Question

我有以下变量，我想将其转换为分钟，因此得到一个数字变量。

目前我正在做以下事情：

BrowserWindow.addDevToolsExtension('path/to/extension')

在“50分钟”的情况下，它转换完美，但在“1小时0分钟”的情况下转换为1分钟。

Answer 1

require(lubridate)
time <- c("1 hour 10 mins", "1 hour 0 mins", "12 mins", "1 hour 0 mins", 
           "1 hour 4 mins", "1 hour 0 mins", "1 hour 0 mins", "1 hour 0 mins", 
           "1 hour 5 mins", "1 hour 0 mins", "34 mins", "1 hour 0 mins", 
           "50 mins", "1 hour 0 mins", "1 hour 0 mins", "1 hour 0 mins", 
           "1 hour 0 mins", "1 hour 0 mins", "1 hour 0 mins", "1 hour 0 mins", 
           "1 hour 0 mins", "1 hour 0 mins", "1 hour 34 mins", "1 hour 0 mins", 
           "43 mins", "1 hour 20 mins", "1 hour 0 mins", "1 hour 0 mins", 
           "1 hour 0 mins", "1 hour 30 mins", "1 hour 0 mins")
time <- gsub(" mins", "", time)
tmp  <- grepl(" hour ", time)
time <- gsub("1 hour ", "", time) # If you sometimes have > 1 hour then use grep or regexpr here, or more gsubs if you want
time <- as.numeric(time)
time[tmp] <- time[tmp]+60

x <- ymd("1900-01-01") # or whenever

minute(x) <- time

x

 [1] "1900-01-01 01:10:00 UTC" "1900-01-01 01:00:00 UTC" "1900-01-01 00:12:00 UTC" "1900-01-01 01:00:00 UTC" "1900-01-01 01:04:00 UTC" "1900-01-01 01:00:00 UTC"
 [7] "1900-01-01 01:00:00 UTC" "1900-01-01 01:00:00 UTC" "1900-01-01 01:05:00 UTC" "1900-01-01 01:00:00 UTC" "1900-01-01 00:34:00 UTC" "1900-01-01 01:00:00 UTC"
[13] "1900-01-01 00:50:00 UTC" "1900-01-01 01:00:00 UTC" "1900-01-01 01:00:00 UTC" "1900-01-01 01:00:00 UTC" "1900-01-01 01:00:00 UTC" "1900-01-01 01:00:00 UTC"
[19] "1900-01-01 01:00:00 UTC" "1900-01-01 01:00:00 UTC" "1900-01-01 01:00:00 UTC" "1900-01-01 01:00:00 UTC" "1900-01-01 01:34:00 UTC" "1900-01-01 01:00:00 UTC"
[25] "1900-01-01 00:43:00 UTC" "1900-01-01 01:20:00 UTC" "1900-01-01 01:00:00 UTC" "1900-01-01 01:00:00 UTC" "1900-01-01 01:00:00 UTC" "1900-01-01 01:30:00 UTC"
[31] "1900-01-01 01:00:00 UTC"

或者如果您只想要原始分钟数，那么您当然已经在time

Answer 2

这是一种方法：

library(chron)

times <- c("1 hour 10 mins", "1 hour 0 mins", "12 mins", "1 hour 0 mins", 
"1 hour 4 mins", "1 hour 0 mins", "1 hour 0 mins", "1 hour 0 mins", 
"1 hour 5 mins", "1 hour 0 mins", "34 mins", "1 hour 0 mins", 
"50 mins", "1 hour 0 mins", "1 hour 0 mins", "1 hour 0 mins", 
"1 hour 0 mins", "1 hour 0 mins", "1 hour 0 mins", "1 hour 0 mins", 
"1 hour 0 mins", "1 hour 0 mins", "1 hour 34 mins", "1 hour 0 mins", 
"43 mins", "1 hour 20 mins", "1 hour 0 mins", "1 hour 0 mins", 
"1 hour 0 mins", "1 hour 30 mins", "1 hour 0 mins")

# Add "0 hour" when only minutes are present
times <- sub(pattern = "^(\\d+ mins)",replacement = "0 hour \\1", times)

# Change format simply to ease next step
times <- sub(pattern = "^(\\d+) hours? (\\d+) mins", replacement = "\\1\\:\\2", times)

# Convert to a real time vector
times.2 <- strptime(x = times, format = "%H:%M")

# Extract number of minutes
minutes <- 60*hours(times.2) + minutes(times.2)

结果

> minutes
 [1] 70 60 12 60 64 60 60 60 65 60 34 60 50 60 60 60 60 60 60 60 60 60 94
[24] 60 43 80 60 60 60 90 60

Answer 3

另一种选择（类似逻辑但不同的方法）：

library(lubridate)

x[!grepl("hour", x)] <- paste("0 hour", x[!grepl('hour', x)], sep = " ")
hour(strptime(x, "%H hour %M mins"))*60 + minute(strptime(x, "%H hour %M mins"))
# [1] 70 60 12 60

数据：

c("1 hour 10 mins", "1 hour 0 mins", "0 hour 12 mins", "1 hour 0 mins" )

时间文本变量为数字

3 个答案:

结果