如何找到百分位数然后在R中分组

Question

我有一个像下面的数据框（DF）。

day  area   hour  time  count
___  ____  _____  ___   ____
 1    1      0     1     10
 1    1      0     2     12
 1    1      0     3     8
 1    1      0     4     12    
 1    1      0     5     15  
 1    1      0     6     18 
 1    1      1     1     10
 1    1      1     2     12
 1    1      1     3     8
 1    1      1     4     12    
 1    1      1     5     15  
 1    1      1     6     18
 1    1      1     7     12    
 1    1      1     8     15  
 1    1      1     9     18
 1    1      2     1     10    
 1    1      2     2     18  
 1    1      2     3     19
 .....
 2    1      0     1     18
 2    1      0     2     12
 2    1      0     3     18
 2    1      0     4     12    
 2    1      1     1     8
 2    1      1     2     12
 2    1      1     3     18
 2    1      1     4     10    
 2    1      1     5     15  
 2    1      1     6     18
 2    1      1     7     12    
 2    1      1     8     15  
 2    1      1     9     18
 2    1      2     1     10    
 2    1      2     2     18  
 2    1      2     3     19
 2    1      2     4     9    
 2    1      2     5     18  
 2    1      2     6     9


..... 
 30    99      23     1     9    
 30    99      23     2     8  
 30    99      23     3     9
 30    99      23     4     19    
 30    99      23     5     18  
 30    99      23     6     9
 30    99      23     7     19    
 30    99      23     8     8  
 30    99      23     9     19

在这里，我有87个区域（1到82，然后有90,93,95,97,99）和每天24小时（0到23）的30天数据，所以该数据大约是穿越时间该地区以及有多少人穿越。

例如：

day  area   hour  time  count
___  ____  _____  ___   ____
 1    1      0     1     10
 1    1      0     2     12
 1    1      0     3     8
 1    1      0     4     12    
 1    1      0     5     15  
 1    1      0     6     18 

这使我在0小时的第1天穿越区域1所花费的时间

time  count   cumulative_count
___    ___    ________________
 1     10           10
 2     12           22
 3     8            30
 4     12           42    
 5     15           57
 6     18           75 
10 vehicles crossed the area in 1 minute.
12 vehicles crossed the area in 2 minutes.
8 vehicles crossed the area in 3 minutes.
12 vehicles crossed the area in 4 minutes.
15 vehicles crossed the area in 5 minutes.
18 vehicles crossed the area in 6 minutes.

据此，我想计算出80％的车辆在1小时0内穿越区域1花费了多少时间，所以车辆总数=（10 + 12 + 8 + 12 + 15 + 18）= 75，因此80％ 75中的60是60，因此80％的车辆（75中的80％为60）在第1天0时通过区域1所花费的时间将介于5到6之间（将接近5）。 因此结果将是这样的：

 day  area   hour    time_taken_for_80%vehicles_to_pass
    ___  ____   ____    ___________________________________
     1    1      0                5.33(approximately)
     1    1      1                7.30
     1    1      2                2.16
    ....
     30   1      23               3.13
     1    2      0                ---
     1    2      1                ---
     1    2      2                ---
     1    2      3                ---

 .......

     30    99     21              ---
     30    99     22              ---
     30    99     23              ---

   I know to I have to take quantile and then group by the area and day and hour.So I tried with 

library(dplyr)
grp <- group_by(df, day,area,hour,quantile(df$count,0.8))

但这是行不通的，请多多帮助

Answer 1

我的解决方案的计算，交叉领域，各车辆的百分比time 。 然后得到第一time的百分比为80％以上：

str <- 'day  area   hour  time  count
1    1      0     1     10
1    1      0     2     12
1    1      0     3     8
1    1      0     4     12    
1    1      0     5     15  
1    1      0     6     18
1    1      1     1     10
1    1      1     2     12
1    1      1     3     8
1    1      1     4     12    
1    1      1     5     15  
1    1      1     6     18
1    1      1     7     12    
1    1      1     8     15  
1    1      1     9     18
1    1      2     1     10    
1    1      2     2     18  
1    1      2     3     19'



file <- textConnection(str)
df <- read.table(file, header = T)

df

library(dplyr)
df %>% group_by(day, area, hour) %>%
  mutate(cumcount = cumsum(count),
         p = cumcount/max(cumcount)) %>%
  filter(p > 0.8) %>%
  summarise(time = min(time))

结果：

    day  area  hour  time
  <int> <int> <int> <int>
1     1     1     0     6
2     1     1     1     8
3     1     1     2     3

或对达到80％的时间进行线性估计：

df %>% group_by(day, area, hour) %>%
  mutate(cumcount = cumsum(count),
         p = cumcount/max(cumcount),
         g = +(p > 0.8),
         order = (g*2-1)*time) %>%
  group_by(day, area, hour,g) %>%
  filter(row_number((g*2-1)*time)==1) %>%
  group_by(day, area, hour) %>%
  summarise(time = min(time)+(0.8-min(p))/(max(p)-min(p)))

结果：

    day  area  hour     time
  <int> <int> <int>    <dbl>
1     1     1     0 5.166667
2     1     1     1 7.600000
3     1     1     2 2.505263

或使用lag和lead得到相同的结果

df %>% group_by(day, area, hour) %>%
  arrange(hour) %>%
  mutate(cumcount = cumsum(count),
         p = cumcount/max(cumcount)) %>%
  filter((p >= 0.8&lag(p)<0.8)|(p < 0.8&lead(p)>=0.8)) %>%
  summarise(time = min(time)+(0.8-min(p))/(max(p)-min(p)))