拥有data.table如下:
station w_1 w_2
1: 1757 ar_2d lm_h_step
2: 2171 lm_h_step lm_h_step
3: 2812 lm_h_step lm_h_step
4: 4501 lm_h_step lm_h_step
5: 4642 ar_2d lm_h_step
6: 5029 ar_2d lm_h_step
7: 5480 lm_h_step lm_h_step
8: 5779 ar_2d ar_2d
9: 5792 ar_1d ar_2d
我想列出每站方法的频率.
所以预期的结果是
1757 2171 2812 ...
lm_h_step 1 2 2
ar_2d 1 0 0
ar_1d 0 0 0 ...
到目前为止我尝试了什么:
apply(dat,1,table)
正在产生正确的结果,但它没有正确地形成.
有任何想法吗？
数据输入:
structure(list(station = c(1757L, 2171L, 2812L, 4501L, 4642L,
5029L, 5480L, 5779L, 5792L), w_1 = c("ar_2d", "lm_h_step", "lm_h_step",
"lm_h_step", "ar_2d", "ar_2d", "lm_h_step", "ar_2d", "ar_2d"),
w_2 = c("lm_h_step", "lm_h_step", "lm_h_step", "lm_h_step",
"lm_h_step", "lm_h_step", "lm_h_step", "ar_2d", "ar_2d")), .Names = c("station",
"w_1", "w_2"), class = c("data.table", "data.frame"), row.names = c(NA,
-9L))
尝试dcast/melt组合
对于data.table v> = 1.9.5,请使用此选项
dcast(melt(dat, "station"), value ~ station, length)
# value 1757 2171 2812 4501 4642 5029 5480 5779 5792
# 1: ar_1d 0 0 0 0 0 0 0 0 1
# 2: ar_2d 1 0 0 0 1 1 0 2 1
# 3: lm_h_step 1 2 2 2 1 1 2 0 0
对于data.table v <1.9.5,您还需要加载reshape2并显式使用dcast.data.table(因为reshape2::dcast它不是通用的,没有dcast.data.table方法).
reshape2::melt另一方面,它是通用的(参见methods(melt))并且有一个melt.data.table方法,所以你不需要告诉它任何东西.它会知道你要取决于使用哪种方法class的dat
require(reshape2)
dcast.data.table(melt(dat, "station"), value ~ station, length)
# value 1757 2171 2812 4501 4642 5029 5480 5779 5792
# 1: ar_1d 0 0 0 0 0 0 0 0 1
# 2: ar_2d 1 0 0 0 1 1 0 2 1
# 3: lm_h_step 1 2 2 2 1 1 2 0 0
如果你没有严格使用data.table方法挑剔,你也可以使用reshape2::recast(参见@shadows注释)这是上面的解决方案的包装,但使用reshape2::dcast而不是,dcast.data.table因此将返回一个data.frame对象,而不是data.table
recast(dat, value ~ station, id.var = "station", length)
# value 1757 2171 2812 4501 4642 5029 5480 5779 5792
# 1 ar_1d 0 0 0 0 0 0 0 0 1
# 2 ar_2d 1 0 0 0 1 1 0 2 1
# 3 lm_h_step 1 2 2 2 1 1 2 0 0