如何将特定时间转换为

Question

我想将字符串列转换为正确的格式。

通常我会这样做：

print(df$Time) 
> "00:00:01"
as.POSIXct(df$Time,format="%H:%M:%S")

但是，我的数据很奇怪。它看起来像这样：

print(df$Time)
850a"  "823a"  NA      "906a"  "321a"  "1154p"

我的解决方案不起作用。因为我首先剥离了字符（在本例中为 "a" 和 "p"）。但是这样做之后，我的时间缺少了一个重要的部分（如果是早上或下午）。

因此我的问题是：如何将这些数据转换为正确的格式？

预期输出：

df$Time_Old
850a"  "823a"  NA      "906a"  "321a"  "1154p"

df$Time_New
08.50   08.23    NA    09.06   03.21   23.54

一些示例数据：

vector_string <- as.vector(tv_Adds[["Time"]])
vector_string = vector_string[1:20]
> vector_string 


[1] "850a"  "823a"  NA      "906a"  "321a"  "1154p" "608p"  "1012a" "354a"  "1121p" "414p"  "1241p" "721p"  "223p"  "316p" 
[16] "345p"  "1145a" "3p"    "937a"  "138p"

> dput(vector_string[1:20])
c("850a", "823a", NA, "906a", "321a", "1154p", "608p", "1012a", 
"354a", "1121p", "414p", "1241p", "721p", "223p", "316p", "345p", 
"1145a", "3p", "937a", "138p")

Answer 1

您必须将小时与分钟分开，因为您输入的内容含糊不清。然后在非 NA 条目的末尾添加 "m"。我想你需要这个：

tvec = c("850a",  "823a",  NA, "906a",  "321a",  "1154p")
notNA <- !is.na(tvec)

#separate hours from minutes with a dot and append m at the end:
tvec[notNA] <- paste0(strtrim(tvec[notNA], nchar(tvec[notNA]) - 3), ".", 
                      substr(tvec[notNA], nchar(tvec[notNA])-2, nchar(tvec[notNA]))
, "m")

as.POSIXct(tvec, format = "%I.%M%p")
[1] "2019-10-25 08:50:00 CEST" "2019-10-25 08:23:00 CEST"
[3] NA                         "2019-10-25 09:06:00 CEST"
[5] "2019-10-25 03:21:00 CEST" "2019-10-25 23:54:00 CEST"

"%I.%M%p"代表

hour(0-12), followed by .,followed by minutes(00-59), followed by "am"(or "pm")

Answer 2

根据您分享的示例，我们似乎需要处理 3 种不同的情况。

当你有 834a 需要变成 8:34am
当你有 1143p 需要变成 11:43pm
当你有 3a 需要变成 3:00am

处理完这些后，在本例中使用简单的 ifelse 语句计算字符数并进行相应修改，然后我们可以通过调用 strptime 以正确的格式简单地转换为 datetime 对象，即

v1[!is.na(v1)] <- paste0(v1[!is.na(v1)], 'm')
v2 <- ifelse(nchar(v1) == 5, gsub('(^[0-9]{1})(.*$)', '\1:\2', v1), 
           ifelse(nchar(v1) == 3, gsub('(^[0-9]{1})(.*$)', '\1:00\2', v1), 
                                                  gsub('(^[0-9]{2})(.*$)', '\1:\2', v1)))

v2
#[1] "8:50am"  "8:23am"  NA        "9:06am"  "3:21am"  "11:54pm" "6:08pm"  "10:12am" "3:54am"  "11:21pm" "4:14pm"  "12:41pm" "7:21pm"  "2:23pm"  "3:16pm"  "3:45pm"  "11:45am" "3:00pm"  "9:37am"  "1:38pm" 

strptime(v2, format = '%I:%M%p')
#[1] "2019-10-29 08:50:00 +03" "2019-10-29 08:23:00 +03" NA                        "2019-10-29 09:06:00 +03" "2019-10-29 03:21:00 +03" "2019-10-29 23:54:00 +03" "2019-10-29 18:08:00 +03" "2019-10-29 10:12:00 +03" "2019-10-29 03:54:00 +03" "2019-10-29 23:21:00 +03"
#[11] "2019-10-29 16:14:00 +03" "2019-10-29 12:41:00 +03" "2019-10-29 19:21:00 +03" "2019-10-29 14:23:00 +03" "2019-10-29 15:16:00 +03" "2019-10-29 15:45:00 +03" "2019-10-29 11:45:00 +03" "2019-10-29 15:00:00 +03" "2019-10-29 09:37:00 +03" "2019-10-29 13:38:00 +03"

已使用数据

dput(v1)
c("850am", "823am", NA, "906am", "321am", "1154pm", "608pm", 
"1012am", "354am", "1121pm", "414pm", "1241pm", "721pm", "223pm", 
"316pm", "345pm", "1145am", "3pm", "937am", "138pm")

如何将特定时间转换为

How can I transform a specific time, to a

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chron

dplyr