Question

我有以下基本格式的文本文件，该文件重复了数千次：

Patient Name- John Smith
Number of dx codes: 123
Number of pr codes: 678
Charges: 910
Lorem ipsum dolor sit amet, consectetur adipiscing elit. 
Duis arcu ipsum, ultrices placerat mattis ac, venenatis eu magna. 
Donec interdum iaculis lacus. Nunc in placerat augue. 
In ut odio et dui aliquam sagittis at id augue. 
Patient Name- Jane Smith
Number of dx codes: 234
Number of pr codes: 567
Charges: 1011

如何最好地将以上文本转换为以下格式

Patient Name    DxCodes    PrCodes    Charges
John Smith      123        678        910
Jane Smith      234        567        1011

我已经能够使用stringi包中的str_extract将所有患者姓名提取到一个数据帧中，并将DxCodes，PrCodes和Charges提取到另一个数据帧中，例如：

Names
John Smith
Jane Smith

和

Number of dx codes: 123
Number of pr codes: 678
Charges: 910
Number of dx codes: 234
Number of pr codes: 567
Charges: 1011

但是不确定如何将上述两个数据帧转换为所需格式吗？我是否应该从一开始就使用其他方法？一定会感谢任何帮助。谢谢！

Answer 1

您可以使用一系列正则表达式，然后将其与data.frame()组装在一起。

inx1 <- grep("Patient Name", txt)
inx2 <- grep("Number of dx codes:", txt)
inx3 <- grep("Number of pr codes:", txt)
inx4 <- grep("Charges", txt)

PatientName <- sub("^Patient Name[- ]*", "", txt[inx1])
DxCodes <- sub("^.*: *([[:digit:]]*)$", "\\1", txt[inx2])
PrCodes <- sub("^.*: *([[:digit:]]*)$", "\\1", txt[inx3])
Charges <- sub("^.*: *([[:digit:]]*)$", "\\1", txt[inx4])

DxCodes <- as.integer(DxCodes)
PrCodes <- as.integer(PrCodes)
Charges <- as.integer(Charges)

result <- data.frame(PatientName, DxCodes, PrCodes, Charges)
result
#  PatientName DxCodes PrCodes Charges
#1  John Smith     123     678     910
#2  Jane Smith     234     567    1011

数据。

conn <- textConnection("
Patient Name- John Smith
Number of dx codes: 123
Number of pr codes: 678
Charges: 910
Lorem ipsum dolor sit amet, consectetur adipiscing elit. 
Duis arcu ipsum, ultrices placerat mattis ac, venenatis eu magna. 
Donec interdum iaculis lacus. Nunc in placerat augue. 
In ut odio et dui aliquam sagittis at id augue. 
Patient Name- Jane Smith
Number of dx codes: 234
Number of pr codes: 567
Charges: 1011
")

txt <- readLines(conn)
close(conn)

Answer 2

这里是一种假设，它假定患者文本块中的消息顺序。

数据：

@SpringBootApplication
public class ServerApplication {
    public static void main(String[] args) {
        MyResourceLoader rLoader = new MyResourceLoader();
        SpringApplication app = new SpringApplicationBuilder().build();
        app.addPrimarySources(Arrays.asList(ServerApplication.class));
        app.setResourceLoader(rLoader);
        app.run(args);
    }
}

@RestController
@RequestMapping("/user")
public class UserController {
    @RequestMapping("/hello")
    public String hello(){
        System.out.println("hello hello hello...");
        return "hello";
    }
}

public class MyResourceLoader extends DefaultResourceLoader {
    private ClassLoader cl = new MyClassLoader();
    @Override
    public Resource getResource(String location) {
        System.out.println("getResource: "+location);
        return super.getResource(location);
    }
    @Override
    public ClassLoader getClassLoader() {
        return cl;
    }
}

public class MyClassLoader extends ClassLoader {
    @Override
    public Class<?> loadClass(String name) throws ClassNotFoundException {
        if (name.equals("com.abc.bootbycustomloader.controller.UserController")) {
            // assump that UserController is the encrypted class
            // i need to load this encrypted class, and decrypted it!
            System.out.println("!!!!!encrypted!!!!! : " + name);

            // load the class from a special place, mock the decrypted processing
            String path = "D:\\_clz\\UserController.class";
            byte[] data = new byte[0];
            try {
                data = Files.readAllBytes(Paths.get(path));
            } catch (IOException e) {
                e.printStackTrace();
            }
            // mock decrypted processing success, return the decrypted class
            Class<?> clz = defineClass(name, data, 0, data.length);  
            return clz;
        } else {
            // assump that other class is not encrypted class
            // just load it as usual
            return super.loadClass(name);
        }
    }
}

将患者分成单独的载体：

txt <- c(
  'Patient Name- John Smith',
  'Number of dx codes: 123',
  'Number of pr codes: 678',
  'Charges: 910',
  'Lorem ipsum dolor sit amet, consectetur adipiscing elit. ',
  'Duis arcu ipsum, ultrices placerat mattis ac, venenatis eu magna. ',
  'Donec interdum iaculis lacus. Nunc in placerat augue. ',
  'In ut odio et dui aliquam sagittis at id augue. ',
  'Patient Name- Jane Smith',
  'Number of dx codes: 234',
  'Number of pr codes: 567',
  'Charges: 1011')

针对每位患者，分析相关部分。假定行的顺序（名称，dx，pr，电荷）是静态的，但可以轻松扩展。

patients <- split(txt, cumsum(grepl("^Patient Name", txt)))
str(patients)
# List of 2
#  $ 1: chr [1:8] "Patient Name- John Smith" "Number of dx codes: 123" "Number of pr codes: 678" "Charges: 910" ...
#  $ 2: chr [1:4] "Patient Name- Jane Smith" "Number of dx codes: 234" "Number of pr codes: 567" "Charges: 1011"

现在组合成一个框架。

patients2 <- lapply(patients, function(pat) {
  nm <- sapply(strsplit(pat[1], "-")[[1]][-1], trimws)
  dx <- as.integer(strsplit(pat[2], ":")[[1]][2])
  pr <- as.integer(strsplit(pat[3], ":")[[1]][2])
  ch <- as.integer(strsplit(pat[4], ":")[[1]][2])
  rest <- paste(pat[-(1:4)], collapse="\n")
  data.frame(name = nm, dx = dx, pr = pr, charges = ch, rest = rest,
             stringsAsFactors = FALSE)
})
str(patients2)
# List of 2
#  $ 1:'data.frame':    1 obs. of  5 variables:
#   ..$ name   : chr "John Smith"
#   ..$ dx     : int 123
#   ..$ pr     : int 678
#   ..$ charges: int 910
#   ..$ rest   : chr "Lorem ipsum dolor sit amet, consectetur adipiscing elit. \nDuis arcu ipsum, ultrices placerat mattis ac, venenatis eu magna. \n"| __truncated__
#  $ 2:'data.frame':    1 obs. of  5 variables:
#   ..$ name   : chr "Jane Smith"
#   ..$ dx     : int 234
#   ..$ pr     : int 567
#   ..$ charges: int 1011
#   ..$ rest   : chr ""

Answer 3

如果您的文本确实如您所展示的那样，是一个连续的块或一个连续的字符串，则可以使用捕获组来做到这一点，假设每个记录都有dx，pr和电荷：

library(stringr)
library(dplyr)
df <- " 
Patient Name- John Smith
Number of dx codes: 123
Number of pr codes: 678
Charges: 910
Lorem ipsum dolor sit amet, consectetur adipiscing elit. 
Duis arcu ipsum, ultrices placerat mattis ac, venenatis eu magna. 
Donec interdum iaculis lacus. Nunc in placerat augue. 
In ut odio et dui aliquam sagittis at id augue. 
Patient Name- Jane Smith
Number of dx codes: 234
Number of pr codes: 567
Charges: 1011"

    df_b <- data.frame(dx=str_match_all(df, "(?<=dx codes:) [[:digit:]]*"), 
              pr=str_match_all(df, "(?<=pr codes:) [[:digit:]]*"),
    charges=str_match_all(df,"(?<=harges:) [[:digit:]]*")) 
    names(df_b) <- c("dx", "pr", "charges")
# it changed names by the structure but you may rename it easily:
df
    dx   pr charges
1  123  678     910
2  234  567    1011

使用R读取文本文件并将提取的数据格式化为表格

3 个答案: