用脏数据格式化 R 时间

Question

我正在使用 R 从数据库生成 CZML 文件。

数据库有脏数据。

我需要一种方法来确保时间的格式为“%H:%M:%S”。

数据可以已经在正确的 %H:%M:%S 中，也可以在小时前丢失零，例如 8:30:00，这是一个无效的 ISO 8601 并完全抛出 CZML 解析.

在 24 小时格式中需要始终像这样 08:30:00 或 07:09:00。

我有错误，因为它像这样 8:30:00 或 7:09:00 仍然是 24 小时格式，我还没有检查分钟或秒是否也不正确，但目前我假设他们是正确的，唯一的问题是时间。

例如，我有一个这样的 csv 文件：

"Date","Time","TZ","Jul.Time","BirdID","Species","Sex","Age","SiteID","Latitude","Longitude"
"4-Mar-13","08:30:00","America/Costa_Rica",2456356.187500,"test2","GREH","M","AHY","56scr25",8.71191178,-82.96866316
"4-Mar-13","8:30:00","America/Costa_Rica",2456356.187500,"test2","GREH","M","AHY","56scr25",8.71191178,-82.96866316

我需要像这样生成 CZML：

"point": {
        "color": { 
            "rgba": [
"2013-03-04T08:30:00Z",225,50,50,196,"2013-03-04T08:30:01Z",50,50,225,196,"2013-03-04T13:30:00Z",225,50,50,196,"2013-03-04T13:30:01Z",50,50,225,196,"2013-03-04T16:00:00Z",225,50,50,196,"2013-03-04T16:00:01Z",50,50,225,196
            ]
        },
        "pixelSize": { "number": 10 }
    }

我的代码是这样的：

        j=1
        numVisits=nrow(visitedTimes)
        while(j<=numVisits){
            date=as.Date(visitedTimes$Date[j], format="%d-%b-%y")
            time=format(visitedTimes$Time[j], format="%H:%M:%S")
            timeOfPassage=paste0(date,"T",time,"Z")
            timeAfter=as.POSIXlt(timeOfPassage, format="%Y-%m-%dT%H:%M:%SZ")
            timeAfter$sec=timeAfter$sec+1
            timeAfter=format(timeAfter, format="%Y-%m-%dT%H:%M:%SZ")
            cat(paste0("\"",timeOfPassage,"\","))
            cat("225,50,50,196,")
            cat(paste0("\"",timeAfter,"\","))
            cat("50,50,225,196")
            if(j<numVisits){
                cat(",")
            }
            j=j+1
        }

但是由于脏数据，它没有产生所需的输出.. 有什么想法吗？

Answer 1

我们可以使用 times 从 chron

library(chron)
times(v1)
#[1] 08:30:00 08:30:00 07:09:00 07:09:00

或使用base R

format(strptime(v2, '%H:%M:%S'), '%H:%M:%S')
#[1] "08:30:00" "08:30:00" "07:09:00" "07:09:00" "07:09:05" "11:10:00"

使用 OP 的更新数据集

df1$Time <- times(df1$Time)
df1$Time
#[1] 08:30:00 08:30:00

或使用regex

sub('^(.:)', '0\1', df1$Time)
gsub('[^:]{2}(*SKIP)(*F)|(\d)', '0\1', v2, perl=TRUE)
#[1] "08:30:00" "08:30:00" "07:09:00" "07:09:00" "07:09:05" "11:10:00"

数据

v1 <- c('8:30:00', '08:30:00', '7:09:00', '7:9:00')
v2 <- c(v1, '7:9:5', '11:10:0')


df1 <- structure(list(Date = c("4-Mar-13", "4-Mar-13"), Time = c("08:30:00", 
"8:30:00"), TZ = c("America/Costa_Rica", "America/Costa_Rica"
), Jul.Time = c(2456356.1875, 2456356.1875), BirdID = c("test2", 
"test2"), Species = c("GREH", "GREH"), Sex = c("M", "M"), Age = c("AHY", 
"AHY"), SiteID = c("56scr25", "56scr25"), Latitude = c(8.71191178, 
8.71191178), Longitude = c(-82.96866316, -82.96866316)), .Names = c("Date", 
"Time", "TZ", "Jul.Time", "BirdID", "Species", "Sex", "Age", 
"SiteID", "Latitude", "Longitude"), class = "data.frame", row.names = c(NA, 
-2L))

用脏数据格式化 R 时间

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