数据

Question

我试图获得总数＆amp;基于日期的一些变量的平均值。我有燃料数据和驾驶几辆车的数据。燃料数据由几个日期组成，而驾驶数据包含更多日期（逻辑上，您只能在x次旅行后加油）。我的最终结果是根据燃油日期获得总和/平均驾驶数据。

燃料数据：

plate = c("AB123", "AB123", "AB123", "AB123", "AC234", "AC234", "AC234", "AC234", "AD345", "AD345") 
date = c("2017-09-08", "2017-09-11", "2017-09-13", "2017-09-20", "2017-09-06", "2017-09-08", "2017-09-15", "2017-09-23", "2017-09-10", "2017-09-18")
liter = c(33, 15, 28, 40, 43, 20, 25, 50, 26, 48) 
df1 = data.frame(plate, date, liter)

驾驶数据：

plate = c("AB123", "AB123", "AB123", "AB123", "AB123", "AB123", "AB123", "AB123", "AB123", "AB123", "AB123", "AB123", "AC234", "AC234", "AC234", "AC234", "AC234", "AC234", "AC234", "AC234", "AC234", "AC234", "AC234", "AC234", "AD345", "AD345", "AD345", "AD345", "AD345", "AD345", "AD345", "AD345", "AD345", "AD345", "AD345") 
date = c("2017-09-01", "2017-01-05", "2017-09-08", "2017-09-10", "2017-09-11", "2017-09-12", "2017-09-13", "2017-09-16", "2017-09-17", "2017-09-20", "2017-09-22", "2017-09-25", "2017-09-02", "2017-09-03", "2017-09-06", "2017-09-07", "2017-09-08", "2017-09-09", "2017-09-13", "2017-09-15", "2017-09-17", "2017-09-20", "2017-09-23", "2017-09-25", "2017-09-01", "2017-09-04", "2017-09-09", "2017-09-12", "2017-09-15", "2017-09-18", "2017-09-19", "2017-09-20", "2017-09-23", "2017-09-27", "2017-09-30")
mileage = c(50, 64, 45, 70, 58, 41, 22, 15, 90, 48, 52, 48, 29, 65, 70, 46, 88, 71, 40, 51, 38, 91, 74, 61, 41, 33, 59, 81, 72, 65, 43, 81, 20, 49, 39)
accx = c(0, 3, 4, 0, 8, 11, 2, 5, 9, 10, 2, 22, 9, 6, 7, 6, 8, 1, 0, 1, 8, 1, 7, 6, 4, 3, 9, 11, 22, 15, 13, 1, 2, 4, 9)
df2 = data.frame(plate, date, mileage,accx)

合并两个数据

df.all = left_join(df2, df1, by.x =c("plate", "date"))

我想根据燃油日期获得总里程（总和）和平均accx。最终结果应如下所示：

Expected data frame

有没有办法使用dplyr来改变所需的结果？仅供参考我只需要他们的板块的变异参数（结果线1,2,4,6,8等不需要）提前谢谢！

Answer 1

可能有更优雅的方式，但这有效：

library(dplyr)

df.all %>% 
  mutate(date = as.Date(date)) %>% 
  group_by(plate) %>% 
  arrange(plate, date) %>% 
  mutate(t.mileage = cumsum(mileage) * !is.na(liter), 
         t.accx = cumsum(accx) * !is.na(liter), 
         n = seq_len(n())) %>% 
  filter(!is.na(liter)) %>% 
  mutate(t.mileage.lag = lag(t.mileage), 
         t.accx.lag = lag(t.accx),  
         n.lag = lag(n),
         t.mileage = ifelse(!is.na(t.mileage.lag), t.mileage - t.mileage.lag, t.mileage), 
         a.accx = ifelse(!is.na(t.accx.lag), (t.accx - t.accx.lag)/(n - n.lag), t.accx/n)) %>% 
  select(-t.mileage.lag, -t.accx.lag, -n.lag, -n, -t.accx)

# A tibble: 9 x 7
# Groups:   plate [3]
#   plate       date mileage  accx liter t.mileage     a.accx
#   <chr>     <date>   <dbl> <dbl> <dbl>     <dbl>      <dbl>
# 1 AB123 2017-09-08      45     4    33       159  2.3333333
# 2 AB123 2017-09-11      58     8    15       128  4.0000000
# 3 AB123 2017-09-13      22     2    28        63  6.5000000
# 4 AB123 2017-09-20      48    10    40       153  8.0000000
# 5 AC234 2017-09-06      70     7    43       164  7.3333333
# 6 AC234 2017-09-08      88     8    20       134  7.0000000
# 7 AC234 2017-09-15      51     1    25       162  0.6666667
# 8 AC234 2017-09-23      74     7    50       203  5.3333333
# 9 AD345 2017-09-18      65    15    48       351 10.6666667

按板分组仅计算以下每个不同的板。然后使用cumsum计算总里程和总计accx，但只保留我们没有丢失升的值。还要计算我们有多少驱动器用n。然后，由于我们只对我们为汽车加油的信息感兴趣，因此我们过滤了不丢失的升。使用lag从每个值中减去先前的总milage和accx（除非没有先前的值，即滞后为NA），然后计算平均accx。

数据

df1 <- 
  data.frame(plate = c("AB123", "AB123", "AB123", "AB123", "AC234", "AC234", "AC234", "AC234", "AD345", "AD345"), 
             date = c("2017-09-08", "2017-09-11", "2017-09-13", "2017-09-20", "2017-09-06", "2017-09-08", 
                      "2017-09-15", "2017-09-23", "2017-09-10", "2017-09-18"), 
             liter = c(33, 15, 28, 40, 43, 20, 25, 50, 26, 48), 
             stringsAsFactors = F)

df2 <-
  data.frame(plate = c("AB123", "AB123", "AB123", "AB123", "AB123", "AB123", "AB123", "AB123", "AB123", "AB123", "AB123", "AB123", "AC234", "AC234", "AC234", "AC234", "AC234", "AC234", "AC234", "AC234", "AC234", "AC234", "AC234", "AC234", "AD345", "AD345", "AD345", "AD345", "AD345", "AD345", "AD345", "AD345", "AD345", "AD345", "AD345"), 
             date = c("2017-09-01", "2017-01-05", "2017-09-08", "2017-09-10", "2017-09-11", "2017-09-12", "2017-09-13", "2017-09-16", "2017-09-17", "2017-09-20", "2017-09-22", "2017-09-25", "2017-09-02", "2017-09-03", "2017-09-06", "2017-09-07", "2017-09-08", "2017-09-09", "2017-09-13", "2017-09-15", "2017-09-17", "2017-09-20", "2017-09-23", "2017-09-25", "2017-09-01", "2017-09-04", "2017-09-09", "2017-09-12", "2017-09-15", "2017-09-18", "2017-09-19", "2017-09-20", "2017-09-23", "2017-09-27", "2017-09-30"),
             mileage = c(50, 64, 45, 70, 58, 41, 22, 15, 90, 48, 52, 48, 29, 65, 70, 46, 88, 71, 40, 51, 38, 91, 74, 61, 41, 33, 59, 81, 72, 65, 43, 81, 20, 49, 39), 
             accx = c(0, 3, 4, 0, 8, 11, 2, 5, 9, 10, 2, 22, 9, 6, 7, 6, 8, 1, 0, 1, 8, 1, 7, 6, 4, 3, 9, 11, 22, 15, 13, 1, 2, 4, 9), 
             stringsAsFactors = F)

df.all <- left_join(df2, df1, by = c("plate", "date"))

Answer 2

虽然@kath提供了一个更方便的解决方案，但这里有一个基础R（如果只是因为我花了一些时间来处理它）：

# generate factor to split on
temp <- which(!is.na(df.all$liter))
vec <- temp - c(0, temp[-length(temp)])
df.all$split <- rep(seq(1, length(temp)+1), c(vec, nrow(df.all)-temp[length(temp)]))

# split df.all and calculate t.mileage and a.accx for each subsample
df.temp <- split(df.all, df.all$split)
t.mileage <- sapply(df.temp, function(x) sum(x[, "mileage"]))
a.accx <- sapply(df.temp, function(x) mean(x[, "accx"]))

# generate new variables and insert calculated values
df.all$t.mileage <- NA
df.all$t.mileage[temp] <- t.mileage[-length(t.mileage)]
df.all$a.accx <- NA
df.all$a.accx[temp] <- a.accx[-length(a.accx)]

# display df.all without splitting factor
df.all <- subset(df.all, select = -split)

> df.all
   plate       date mileage accx liter t.mileage     a.accx
1  AB123 2017-09-01      50    0    NA        NA         NA
2  AB123 2017-01-05      64    3    NA        NA         NA
3  AB123 2017-09-08      45    4    33       159  2.3333333
4  AB123 2017-09-10      70    0    NA        NA         NA
5  AB123 2017-09-11      58    8    15       128  4.0000000
6  AB123 2017-09-12      41   11    NA        NA         NA
7  AB123 2017-09-13      22    2    28        63  6.5000000
8  AB123 2017-09-16      15    5    NA        NA         NA
9  AB123 2017-09-17      90    9    NA        NA         NA
10 AB123 2017-09-20      48   10    40       153  8.0000000
11 AB123 2017-09-22      52    2    NA        NA         NA
12 AB123 2017-09-25      48   22    NA        NA         NA
13 AC234 2017-09-02      29    9    NA        NA         NA
14 AC234 2017-09-03      65    6    NA        NA         NA
15 AC234 2017-09-06      70    7    43       264  9.2000000
16 AC234 2017-09-07      46    6    NA        NA         NA
17 AC234 2017-09-08      88    8    20       134  7.0000000
18 AC234 2017-09-09      71    1    NA        NA         NA
19 AC234 2017-09-13      40    0    NA        NA         NA
20 AC234 2017-09-15      51    1    25       162  0.6666667
21 AC234 2017-09-17      38    8    NA        NA         NA
22 AC234 2017-09-20      91    1    NA        NA         NA
23 AC234 2017-09-23      74    7    50       203  5.3333333
24 AC234 2017-09-25      61    6    NA        NA         NA
25 AD345 2017-09-01      41    4    NA        NA         NA
26 AD345 2017-09-04      33    3    NA        NA         NA
27 AD345 2017-09-09      59    9    NA        NA         NA
28 AD345 2017-09-12      81   11    NA        NA         NA
29 AD345 2017-09-15      72   22    NA        NA         NA
30 AD345 2017-09-18      65   15    48       412 10.0000000
31 AD345 2017-09-19      43   13    NA        NA         NA
32 AD345 2017-09-20      81    1    NA        NA         NA
33 AD345 2017-09-23      20    2    NA        NA         NA
34 AD345 2017-09-27      49    4    NA        NA         NA
35 AD345 2017-09-30      39    9    NA        NA         NA

顺便说一下，必须有一种更简单的方法来生成上面第1步的因子，有谁知道怎么做？

R：基于来自2个不同长度的数据帧的2个条件的和值

2 个答案:

数据