我是R的新手。我有以下数据集超过500万行，我需要重塑。我需要为每个perc_full
hour个station.id的平均值
此时我精通的是每小时和每个小时的子集，这需要很长时间。有没有办法加快这个过程？
dim(data)
[1] 5116857 12
head(data, n = 10)
id station_id status available_bike_count available_dock_count created_at
1 21141047 1 Active 12 23 2014-10-01 00:00:05
2 21141048 2 Active 1 32 2014-10-01 00:00:05
3 21141049 3 Active 8 17 2014-10-01 00:00:05
4 21141050 4 Active 23 39 2014-10-01 00:00:05
5 21141051 5 Active 6 31 2014-10-01 00:00:05
6 21141052 6 Active 5 14 2014-10-01 00:00:05
7 21141053 7 Active 2 17 2014-10-01 00:00:05
8 21141054 8 Active 20 8 2014-10-01 00:00:05
9 21141055 9 Active 3 27 2014-10-01 00:00:05
10 21141056 10 Active 0 45 2014-10-01 00:00:05
station_summary_id month year hour tot_docks perc_full
1 64087 10 2014 0 35 0.34285714
2 64087 10 2014 0 33 0.03030303
3 64087 10 2014 0 25 0.32000000
4 64087 10 2014 0 62 0.37096774
5 64087 10 2014 0 37 0.16216216
6 64087 10 2014 0 19 0.26315789
7 64087 10 2014 0 19 0.10526316
8 64087 10 2014 0 28 0.71428571
9 64087 10 2014 0 30 0.10000000
10 64087 10 2014 0 45 0.00000000
最后，我应该有25个列的结果 - 每个hour 24个，station.id个
output
id 1 2 3 4 5 6 7 8 9
1 1 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
2 2 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
3 3 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
4 4 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
5 5 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
6 6 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
7 7 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
8 8 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
9 9 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
10 10 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
10 11 12 13 14 15 16 17 18
1 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
2 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
3 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
4 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
5 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
6 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
7 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
8 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
9 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
10 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
19 20 21 22 23 24
1 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
2 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
3 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
4 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
5 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
6 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
7 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
8 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
9 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
10 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362 0.5554362
sapply(data, class)
$id
[1] "integer"
$station_id
[1] "integer"
$status
[1] "factor"
$available_bike_count
[1] "integer"
$available_dock_count
[1] "integer"
$created_at
[1] "POSIXlt" "POSIXt"
$station_summary_id
[1] "integer"
$month
[1] "integer"
$year
[1] "integer"
$hour
[1] "integer"
$tot_docks
[1] "integer"
$perc_full
[1] "numeric"
这是第二个数据集，我想要完全相同的矩阵，只有这一次通过求和每小时start.station.id的数量
> head(test, n = 10) bikeid end.station.id start.station.id diff.time hour 1 16052 244 322 6544 14 2 16052 284 432 3406 21 3 16052 461 519 33416 3 4 16052 228 519 26876 13 5 16052 72 435 388 17 6 16052 319 127 27702 11 7 16052 282 2002 33882 4 8 16052 524 2021 2525 10 9 16052 387 351 2397 12 10 16052 388 526 32507 13我应该使用这样的东西吗？
matrix <- test %>%
group_by(start.station.id, hour)%>%
summarise(sum = nrow) %>%
spread(hour, nrow)
答案 0 :(得分：2)
试试这个：
String findProcess = "chrome.exe";
String filenameFilter = "/nh /fi \"Imagename eq "+findProcess+"\"";
String tasksCmd = System.getenv("windir") +"/system32/tasklist.exe "+filenameFilter;
Process p = Runtime.getRuntime().exec(tasksCmd);
BufferedReader input = new BufferedReader(new InputStreamReader(p.getInputStream()));
ArrayList<String> procs = new ArrayList<String>();
String line = null;
while ((line = input.readLine()) != null)
procs.add(line);
input.close();
Boolean processFound = procs.stream().filter(row -> row.indexOf(findProcess) > -1).count() > 0;
// Head-up! If no processes were found - we still get:
// "INFO: No tasks are running which match the specified criteria."