问题描述
这个问题似乎很容易删除R中的字符串中的空格字符.但是,当我加载下表时,我无法删除两个数字之间的空格(例如11 846.4
):
This question seems to make it easy to remove space characters in a string in R. However when I load the following table I'm not able to remove a space between two numbers (eg.11 846.4
):
require(XML)
require(RCurl)
require(data.table)
link2fetch = 'https://www.destatis.de/DE/Themen/Branchen-Unternehmen/Landwirtschaft-Forstwirtschaft-Fischerei/Feldfruechte-Gruenland/Tabellen/ackerland-hauptnutzungsarten-kulturarten.html'
theurl = getURL(link2fetch, .opts = list(ssl.verifypeer = FALSE) ) # important!
area_cult10 = readHTMLTable(theurl, stringsAsFactors = FALSE)
area_cult10 = rbindlist(area_cult10)
test = sub(',', '.', area_cult10$V5) # change , to .
test = gsub('(.+)\\s([A-Z]{1})*', '\\1', test) # remove LETTERS
gsub('\\s', '', test[1]) # remove white space?
为什么不能删除test[1]
中的空格?感谢您的任何建议!除了空格字符,这还可以吗?也许答案真的很简单,但我忽略了某些事情.
Why can't I remove the space in test[1]
?Thanks for any advice! Can this be something else than a space character? Maybe the answer is really easy and I'm overlooking something.
推荐答案
您可以将test
创建缩短到2个步骤,并且仅使用1个 PCRE 正则表达式(请注意perl=TRUE
参数) :
You may shorten the test
creation to just 2 steps and using just 1 PCRE regex (note the perl=TRUE
parameter):
test = sub(",", ".", gsub("(*UCP)[\\s\\p{L}]+|\\W+$", "", area_cult10$V5, perl=TRUE), fixed=TRUE)
结果:
[1] "11846.4" "6529.2" "3282.7" "616.0" "1621.8" "125.7" "14.2"
[8] "401.6" "455.5" "11.7" "160.4" "79.1" "37.6" "29.6"
[15] "" "13.9" "554.1" "236.7" "312.8" "4.6" "136.9"
[22] "1374.4" "1332.3" "1281.8" "3.7" "5.0" "18.4" "23.4"
[29] "42.0" "2746.2" "106.6" "2100.4" "267.8" "258.4" "13.1"
[36] "23.5" "11.6" "310.2"
gsub
正则表达式值得特别注意:
The gsub
regex is worth special attention:
-
(*UCP)
-将模式强制为Unicode识别的PCRE动词 -
[\\s\\p{L}]+
-匹配1+个空格或字母字符 -
|
-或(交替运算符) -
\\W+$
-字符串末尾有1个以上的非单词字符.
(*UCP)
- the PCRE verb that enforces the pattern to be Unicode aware[\\s\\p{L}]+
- matches 1+ whitespace or letter characters|
- or (an alternation operator)\\W+$
- 1+ non-word chars at the end of the string.
然后,sub(",", ".", x, fixed=TRUE)
会将第一个,
替换为.
作为文字字符串,由于不必编译正则表达式,因此fixed=TRUE
可以节省性能.
Then, sub(",", ".", x, fixed=TRUE)
will replace the first ,
with a .
as literal strings, fixed=TRUE
saves performance since it does not have to compile a regex.
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