan 2001"
[15] "Jan 2001" "Jan 2001" "Jan 2001" "Jan 2001" "Jan 2001" "Jan 2001" "Jan 2001"
[22] "Jan 2001"
我尝试使用以下代码并返回所有12个月的同一个月。
.grouped-link {
cursor: pointer;
}
.slideout-link {
span.show:hover {
color: $darkgrey;
}
}
.section-subtitle a, a.section-subtitle {
color: $darkgrey;
float: right;
text-transform: uppercase;
font-size: 16px;
font-weight: 400;
letter-spacing: 2px;
margin-top: 65px;
text-decoration: none;
border-bottom: 1px solid $darkgrey;
&:hover {
border-color: $midgrey;
color: $midgrey;
text-decoration: none;
}
}
.underline-link {
border-bottom: 1px solid $lightgrey;
cursor: pointer;
padding-bottom: 5px;
text-transform: uppercase;
width: 150px;
&.dark {
border-color: $grey;
color: $grey;
}
&:hover {
border-color: $midgrey;
color: $midgrey;
text-decoration: none;
}
&.light {
border-color: $upmidgrey;
color: $upmidgrey;
}
&.lightest {
color: $lightgrey;
}
}
.inline-link {
color: $grey;
cursor: text;
&:hover {
text-decoration: none;
cursor: text;
}
}
a[target="_blank"]:not(.inline-link):hover {
text-decoration: none;
cursor: pointer;
}
任何人都可以帮助我,让它正确。谢谢
答案 0 :(得分:4)
试试这个:
require(zoo)
as.yearmon(dta$time, format="%d/%m/%Y")
输出:
[1] "Jan 2000" "Feb 2000" "Mar 2000" "Apr 2000" "May 2000" "Jun 2000"
[7] "Jul 2000" "Aug 2000" "Sep 2000" "Oct 2000" "Nov 2000" "Dec 2000"
[13] "Jan 2001" "Feb 2001" "Mar 2001" "Apr 2001" "May 2001" "Jun 2001"
[19] "Jul 2001" "Aug 2001" "Sep 2001" "Oct 2001"