使用R将数据帧中的列重新形成几列 [英] Reshaping a column from a data frame into several columns using R

问题描述

我有一个数据库，如下所示：

I have a database that looks like this:

start<-as.POSIXct("2012-01-15")
interval<-60
end<-start+as.difftime(31,units="days")
date<-seq(from=start,by=interval*60, to=end) # date/time information
l<-length(date)

stations<-as.factor(rep(1:3,len=l)) # stations
df<-data.frame(date,stations) # data frame

我想要将站点列从这个数据框架重新形成几列（在这个例子中将是3列），并计算每个站点在每个日期/时间排。但是，我想保留数据库中的原始日期/时间列。如果一个电台没有记录在一个特定的日期/时间，那么我想分配一个值为零。

What I would like is to reshape the station column from this data frame into several columns (in this example it will be 3 columns) and calculate the number of time each station was recorded in each date/time row. However, I would like to keep the original date/time column from the data base. If a station was not recorded in one specific date/time, then I want to assign a value of zero.

理想情况下，我想要一个这样的输出： p>

Ideally, I would like an output like this:

date              1   2   3
2012-01-15 0:00   1   0   0
2012-01-15 1:00   0   1   0
2012-01-15 2:00   0   0   1
2012-01-15 3:00   1   0   0
2012-01-15 4:00   0   1   0
2012-01-15 5:00   0   0   1
2012-01-15 6:00   1   0   0
2012-01-15 7:00   0   1   0
2012-01-15 8:00   0   0   1
2012-01-15 9:00   1   0   0
2012-01-15 10:00  0   1   0

推荐答案

您可以尝试使用函数 dcast（）从库 reshape2 。

You can try to use function dcast() from library reshape2.

library(reshape2)
dcast(df,date~stations,length)
                   date 1 2 3
1   2012-01-15 00:00:00 1 0 0
2   2012-01-15 01:00:00 0 1 0
3   2012-01-15 02:00:00 0 0 1
4   2012-01-15 03:00:00 1 0 0

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