R 中故障率和日期时间操作的计算

Question

我有一个正在使用的示例数据框

Datetime <- c("2015-09-29 08:22:00", "2015-09-29 09:45:00", "2015-09-29 09:53:00", "2015-09-29 10:22:00", "2015-09-29 10:42:00",
                  "2015-09-29 11:31:00", "2015-09-29 11:47:00", "2015-09-29 12:45:00", "2015-09-29 13:11:00", "2015-09-29 13:44:00",
                  "2015-09-29 15:24:00", "2015-09-29 16:28:00", "2015-09-29 20:22:00", "2015-09-29 21:38:00", "2015-09-29 23:34:00")
Measurement <- c("Length","Length","Width","Height","Width","Height","Length","Width","Width","Height","Width","Length",
                     "Length","Height","Height")
PASSFAIL <- c("PASS","PASS","FAIL","PASS","PASS","FAIL_AVG_HIGH","FAIL#Pts","FAIL","FAIL_AVG_LOW","FAIL","PASS","PASS","FAIL#RNG#HIGH","PASS","FAIL")

df1 <- data.frame(Datetime,Measurement,PASSFAIL)

df1

              Datetime Measurement      PASSFAIL
1  2015-09-29 08:22:00      Length          PASS
2  2015-09-29 09:45:00      Length          PASS
3  2015-09-29 09:53:00       Width          FAIL
4  2015-09-29 10:22:00      Height          PASS
5  2015-09-29 10:42:00       Width          PASS
6  2015-09-29 11:31:00      Height FAIL_AVG_HIGH
7  2015-09-29 11:47:00      Length      FAIL#Pts
8  2015-09-29 12:45:00       Width          FAIL
9  2015-09-29 13:11:00       Width  FAIL_AVG_LOW
10 2015-09-29 13:44:00      Height          FAIL
11 2015-09-29 15:24:00       Width          PASS
12 2015-09-29 16:28:00      Length          PASS
13 2015-09-29 20:22:00      Length FAIL#RNG#HIGH
14 2015-09-29 21:38:00      Height          PASS
15 2015-09-29 23:34:00      Height          FAIL

我正在研究一个有趣的问题，以找出一天中 12 AM-12 PM 和 12 PM-12 AM（第二天）的每次测量的失败率。

注意：在 df1 中，任何在 PASSFAIL 列中有 FAIL 的都被视为失败。

Fail Rate = (Number of Fails)/(Number of Fails + Number of Pass)

我想要的输出是这样的

                Datetime FailRate_length Total_length FailRate_Width Total_Width FailRate_Height Total_Height
1 2015-09-29 00:00:00 AM            0.33            3           0.50           2            0.50            2
2 2015-09-29 12:00:00 PM            0.50            2           0.66           3            0.66            3

我正在尝试使用 dplyr 和 data.table 包来解决这个问题，但我只是不知道如何划分 df1 中的时间间隔以获得具有 2 个值的 df2 -> 12AM（前 7 个观察值df1) 和 12PM（df1 中接下来的 8 个观测值）。有人可以帮我吗？

Answer 1

使用data.table...

library(data.table)

# thanks to @DavidArenburg for suggesting this approach:

df1[, `:=`( 
  d        = as.IDate(Datetime), 
  antepost = c("am","pm")[1+(hour(Datetime) >= 12)] ) 
]

res <- setDT(df1)[ , .( 
  failrate    = sum(PASSFAIL != "PASS")/.N,
  N           = .N
), by = .(d, antepost, Measurement)]

这给出

            d antepost Measurement  failrate N
1: 2015-09-29       am      Length 0.3333333 3
2: 2015-09-29       am       Width 0.5000000 2
3: 2015-09-29       am      Height 0.5000000 2
4: 2015-09-29       pm       Width 0.6666667 3
5: 2015-09-29       pm      Height 0.6666667 3
6: 2015-09-29       pm      Length 0.5000000 2

语法为DT[i,j,by]，其中by用于对变量进行分组； j 用于处理列。 := 在 j 中创建新列。

重塑为 OP 所需的输出...

dcast(res, d + antepost ~ Measurement, value.var = c("failrate", "N"))

这给出了

            d antepost failrate_Height failrate_Length failrate_Width N_Height N_Length N_Width
1: 2015-09-29       am       0.5000000       0.3333333      0.5000000        2        3       2
2: 2015-09-29       pm       0.6666667       0.5000000      0.6666667        3        2       3

感谢@Arun，这是一种一步完成所有操作的方法：

dcast(setDT(df1), 
  as.IDate(Datetime) + c("am","pm")[1+(hour(Datetime) >= 12)] ~ Measurement, 
  value.var = "PASSFAIL", 
  fun.agg = list(function(x) sum(x != "PASS")/length(x), length)
)

这给出了

     Datetime Datetime_1 PASSFAIL_function_Height PASSFAIL_function_Length PASSFAIL_function_Width PASSFAIL_length_Height PASSFAIL_length_Length PASSFAIL_length_Width
1: 2015-09-29         am                0.5000000                0.3333333               0.5000000                      2                      3                     2
2: 2015-09-29         pm                0.6666667                0.5000000               0.6666667                      3                      2                     3

列名是根据 ~ 部分的根变量和每个函数定义的第一个单词自动生成的。

Answer 2

一个 dplyr + tidyr 等价物（装箱略有不同，尽管上面那个很优雅）：

library(plyr)
library(dplyr)
library(tidyr)

df1 %>%
  mutate(
    half_day = 
      Datetime %>%
      as.POSIXct(tz = "UTC") %>%
      round_any(60*60*12, f = floor) ) %>%
  group_by(half_day, Measurement) %>%
  summarize(Total = n(),
         FailRate = sum(PASSFAIL != "PASS")/Total) %>%
  gather(variable, value, FailRate, Total) %>%
  unite(variable_new, variable, Measurement, sep = "_") %>%
  spread(variable_new, value)

gather、unite、spread 序列是 dcast 的 tidyr 等价物。注意

half day * (12 hour/half day) * (60 min / hour) * (60 seconds/min) = 60*60*12 seconds

R 中故障率和日期时间操作的计算

Calculation of failure rate and date time manipulation in R

r

reshape2

dplyr

data.table