如何从累积的每日数据中获取每小时的差异

I have data in two columns:

我有两个栏目的数据:

DateTime    Profit
20130319T01  5
20130319T02  135
20130319T03  245
20130320T01  10
20130320T02  115

and I want to create a column that has the diff for each hour BUT the problem is that Profit resets to zero each day. I want to get the following:

我想创建一个每小时都有差异的列，但问题是利润每天都是零。我想要得到以下信息:

DateTime     Diff
20130319T01  5
20130319T02  130
20130319T03  110
20130320T01  10
20130320T02  105

1 个解决方案

#1

Assuming the format of your DateTime character vector is always "YYYYMMDD" then you can use the ddply function from plyr to get what you want:

假设您的DateTime字符向量的格式总是“YYYYMMDD”，那么您可以使用plyr中的ddply函数来获得您想要的:

require(plyr)
df$Date <- substr( df$DateTime , 1 , 8 )
ddply( df ,  .(Date)  , summarise , Diff = diff(c(0,Profit)) )
#     Date Diff
#1 20130319    5
#2 20130319  130
#3 20130319  110
#4 20130320   10
#5 20130320  105

Another way using base's ave:

使用base's ave的另一种方式是:

within(df, { Profit_diff <- ave(Profit, list(gsub("T.*$", "", DateTime)), 
             FUN=function(x) c(x[1], diff(x)))})

#      DateTime Profit Profit_diff
# 1 20130319T01      5           5
# 2 20130319T02    135         130
# 3 20130319T03    245         110
# 4 20130320T01     10          10
# 5 20130320T02    115         105

#1