R:计算特定事件在未来指定时间发生的次数 [英] R: calculate the number of occurrences of a specific event in a specified time future

问题描述

我的简化数据如下所示:

my simplified data looks like this:

set.seed(1453); x = sample(0:1, 10, TRUE)
date = c('2016-01-01', '2016-01-05', '2016-01-07',  '2016-01-12',  '2016-01-16',  '2016-01-20',
             '2016-01-20',  '2016-01-25',  '2016-01-26',  '2016-01-31')


df = data.frame(x, date = as.Date(date))


df 
x       date
1 2016-01-01
0 2016-01-05
1 2016-01-07
0 2016-01-12
0 2016-01-16
1 2016-01-20
1 2016-01-20
0 2016-01-25
0 2016-01-26
1 2016-01-31

我想计算指定时间段内 x == 1 的出现次数，例如从当前日期开始的 14 天和 30 天(但不包括当前条目，如果它是 x == 1.所需的输出将如下所示:

I'd like to calculate the number of occurrences for x == 1 within a specified time period, e.g. 14 and 30 days from the current date (but excluding the current entry, if it is x == 1. The desired output would look like this:

solution
x       date x_plus14 x_plus30
1 2016-01-01        1        3
0 2016-01-05        1        4
1 2016-01-07        2        3
0 2016-01-12        2        3
0 2016-01-16        2        3
1 2016-01-20        2        2
1 2016-01-20        1        1
0 2016-01-25        1        1
0 2016-01-26        1        1
1 2016-01-31        0        0

理想情况下，我希望它位于 dplyr 中，但这不是必须的.任何想法如何实现这一目标?非常感谢您的帮助！

Ideally, I'd like this to be in dplyr, but it is not a must. Any ideas how to achieve this? Thanks a lot for your help!

推荐答案

添加另一种基于 findInterval 的方法:

Adding another approach based on findInterval:

cs = cumsum(df$x) # cumulative number of occurences
data.frame(df, 
           plus14 = cs[findInterval(df$date + 14, df$date, left.open = TRUE)] - cs, 
           plus30 = cs[findInterval(df$date + 30, df$date, left.open = TRUE)] - cs)
#   x       date plus14 plus30
#1  1 2016-01-01      1      3
#2  0 2016-01-05      1      4
#3  1 2016-01-07      2      3
#4  0 2016-01-12      2      3
#5  0 2016-01-16      2      3
#6  1 2016-01-20      2      2
#7  1 2016-01-20      1      1
#8  0 2016-01-25      1      1
#9  0 2016-01-26      1      1
#10 1 2016-01-31      0      0

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