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如何计算特定时间段内的事件数量

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我试图在“df1”定义的时间段内计算“df2”中的事件数（每行是一个事件） . 我能够在大约5分钟的整个时间段内完成此操作，但我想将时间段分解为更小的块（1分钟）并进行相同的计算

df1<- structure(list(Location = 1:10, Lattitude = c(57.140532, 57.140527, 
57.13959, 57.13974, 57.14059, 57.14058, 57.1398, 57.13989, 57.14158, 
57.14386), t_in = structure(c(1455626730, 1455627326, 1455628122, 
1455628644, 1455629174, 1455629708, 1455630230, 1455630765, 1455631396, 
1455631931), class = c("POSIXct", "POSIXt"), tzone = ""), t_out = structure(c(1455627047, 
1455627615, 1455628462, 1455628933, 1455629486, 1455630015, 1455630552, 
1455631070, 1455631719, 1455632242), class = c("POSIXct", "POSIXt"
), tzone = "")), .Names = c("Location", "Lattitude", "t_in", 
"t_out"), class = "data.frame", row.names = c(NA, -10L))

df2<- structure(list(date.time = structure(c(1455630964, 1455630976, 
1455630987, 1455630998, 1455631009, 1455631021, 1455631032, 1455631043, 
1455631054, 1455631066, 1455631077, 1455631088, 1455631099, 1455631111, 
1455631423, 1455631446, 1455631479, 1455631502, 1455631569, 1455631772
), class = c("POSIXct", "POSIXt"), tzone = ""), code = structure(c(2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L), .Label = c("1003", "32221"), class = "factor"), 
rec_id = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("301976", 
"301978", "301985", "301988"), class = "factor"), Lattitude = c("57.14066", 
"57.14066", "57.14066", "57.14066", "57.14066", "57.14066", 
"57.14066", "57.14066", "57.14066", "57.14066", "57.14066", 
"57.14066", "57.14066", "57.14066", "57.141869", "57.141869", 
"57.141869", "57.141869", "57.141869", "57.141869"), Longitude = c("2.075702", 
"2.075702", "2.075702", "2.075702", "2.075702", "2.075702", 
"2.075702", "2.075702", "2.075702", "2.075702", "2.075702", 
"2.075702", "2.075702", "2.075702", "2.081576", "2.081576", 
"2.081576", "2.081576", "2.081576", "2.081576"), Location = list(
    8, 8, 8, 8, 8, 8, 8, 8, 8, 8, NA, NA, NA, NA, 9, 9, 9, 
    9, 9, NA)), .Names = c("date.time", "code", "rec_id", 
"Lattitude", "Longitude", "Location"), row.names = 94:113, class = "data.frame")

如果df2中的date.time位于df1 $ t_in和df1 $ t_out之间，函数将从df1返回位置 . 这可能看起来很圆，但可以使用此代码进行后续计算

ids <- as.numeric(df1$Location)
f <- function(x){
  a <- ids[ (df1$t_in < x) & (x < df1$t_out) ]
  if (length(a) == 0) NA else a
}   

df2$Location <- lapply(df2$date.time, f)

以上返回一个列表，因此需要将其转换为数字 . 有点偏僻，但不能绕过它

df2$Location<- paste(df2$Location)
df2$Location<- as.numeric(df2$Location)

然后移除NA，因为它们位于df1中定义的时间段之外，因此无关紧要 .

df2<-df2[!is.na(df2$Location),]

然后计算每个rec_id和Location的事件数（即每行）

library (plyr)
df3 <- ddply(df2, c("rec_id","Location"), function(df){data.frame (detections=nrow(df))})

  rec_id Location detections
1 301976        9          5
2 301978        8         10

...完善！

但是我想在较短的时间内这样做 . 每分钟都是准确的 . 并且周期应该从每个位置的t_in（df1）开始直到t_out（df1） . 我可以在excel中做很多工作，但肯定可以在R中自动化（它是一个大型数据集） .

所以最终我可以计算df1中t_in和t_out之间每1分钟时间段内每个位置的事件数（nrow）

例如（仅视觉示例而非实际数据）：

rec_id Location  minute(or period) detections
 301976        9             1           1
 301976        9             2           2
 301976        9             3           0
 301976        9             4           0
 301976        9             5           2
 301978        8             1           4
 301978        8             2           3
 301978        8             3           1
 301978        8             4           0
 301978        8             5           2

我可以从第一个位置创建间隔，但我不知道如何进一步应用它

seq(from = head(df1$t_in,1), to = head(df1$t_out,1) , by = "mins")
r datetime plyr seq

1 回答

  • 1

    我认为以下内容可用于生成具有序列分割输出的新 df1 数据帧，然后您可以使用新的 df1 应用上面的步骤 .

    他们可能会合并，但我只是想确保它实际上能满足你的需求 .

    首先，我们扩展原始数据框中的时间间隔，并生成扩展期间的列表 . df1 中的每一行都成为列表中的一个元素 .

    res1 <- sapply(1:nrow(df1), function(i) {
                 seq(from = df1$t_in[i], to = df1$t_out[i] , by = "mins")})

    然后我们将序列列表转换为数据帧（两列）

    res2 <- lapply(res1, function(x) { 
                 data.frame(t_in = x[1:(length(x)-1)], t_out=x[2:length(x)]) })

    最后我们将所有内容合并在一起

    df1v2 <- Reduce(function(...) merge(..., all=T), res2)

    然后（调整你的代码）

    ids <- seq_len(nrow(df1v2))
f <- function(x){
  a <- ids[ (df1v2$t_in < x) & (x < df1v2$t_out) ]
  if (length(a) == 0) NA else a
}   

df2$Location <- lapply(df2$date.time, f)

    产量

    date.time  code rec_id Lattitude Longitude Location
94  2016-02-16 14:56:04 32221 301978  57.14066  2.075702       37
95  2016-02-16 14:56:16 32221 301978  57.14066  2.075702       37
96  2016-02-16 14:56:27 32221 301978  57.14066  2.075702       37
97  2016-02-16 14:56:38 32221 301978  57.14066  2.075702       37
98  2016-02-16 14:56:49 32221 301978  57.14066  2.075702       38
99  2016-02-16 14:57:01 32221 301978  57.14066  2.075702       38
100 2016-02-16 14:57:12 32221 301978  57.14066  2.075702       38
101 2016-02-16 14:57:23 32221 301978  57.14066  2.075702       38
102 2016-02-16 14:57:34 32221 301978  57.14066  2.075702       38
103 2016-02-16 14:57:46 32221 301978  57.14066  2.075702       NA
104 2016-02-16 14:57:57 32221 301978  57.14066  2.075702       NA
105 2016-02-16 14:58:08 32221 301978  57.14066  2.075702       NA
106 2016-02-16 14:58:19 32221 301978  57.14066  2.075702       NA
107 2016-02-16 14:58:31 32221 301978  57.14066  2.075702       NA
108 2016-02-16 15:03:43 32221 301976 57.141869  2.081576       39
109 2016-02-16 15:04:06 32221 301976 57.141869  2.081576       39
110 2016-02-16 15:04:39 32221 301976 57.141869  2.081576       40
111 2016-02-16 15:05:02 32221 301976 57.141869  2.081576       40
112 2016-02-16 15:06:09 32221 301976 57.141869  2.081576       41
113 2016-02-16 15:09:32 32221 301976 57.141869  2.081576       NA

    我不确定边界检查是否正确（修改 f ），但看起来好像你得到了你所追求的 . 加速有多重要？

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