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R 正则表达式/gsub : extract part of pattern

转载 作者:行者123 更新时间:2023-12-04 16:40:12 24 4
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我有一个按纬度和经度列出的气象站及其位置列表。存在格式问题,其中一些有小时和分钟,而其他有小时、分钟和秒。我可以使用正则表达式找到模式,但我无法提取各个部分。

这里的数据:

> head(wthrStat1 )
Station lat lon
1940 K01R 31-08N 092-34W
1941 K01T 28-08N 094-24W
1942 K03Y 48-47N 096-57W
1943 K04V 38-05-50N 106-10-07W
1944 K05F 31-25-16N 097-47-49W
1945 K06D 48-53-04N 099-37-15W

我想要这样的东西:
   Station       latHr latMin   latSec  latDir   lonHr lonMin  lonSec lonDir
1940 K01R 31 08 00 N 092 34 00 W
1941 K01T 28 08 00 N 094 24 00 W
1942 K03Y 48 47 00 N 096 57 00 W
1943 K04V 38 05 50 N 106 10 07 W
1944 K05F 31 25 16 N 097 47 49 W
1945 K06D 48 53 04 N 099 37 15 W

我可以匹配到这个正则表达式:
data.format <- "\\d{1,3}-\\d{1,3}(?:-\\d{1,3})?[NSWE]{1}"
grep(data.format, wthrStat1$lat)

但我不确定如何将各个部分放入列中。我尝试了一些事情,例如:
wthrStat1$latHr <- ifelse(grepl(data.format, wthrStat1$lat), gsub(????), NA)

但没有运气。

这是一个 dput():
> dput(wthrStat1[1:10,] )
structure(list(Station = c("K01R", "K01T", "K03Y", "K04V", "K05F",
"K06D", "K07G", "K07S", "K08D", "K0B9"), lat = c("31-08N", "28-08N",
"48-47N", "38-05-50N", "31-25-16N", "48-53-04N", "42-34-28N",
"47-58-27N", "48-18-03N", "43-20N"), lon = c("092-34W", "094-24W",
"096-57W", "106-10-07W", "097-47-49W", "099-37-15W", "084-48-41W",
"117-25-42W", "102-24-23W", "070-24W")), .Names = c("Station",
"lat", "lon"), row.names = 1940:1949, class = "data.frame")

有什么建议?

最佳答案

strapplyc在 gsubfn 包中将提取正则表达式中括号括起来的每个组:

library(gsubfn)
data.format <- "(\\d{1,3})-(\\d{1,3})-?(\\d{1,3})?([NSWE]{1})"
parts <- strapplyc(wthrStat1$lat, data.format, simplify = rbind)
parts[parts == ""] <- "00"

这使:
> parts
[,1] [,2] [,3] [,4]
[1,] "31" "08" "00" "N"
[2,] "28" "08" "00" "N"
[3,] "48" "47" "00" "N"
[4,] "38" "05" "50" "N"
[5,] "31" "25" "16" "N"
[6,] "48" "53" "04" "N"
[7,] "42" "34" "28" "N"
[8,] "47" "58" "27" "N"
[9,] "48" "18" "03" "N"
[10,] "43" "20" "00" "N"

关于R 正则表达式/gsub : extract part of pattern,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/14700799/

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