使用here中的信息，我可以使用时间戳信息汇总我的数据框。
data <- read.csv("data.csv", header=T)
data$TIMESTAMP <- strptime(data$TIMESTAMP, "%m/%d/%Y %H:%M") # make unambigous
data$TIMESTAMP <- as.POSIXct(data$TIMESTAMP) # plyr/ddply does not seem to like POSIXlt
data$DAY <- as.factor(format(data$TIMESTAMP,'%d'))
data$MONTH <- as.factor(format(data$TIMESTAMP,'%m'))
ddply(data,.(MONTH,DAY),summarise, V1 = mean(P), V2 = max(WS)) # summarize by month by day
但是，我只想使用一天中的某些小时，例如13:00:00到16:00:00。如果我在as.POSIXct行执行之前这样做，我可以子集数据帧（例如，创建一个新的数据帧，它是原始df的子集）。但是，当我尝试以下序列时：
data.1.4 <- data[format(data$TIMESTAMP, "%H") >= 13 & format(data$TIMESTAMP, "%H") < 17,] # subset the data
data.1.4$TIMESTAMP <- as.POSIXct(data.1.4$TIMESTAMP)
data.1.4$DAY <- as.factor(format(data.1.4$TIMESTAMP,'%d'))
data.1.4$MONTH <- as.factor(format(data.1.4$TIMESTAMP,'%m'))
ddply(data.1.4,.(MONTH,DAY),summarise, V1 = mean(P), V2 = max(WS)) # summarize by month by day
我收到以下错误Error in eval(expr, envir, enclos) : object 'MONTH' not found。我在data和data.1.4之间看到的唯一区别是row.names是'visible'和非顺序data.1.4数据帧。
那么，我如何从数据框中分配或选择条目来创建一个我可以概括的新df？
# dput of subset of data
data <- structure(list(TIMESTAMP = structure(1:15, .Label = c("1/1/2012 11:00",
"1/1/2012 12:00", "1/1/2012 13:00", "1/1/2012 14:00", "1/1/2012 15:00",
"1/2/2012 11:00", "1/2/2012 12:00", "1/2/2012 13:00", "1/2/2012 14:00",
"1/2/2012 15:00", "4/7/2012 11:00", "4/7/2012 12:00", "4/7/2012 13:00",
"4/7/2012 14:00", "4/7/2012 15:00"), class = "factor"), P = c(992.4,
992.4, 992.4, 992.4, 992.4, 992.4, 992.4, 992.4, 992.4, 992.4,
239, 239, 239, 239, 239), WS = c(4.023, 3.576, 4.023, 6.259,
4.47, 3.576, 3.576, 2.682, 4.023, 3.576, 2.682, 3.129, 2.682,
2.235, 2.682), WD = c(212L, 200L, 215L, 213L, 204L, 304L, 276L,
273L, 307L, 270L, 54L, 24L, 304L, 320L, 321L), AT = c(16.11,
18.89, 20, 20, 19.44, 10.56, 11.11, 11.67, 12.22, 11.11, 17.22,
18.33, 19.44, 20.56, 21.11), FT = c(17.22, 22.22, 22.78, 22.78,
20, 11.11, 15.56, 17.22, 17.78, 15.56, 24.44, 25.56, 29.44, 30.56,
29.44), H = c(50L, 38L, 38L, 39L, 48L, 24L, 19L, 18L, 16L, 18L,
23L, 20L, 18L, 17L, 15L), B = c(1029L, 1027L, 1026L, 1024L, 1023L,
1026L, 1025L, 1024L, 1023L, 1023L, 1034L, 1033L, 1032L, 1031L,
1030L), FM = c(14.9, 14.4, 14, 13.7, 13.6, 13.1, 12.8, 12.3,
12, 11.7, 12.8, 12, 11.4, 10.9, 10.4), GD = c(204L, 220L, 227L,
222L, 216L, 338L, 311L, 326L, 310L, 273L, 62L, 13L, 312L, 272L,
281L), MG = c(8.047, 9.835, 10.28, 13.41, 11.18, 9.388, 8.941,
8.494, 9.835, 10.73, 6.706, 7.153, 8.047, 8.047, 7.6), SR = c(522L,
603L, 604L, 526L, 248L, 569L, 653L, 671L, 616L, 487L, 972L, 1053L,
1061L, 1002L, 865L), WS2 = c(2.235, 3.576, 4.47, 4.47, 5.364,
4.023, 2.682, 3.576, 3.576, 4.023, 3.129, 3.129, 3.576, 2.682,
3.129), WD2 = c(200L, 201L, 206L, 210L, 211L, 319L, 315L, 311L,
302L, 290L, 49L, 39L, 15L, 348L, 329L)), .Names = c("TIMESTAMP",
"P", "WS", "WD", "AT", "FT", "H", "B", "FM", "GD", "MG", "SR",
"WS2", "WD2"), class = "data.frame", row.names = c(NA, -15L))
答案 0 :(得分：3)
nofunsally，这是您可能想要使用data.table考虑的解决方案。如果您想取{em>所有列的平均值hour >= 13和hour <= 15
# changing the name from `TIMESTAMP` to `ts` : too long for me!
names(data)[1] <- "ts"
Step 1:获取data.table并将ts转换为POSIXct
require(data.table)
dt <- data.table(data) # convert to data.table
dt[, ts := as.POSIXct(ts, format="%m/%d/%Y %H:%M")] # convert to POSIXct
Step 2:获取hour，day和month
# get hour, day, month
dt[, `:=`(day=as.POSIXlt(ts)$mday + 1,
month=as.POSIXlt(ts)$mon + 1,
hour=as.POSIXlt(ts)$hour)]
# set keys for blazing fast access
setkey(dt, "day","month")
step 3:获取所有必要的意思（单行代码）
# the `by` part groups your data.table by day and month
# The group can be accessed by `.SD` and we filter there for hours
# then we select all the columns that we want using `with=FALSE` that makes
# data.table behave like a data.frame only for that operation
# then we use `lapply` to take the mean of all the columns.
# all this can be done in one line! That's the sweetest part!
dt[, lapply(.SD[hour >= 13 & hour <= 15][, 2:13, with=FALSE], mean), by = "day,month"]
# day month P WS WD AT FT H B FM GD MG SR WS2
# 1: 2 1 992.4 4.917333 210.6667 19.81333 21.85333 41.66667 1024.333 13.76667 221.6667 11.623333 459.3333 4.768
# 2: 3 1 992.4 3.427000 283.3333 11.66667 16.85333 17.33333 1023.333 12.00000 303.0000 9.686333 591.3333 3.725
# 3: 8 4 239.0 2.533000 315.0000 20.37000 29.81333 16.66667 1031.000 10.90000 288.3333 7.898000 976.0000 3.129
或者，您可以先subset hour，然后选择mean获取columns：
dt1 <- dt[, .SD[hour >= 13 & hour <= 15], by="day,month"]
dt1[, lapply(.SD, mean), by="day,month", .SDcols=4:17]