Question

我想计算一列，该列计算特定ID在预定时间间隔（例如2天）内向后看的出现次数。我在R中有以下数据结构（参见下面的代码），并希望自动计算列countLast2d：

userID <- c(1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,3,3,3)

datetime <-c("2015-07-02 13:20:00", "2015-07-03 13:20:00", "2015-07-04 01:20:00", 
"2015-07-10 01:20:00", "2015-07-23 01:20:00", "2015-07-23 06:08:00", "2015-07-24 06:08:00", 
"2015-09-02 09:01:00", "2015-08-19 11:41:00", "2015-08-19 14:38:00", "2015-08-19 17:36:00", 
"2015-08-19 20:33:00", "2015-08-19 23:30:00", "2015-08-19 23:46:00", "2015-08-19 05:19:00", 
"2015-09-13 17:02:00", "2015-10-01 00:32:00", "2015-10-01 00:50:00")

结果应该采用这些价值观：

countLast2d <- c(0,1,2,0,0,1,2,0,0,1,0,0,0,1,0,0,0,1)

df <- data.frame(userID, countLast2d, datetime)
df$datetime = as.POSIXct(strptime(df$datetime, format = "%Y-%m-%d %H:%M:%S"))

在Excel中，我会使用以下公式：

=countifs([datecolumn],"<"&[date cell in that row],[datecolumn],"<"&[date cell in that row]-2,[idcolumn],[id cell in that row])

（例如[C2]=+COUNTIFS($B:$B,"<"&$B2,$B:$B,">="&$B2-2,$A:$A,$A2)，如果列A包含id，列B包含日期）

我之前已经问了一次这个问题（https://stackoverflow.com/questions/30998596/r-count-number-of-occurences-by-id-in-the-last-48h），但在我的问题中没有包含一个例子。很抱歉再次询问。

Answer 1

这是一个解决方案：

df <- data.frame(userID=c(1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,3,3,3),datetime=as.POSIXct(c('2015-07-02 13:20:00','2015-07-03 13:20:00','2015-07-04 01:20:00','2015-07-10 01:20:00','2015-07-23 01:20:00','2015-07-23 06:08:00','2015-07-24 06:08:00','2015-09-02 09:01:00','2015-08-19 11:41:00','2015-08-19 14:38:00','2015-08-19 17:36:00','2015-08-19 20:33:00','2015-08-19 23:30:00','2015-08-19 23:46:00','2015-08-19 05:19:00','2015-09-13 17:02:00','2015-10-01 00:32:00','2015-10-01 00:50:00')));
window <- as.difftime(2,units='days');
df$countLast2d <- sapply(1:nrow(df),function(r) sum(df$userID==df$userID[r] & df$datetime<df$datetime[r] & df$datetime>=df$datetime[r]-window));
df;
##    userID            datetime countLast2d
## 1       1 2015-07-02 13:20:00           0
## 2       1 2015-07-03 13:20:00           1
## 3       1 2015-07-04 01:20:00           2
## 4       1 2015-07-10 01:20:00           0
## 5       1 2015-07-23 01:20:00           0
## 6       1 2015-07-23 06:08:00           1
## 7       1 2015-07-24 06:08:00           2
## 8       1 2015-09-02 09:01:00           0
## 9       2 2015-08-19 11:41:00           1
## 10      2 2015-08-19 14:38:00           2
## 11      2 2015-08-19 17:36:00           3
## 12      2 2015-08-19 20:33:00           4
## 13      2 2015-08-19 23:30:00           5
## 14      2 2015-08-19 23:46:00           6
## 15      2 2015-08-19 05:19:00           0
## 16      3 2015-09-13 17:02:00           0
## 17      3 2015-10-01 00:32:00           0
## 18      3 2015-10-01 00:50:00           1

请注意，这与您预期的输出不同，因为userID==2的预期输出不正确。

无论df的排序如何，此解决方案都会有效，这对于您的示例df至关重要，因为它userID==2无序（或至少没有完美排序）。

修改这里有可能，使用by()按userID分组并仅将每个元素与较小索引元素进行比较，假设只有那些元素可以在回顾窗口中：

df2 <- df[order(df$userID,df$datetime),];
df2$countLast2d <- do.call(c,by(df2$datetime,df$userID,function(x) c(0,sapply(2:length(x),function(i) sum(x[1:(i-1)]>=x[i]-window)))));
df2;
##    userID            datetime countLast2d
## 1       1 2015-07-02 13:20:00           0
## 2       1 2015-07-03 13:20:00           1
## 3       1 2015-07-04 01:20:00           2
## 4       1 2015-07-10 01:20:00           0
## 5       1 2015-07-23 01:20:00           0
## 6       1 2015-07-23 06:08:00           1
## 7       1 2015-07-24 06:08:00           2
## 8       1 2015-09-02 09:01:00           0
## 15      2 2015-08-19 05:19:00           0
## 9       2 2015-08-19 11:41:00           1
## 10      2 2015-08-19 14:38:00           2
## 11      2 2015-08-19 17:36:00           3
## 12      2 2015-08-19 20:33:00           4
## 13      2 2015-08-19 23:30:00           5
## 14      2 2015-08-19 23:46:00           6
## 16      3 2015-09-13 17:02:00           0
## 17      3 2015-10-01 00:32:00           0
## 18      3 2015-10-01 00:50:00           1

R：按id计算，在预定义的时间间隔内出现的次数

1 个答案: