Question

我正在处理每月花旗自行车旅行数据，如下所示：

 
> head(data)
  tripduration           starttime            stoptime start.station.id       start.station.name end.station.id 
1          732 2015-07-01 00:00:03 2015-07-01 00:12:16              489         10 Ave & W 28 St            368
2          322 2015-07-01 00:00:06 2015-07-01 00:05:29              304    Broadway & Battery Pl           3002
3          790 2015-07-01 00:00:17 2015-07-01 00:13:28              447          8 Ave & W 52 St            358
4         1228 2015-07-01 00:00:23 2015-07-01 00:20:51              490          8 Ave & W 33 St            250
5         1383 2015-07-01 00:00:44 2015-07-01 00:23:48              327 Vesey Pl & River Terrace             72
6          603 2015-07-01 00:01:00 2015-07-01 00:11:04              455          1 Ave & E 44 St            367
               end.station.name bikeid   usertype birth.year gender
1            Carmine St & 6 Ave  18669 Subscriber       1970      1
2    South End Ave & Liberty St  14618 Subscriber       1984      1
3 Christopher St & Greenwich St  18801 Subscriber       1992      1
4      Lafayette St & Jersey St  19137 Subscriber       1990      1
5              W 52 St & 11 Ave  15808 Subscriber       1988      1
6       E 53 St & Lexington Ave  17069 Subscriber       1953      1

每次旅行都有自己独特的记录，仅7月份就有1,085,676次旅行。可以在此处找到.csv格式的原始数据on the Citi Bike website。通常情况下，自行车从前一次旅行结束的车站开始。然而，这并非总是如此。有时自行车将在不同的车站开始，而不是结束，这表明自行车已经重新平衡了＃34;或者用卡车从一个车站移动到另一个车站以满足车站要求。我想过滤掉所有＆＃34;普通＆＃34;旅行并隔离自行车在不同站点开始的所有实例（例如start.station.id不等于之前的end.station.id。）自行车的唯一识别因素是bikeid ，这必须使用。以下是一个自行车（最常骑的自行车）的月份数据的子集：

head(onebike)
     tripduration           starttime            stoptime start.station.id     start.station.name
1952          691 2015-07-01 07:23:24 2015-07-01 07:34:56              161  LaGuardia Pl & W 3 St
2369          332 2015-07-01 07:38:49 2015-07-01 07:44:22              379        W 31 St & 7 Ave
3879          259 2015-07-01 08:14:34 2015-07-01 08:18:54              472     E 32 St & Park Ave
4310         1112 2015-07-01 08:22:53 2015-07-01 08:41:25              498     Broadway & W 32 St
5795         1509 2015-07-01 08:47:18 2015-07-01 09:12:27              345        W 13 St & 6 Ave
7857         1361 2015-07-01 09:23:50 2015-07-01 09:46:32              348 W Broadway & Spring St
     end.station.id       end.station.name bikeid   usertype birth.year gender
1952            379        W 31 St & 7 Ave  22075 Subscriber       1985      1
2369            472     E 32 St & Park Ave  22075 Subscriber       1986      1
3879            498     Broadway & W 32 St  22075 Subscriber       1986      1
4310            345        W 13 St & 6 Ave  22075   Customer         NA      0
5795            348 W Broadway & Spring St  22075   Customer         NA      0
7857            386   Centre St & Worth St  22075   Customer         NA      0

现在的任务是选择行的start.station.id不等于前一行end.station.id的实例。

结果应包含bikeid，end.station.id，start.station.id以及自行车下降和拾取之间的时差（表示大约何时移动）。

这是使用shift()功能的最佳方法吗？

如何循环遍历第一个数据集中的每个bike.id（大约有7000个）来揭示所有隐藏的动作？

Answer 1

我确信有数百种方法可以做到这一点，但是因为数据只有大约100MB，所以一个简单的for循环非常有能力，并且在这种情况下非常灵活地修改和扩展，所以这里是（在几秒钟内完成））：

raw_data = read.csv("201511-citibike-tripdata.csv")
bikeid <-22075
onebike <- raw_data[ which(raw_data$bikeid== bikeid), ]
output <- data.frame("bikeid"= integer(0), "end.station.id"= integer(0), "start.station.id" = integer(0), "diff.time" = numeric(0))

for(i in 2:nrow(onebike)) {
  if(onebike[i-1,"end.station.id"] != onebike[i,"start.station.id"]){
    diff_time <- as.double(difftime(strptime(onebike[i-1,"stoptime"], "%m/%d/%Y %H:%M:%S"),
                                    strptime(onebike[i,"starttime"], "%m/%d/%Y %H:%M:%S"),units = "mins"))
    new_row <- c(bikeid, onebike[i-1,"end.station.id"], onebike[i,"start.station.id"], diff_time)
    output[nrow(output) + 1,] = new_row
  }
}
output

bikeid end.station.id start.station.id diff.time
1  22075            514              520  181.5667
2  22075            356              502  628.8833

编辑：这是为了进一步回答评论中的问题。这是一个简单的扩展，包括所有的自行车：

raw_data = read.csv("201511-citibike-tripdata.csv")
unique_id = unique(raw_data$bikeid)
#bikeid <-22075

output <- data.frame("bikeid"= integer(0), "end.station.id"= integer(0), "start.station.id" = integer(0), "diff.time" = numeric(0),  "stoptime" = character(),"starttime" = character(), stringsAsFactors=FALSE)

for (bikeid in unique_id)
{
onebike <- raw_data[ which(raw_data$bikeid== bikeid), ]

if(nrow(onebike) >=2 ){
for(i in 2:nrow(onebike )) {
  if(is.integer(onebike[i-1,"end.station.id"]) & is.integer(onebike[i,"start.station.id"]) &
     onebike[i-1,"end.station.id"] != onebike[i,"start.station.id"]){
    diff_time <- as.double(difftime(strptime(onebike[i,"starttime"], "%m/%d/%Y %H:%M:%S"),
                                    strptime(onebike[i-1,"stoptime"], "%m/%d/%Y %H:%M:%S")
                                    ,units = "mins"))
    new_row <- c(bikeid, onebike[i-1,"end.station.id"], onebike[i,"start.station.id"], diff_time, as.character(onebike[i-1,"stoptime"]), as.character(onebike[i,"starttime"]))
    output[nrow(output) + 1,] = new_row
  }
}
}
}
dim(output)
[1] 32589     6
head(output)
  bikeid end.station.id start.station.id        diff.time           stoptime          starttime
1  22545            520              529 24.8166666666667 11/2/2015 08:38:22 11/2/2015 09:03:11
2  22545            520              517 537.483333333333 11/2/2015 09:39:19 11/2/2015 18:36:48
3  22545           2004             3230 563.066666666667 11/2/2015 22:06:27 11/3/2015 07:29:31
4  22545            296             3236 471.783333333333 11/4/2015 23:40:29 11/5/2015 07:32:16
5  22545            520              449 43.4166666666667 11/9/2015 08:24:06 11/9/2015 09:07:31
6  22545            359              519 30.7166666666667 11/9/2015 09:14:46 11/9/2015 09:45:29

使用shift（）函数比较上一行中的数据

1 个答案: