在R中用data.frame制作表格

Question

如何使我的数据框架或表

   TIME_PERIOD MARRIAGE_LICENSES
1      2011-01               742
2      2011-02               796
3      2011-03              1210
4      2011-04              1376
....

看上去像是

      01  02  03   04  05  06  07  08  09  10  11 11
2011 742 796 1210 1376 
2012
2013
2014
2015

诸若此类?

Answer 1

您可以在data.table中这样做，如：

library(data.table)
setDT(dat)
dcast(dat, format(TIME_PERIOD, "%Y") ~ format(TIME_PERIOD, "%m"),
      value.var = "MARRIAGE_LICENSES")

(注意:需要将MARRIAGE_LICENSE存储为具有适当format方法的Date或其他对象)

Answer 2

使用reshape2，您可以执行以下操作(这将为您提供一个data.frame)

require(reshape2)
dat$year <- as.numeric(substr(dat$TIME_PERIOD,1,4))
dat$month <- as.numeric(substr(dat$TIME_PERIOD,6,8))

require(reshape2)
dcast(dat, year~month, value.var = "MARRIAGE_LICENSES")

这给了你

  year   1   2    3    4
1 2011 742 796 1210 1376

如果您想像行名那样使用您的格式和年份：

df <- dcast(dat, year~month, value.var = "MARRIAGE_LICENSES")
rownames(df)  <- df$year
df[,-1]

导致

       1   2    3    4
2011 742 796 1210 1376

Answer 3

你可以做这样的事

library(tidyr) ## for separate()
xtabs(MARRIAGE_LICENSES ~ ., separate(df, TIME_PERIOD, c("year", "month"), "-"))
#       month
# year     01   02   03   04
#   2011  742  796 1210 1376

数据：

df <- structure(list(TIME_PERIOD = structure(1:4, .Label = c("2011-01", 
"2011-02", "2011-03", "2011-04"), class = "factor"), MARRIAGE_LICENSES = c(742L, 
796L, 1210L, 1376L)), .Names = c("TIME_PERIOD", "MARRIAGE_LICENSES"
), class = "data.frame", row.names = c("1", "2", "3", "4"))