(高中表格),這是我的原始數據的子集,我有工作:一天的時間減去:列
dput(datumi)
structure(c("21:26", "21:33", "21:38", "23:02", "23:03", "21:27",
"21:34", "21:39", "23:03", "23:04", "21:26", "21:33", "21:38",
"23:02", "23:04", "21:26", "21:34", "21:38", "23:02", "23:04",
"21:27", "21:34", "21:39", "23:02", "23:04"), .Dim = c(5L, 5L
), .Dimnames = list(c("2", "3", "4", "5", "6"), c("Datum_1",
"Datum_2", "Datum_3", "Datum_4", "Datum_5")))
所以我有時間,例如在那裏工作,21:26意味着時間的一天。
現在我想從第一個和第二從等第三減去第二列中,這意味着,我會從Datum_2
和Datum_4
從Datum_3
減去Datum_1
Datum_2
柱和柱Datum_3
。 我的輸出將是新的列與
我已經創建函數/循環,這是否在幾秒鐘的差別,如果我的數據會numeric
,因此,例如在數字數據的情況下,我會做到這一點,得到了所需的輸出:
dat <- data.frame(
column1 = round(runif(n = 10, min=0, max=5),0),
column2 = round(runif(n = 10, min=0, max=5),0),
column3 = round(runif(n = 10, min=0, max=5),0),
column4 = round(runif(n = 10, min=0, max=5),0)
)
results <- list()
for(i in 1:length(dat)) {
if (i==length(dat)){
results[[i]] <-dat[,i]
} else {results[[i]] <-dat[,i+1] - dat[,i]}
}
results <- t(do.call(rbind,results))
results <- data.frame(results)
但我想不出它的時間格式,我曾嘗試strptime
和as.POSIXct
x1 <- strptime(datumi, "%H:%M")
as.numeric(x1,units="secs")
和
as.POSIXct(datumi,format="%H:%M")
而且還搜索這
Subtracting Two Columns Consisting of Both Date and Time in R
convert character to time in R