Question

我的data.table包括每小时对发动机（output）产生的功率的观察以及系统状态描述符tag，该状态描述符指示发动机的所有组件均已打开。

数据

structure(list(time = structure(c(1517245200, 1517247000, 1517248800, 
1517250600, 1517252400, 1517254200, 1517256000, 1517257800, 1517259600, 
1517261400, 1517263200, 1517265000, 1517266800, 1517268600, 1517270400, 
1517272200, 1517274000, 1517275800, 1517277600, 1517279400, 1517281200, 
1517283000, 1517284800, 1517286600), class = c("POSIXct", "POSIXt"
), tzone = ""), output1 = c(160.03310020928, 159.706274495615, 
159.803834736236, 159.753928429527, 159.54807802046, 159.21298848298, 
158.904290018581, 158.683643772917, 158.670475839199, 158.793901799427, 
158.886487460894, 159.167829223303, 159.66751884913, 159.1288534448, 
159.141463186901, 160.116892086363, 160.517879769862, 160.615925580417, 
160.915687799509, 161.590897854561, 161.568455821241, 161.411642091721, 
161.811137570257, 162.193040254917), tag1 = c("evap only", "evap only", 
"fog & evap", "fog & evap", "evap only", "evap only", "evap only", 
"neither fog nor evap", "neither fog nor evap", "fog & evap", "evap only", "evap only", 
"evap only", "fog & evap", "evap only", "fog & evap", "evap only", 
"evap only", "evap only", "evap only", "fog & evap", "fog & evap", 
"bad data", "neither fog nor evap")), row.names = c(NA, -24L
), class = c("data.table", "data.frame"))

您还可以使用以下方法生成一些示例数据：

sample_data <- data.table(time = seq.POSIXt(from = Sys.time(), by = 60*60*3, length.out = 100), 
           output = runif(n = 100, min = 130, max = 172), 
           tag = sample(x = c('evap only', 'bad data', 'neither fog nor evap', 'fog and evap'), 
                        size = 100, replace = T))

我想按天分组（上面的样本数据只有两天，而实际数据只有3年的数据），然后找到与每个tag相对应的平均功效。我希望输出为：

          time  evap only fog & evap  neither fog nor evap bad data
 1: 2018-01-29  159.8391  160.0825    159.8491             161.8111

我尝试了以下代码，但结果不是我想要的形式。我使用.SDcols是因为实际的数据集还有大量其他列。

sample_data[, lapply(.SD, function(z){mean(z, na.rm = T)}), .SDcols = c('output1'), by = .(round_date(time, 'day'), tag1)]
   round_date                 tag1  output1
1: 2018-01-30            evap only 159.8391
2: 2018-01-30           fog & evap 160.0825
3: 2018-01-30 neither fog nor evap 159.8491
4: 2018-01-30             bad data 161.8111

我已经在堆栈溢出中看到了以下问题。

是否有data.table的方式来实现？

Answer 1

这是一种数据表方法

#explanation of mean(.SD[[1]] ..), see akrun's comment here:
# https://stackoverflow.com/questions/29568732/using-mean-with-sd-and-sdcols-in-data-table#comment47286876_29568732
ans <- DT[, .(mean_output1 = mean(.SD[[1]], na.rm = TRUE )), 
          by = .( date = as.Date( time ), tag1 ), 
          .SDcols = c("output1") ]

dcast( ans, date~tag1, value.var = "mean_output1" )

#          date bad data evap only fog & evap neither fog nor evap
# 1: 2018-01-29       NA  159.3908   159.3701             158.6771
# 2: 2018-01-30 161.8111  160.5564   161.0323             162.1930

Answer 2

library(dplyr)
library(lubridate)
# test is the dataframe provided in question
test1 = test %>% group_by(date = date(time), tag1) %>% 
          summarise(mean_power = mean(output1))

将上述代码产生的tibble转换为dataframe

test1_df = data.frame(test1)

将数据重塑为宽格式

reshape(test1_df, idvar = "date", timevar = 
            "tag1", direction = "wide")

输出：

> output
        date evap only fog & evap bad data neither fog nor evap
1 2018-01-29  159.8697   159.8038       NA                   NA
3 2018-01-30  159.8335   160.1289 161.8111             159.8491

自日期2018年1月30日首次出现在test1_df的第三行起，行号就显示为1以后的3。

根据现有列创建新的data.table列

2 个答案: