如何以矢量化方式遍历 R 中的字符串答案

【问题标题】：How do I loop through strings in R in a vectorized way如何以矢量化方式遍历 R 中的字符串
【发布时间】：2014-07-06 06:45:20
【问题描述】：

我现在正在学习 R，我无法以有效的方式循环 R，虽然我可以使用 for 循环以非常复杂的方式进行字符串解析，但我对如何在矢量化方式。

例如

#Social security numbers in the United States are represented by
# numbers conforming to the following format:
#
# a leading 0 followed by two digits
# followed by a dash
# followed by two digits
# followed by a dash
# finally followed by four digits
#
# For example 023-45-7890 would be a valid value,
# but 05-09-1995 and 059-2-27 would not be.
#
# Implement the body of the function 'extractSecuNum' below so that it
# returns a numeric vector whose elements are Social Security numbers
# extracted from a text, i.e., a vector of strings representing the text lines,
# passed to the function as its 'text' argument.
# (You can assume that each string in 'text' contains
# either zero or one Social Security numbers.)


extractSecuNum = function(text){
# Write your code here!

x = 1:length(text)
list_of_input = rep(0, length(text))


for (ind in x){
  list_of_input[ind] = sub(' .*', '', sub('^[^0-9]*', '', text[ind]))
}

temp = c()

for (ind in x){
  if(list_of_input[ind] != ''){
    temp = c(temp, list_of_input[ind])
  }
}

temp2 = c()
for (ind in 1:length(temp)){
  temp3 = strsplit(temp[ind], '-')
  temp2 = c(temp2, temp3)
}

final = c()

for(ind in 1:length(temp2)){
  if (sub('0[0-9][0-9]', '', temp2[[ind]][1]) == ''){
    if (sub('[0-9][0-9]', '', temp2[[ind]][2]) == ''){
      if (sub('[0-9]{4}', '', temp2[[ind]][3]) == '')
      { final = c(final, paste(temp2[[ind]][1], temp2[[ind]][2], temp2[[ind]][3], sep='-')) }
    }
    }
  }

return(final)
}

这些是其他类似问题的问题，如果你仔细研究一下，你会发现第二个问题非常复杂和不优雅

https://gist.github.com/anonymous/c1c68121323af19c766c

我认为问题是R中的原子变量是一个数组，我无法访问字符串中的字符

任何建议将不胜感激

【问题讨论】：

你能给你的extractSecuNum提供样本输入吗？
if (nchar(x <- gsub('[^0-9]', '', text)) == 9) paste(substr(x, 1, 3), substr(x, 4, 5), substr(x, 6, 9), sep = '-') 但如果有其他数字恰好包含九位数字，则不可靠。这是可行的，因为该问题假设 SSN 为零或一个。
@extracSecuNum testInput = c( 'For example', '023-45-7890', 'would be a valid value', 'but 05-09-1995', 'and 059-2-27 would not be.', 'Also 011-99-2234 is okay.' ) correctOutput = c( '023-45-7890', '011-99-2234' ) 这些是我的测试输入，它们有效
@rawr 我知道我的解决方案并不严谨，我现在只是想学习在 R 中进行解析的一般方法，在 R 中解析与在 python 中解析确实不同，这有点令人困惑对我来说

标签： string r parsing

【解决方案1】：

extractSecuNum = function(text){
  pattern <- "0\\d{2}-\\d{3}-\\d{4}"
  unlist(regmatches(text,gregexpr(pattern,text)))
}

text <- paste0("fdkmsal ",
               "0",sample(10:99,10),"-",
               sample(100:999,10),"-",
               sample(1000:9999,10), " vaklra")
result <- extractSecuNum(text)

head(text)
# [1] "fdkmsal 034-965-3362 vaklra" "fdkmsal 029-190-2488 vaklra"
# [3] "fdkmsal 055-785-3898 vaklra" "fdkmsal 033-950-5589 vaklra"
# [5] "fdkmsal 025-833-9312 vaklra" "fdkmsal 054-375-5596 vaklra"
result
# [1] "034-965-3362" "029-190-2488" "055-785-3898" "033-950-5589" "025-833-9312"
# [6] "054-375-5596" "057-680-3317" "020-951-1417" "031-996-4757" "068-402-8678"

【讨论】：

pattern <- "0\\d{2}-\\d{3}-\\d{4}" unlist(regmatches(text,gregexpr(pattern,text))) 通常是用字符串解析的解决方案吗（与模式正则表达式略有不同）？