如何在R中将播放时间转换为秒答案

【问题标题】：How to convert playtime to seconds in R如何在R中将播放时间转换为秒
【发布时间】：2021-09-07 20:11:25
【问题描述】：

我有一个向量如下：

playtimes <- c("1H18M20S", "1H27M5S", "18M27S", "56M38S", "21S")

我想在秒内将这些转换为播放时间。例如，生成的向量会是这样的：

playtimeInSeconds <- c(4700, 5225, 1107, 3398, 21)

我无法根据 H、M 和 S 正确分离字符串。我写了以下内容，适用于 1 小时以下的播放时间

minutes <- gsub("M.*", "", playtime)
seconds <- gsub(".*M", "", playtime) %>%
    gsub("S", "", .)

totalPlaytime <- as.numeric(minutes)*60 + as.numeric(seconds)

但我不确定如何处理某些字符串的 H 部分。

【问题讨论】：

相关：Calculate character string "days, hours, minutes, seconds" to numeric total days。例如。 period_to_seconds(period(playtimes))

标签： r

【解决方案1】：

您可以strsplit 并将列表元素reversely 的length 调整为3，这允许您使用sapply 来获得一个矩阵，您可以在其中应用矩阵乘积%*%。

m <- sapply(strsplit(p, 'H|M|S'), \(x) as.double(rev(`length<-`(rev(x), 3))))
res <- as.vector(t(replace(m, is.na(m), 0)) %*% rbind(3600, 60, 1))
res
# [1] 4700 5225 1107 3398   21

【讨论】：

【解决方案2】：

有趣的问题。这是一个可能更有效但可以完成工作的解决方案

# function from https://www.statworx.com/de/blog/strsplit-but-keeping-the-delimiter/
strsplit <- function(x,
                     split,
                     type = "remove",
                     perl = FALSE,
                     ...) {
  if (type == "remove") {
    # use base::strsplit
    out <- base::strsplit(x = x, split = split, perl = perl, ...)
  } else if (type == "before") {
    # split before the delimiter and keep it
    out <- base::strsplit(x = x,
                          split = paste0("(?<=.)(?=", split, ")"),
                          perl = TRUE,
                          ...)
  } else if (type == "after") {
    # split after the delimiter and keep it
    out <- base::strsplit(x = x,
                          split = paste0("(?<=", split, ")"),
                          perl = TRUE,
                          ...)
  } else {
    # wrong type input
    stop("type must be remove, after or before!")
  }
  return(out)
}


# convert to seconds
to_seconds <- c(H = 60 * 60,
                M = 60,
                S = 1)

get_seconds <- function(value, unit) {
  value * to_seconds[unit]
}


# example vector
playtimes <- c("1H18M20S", "1H27M5S", "18M27S", "56M38S", "21S")

# extract time parts
times <- strsplit(playtimes, 
                  split = "[A-Z]",
                  type = "after")
times
#> [[1]]
#> [1] "1H"  "18M" "20S"
#> 
#> [[2]]
#> [1] "1H"  "27M" "5S" 
#> 
#> [[3]]
#> [1] "18M" "27S"
#> 
#> [[4]]
#> [1] "56M" "38S"
#> 
#> [[5]]
#> [1] "21S"

# calculate each time in seconds
sapply(times,
       function(t) {
         # split numeric and unit part
         t_split <- strsplit(x = t,
                             split = "[A-Z]",
                             type = "before")
         
         # calculate seconds for each unit part
         times_in_seconds <- get_seconds(value = as.numeric(sapply(t_split, `[`, 1)),
                                         unit = sapply(t_split, `[`, 2))
         
         # sum of all parts
         sum(times_in_seconds)
       })
#> [1] 4700 5225 1107 3398   21

【讨论】：

【解决方案3】：

我按照第三个答案here中给出的示例进行了以下操作

playtime <- sapply(playtime, function(x){paste(paste(rep(0, 3 - str_count(x, '[0-9]+')), collapse = ' '), x)})
totalPlaytime <- time_length(hms(playtime))

短小精悍，并检查播放时间少于 1 小时或少于 1 分钟的潜在错误。

【讨论】：