矩阵中的R列表到数据帧转换

Question

R挣扎。我正在使用以下内容从文本中提取引文，在大型数据集上有多个结果。我试图让输出成为数据帧中的字符串，因此我可以轻松地将其作为 csv 与其他人共享。

示例数据：

normalCase <- 'He said, "I am a test," very quickly.'
endCase <- 'This is a long quote, which we said, "Would never happen."'
shortCase <- 'A "quote" yo';
beginningCase <- '"I said this," he said quickly';
multipleCase <- 'When asked, "No," said Sam "I do not like green eggs and ham."'
testdata = c(normalCase,endCase,shortCase,beginningCase,multipleCase)

使用以下内容提取引号和字符缓冲区：

result <-function(testdata) {
  str_extract_all(testdata, '[^\"]?{15}"[^\"]+"[^\"]?{15}')
}
extract <- sapply(testdata, FUN=result)

摘录是矩阵中的列表。但是，我希望提取物是一个字符串，稍后我可以将其作为列合并到数据框中。我该如何转换？

Answer 1

代码

normalCase <- 'He said, "I am a test," very quickly.'
endCase <- 'This is a long quote, which we said, "Would never happen."'
shortCase <- 'A "quote" yo';
beginningCase <- '"I said this," he said quickly';
multipleCase <- 'When asked, "No," said Sam "I do not like green eggs and ham."'
testdata = c(normalCase,endCase,shortCase,beginningCase,multipleCase)

# extract quotations
gsub(pattern = "[^\"]*((?:\"[^\"]*\")|$)", replacement = "\1 ", x = testdata)

输出

[1] "\"I am a test,\"  "                            
[2] "\"Would never happen.\" "                      
[3] "\"quote\"  "                                   
[4] "\"I said this,\"  "                            
[5] "\"No,\" \"I do not like green eggs and ham.\" "

说明

pattern = "[^\"]" 将匹配除双引号外的任何字符
pattern = "[^\"]*" 将与除双引号之外的任何字符匹配 0 次或更多次
pattern = "\"[^\"]*\"" 将匹配双引号，然后是任何除双引号 0 次或多次外的字符，然后是另一个双引号引用（即）引用
pattern = "(?:\"[^\"]*\")" 将匹配引号，但不会捕获它
pattern = "((?:\"[^\"]*\")|$)" 将匹配引号或 endOfString，并捕获它。请注意，这是我们捕获的第一组
replacement = "\1 " 将替换为我们捕获的第一组，然后是 space

矩阵中的R列表到数据帧转换

R list within matrix to dataframe conversion

r

list

dataframe

代码

输出

说明