Question

我目前正在处理日志文件中的数据，如下所示：

Subject: MS1989, Characters: G3, Position: TL, Block: 2, Reaction Time: 957 ms

Subject: MS1989, Characters: 7A, Position: TR, Block: 2, Reaction Time: 220 ms

如何对此进行转换，使其看起来更像这样：

Subject Characters Position Block Reaction Time

MS1989 G3 ...

之前可能已经回答了，但我真的不知道怎么说这个，所以搜索很难。也许重塑可能在这里很有用，但这似乎不是长格式或宽格式。

Answer 1

我确信可以有更好的解决方案，

library(tidyr)
df = data.frame(x = c("Subject: MS1989, Characters: G3, Position: TL, Block: 2,  Reaction Time: 957 ms",
  "Subject: MS1989, Characters: 7A, Position: TR, Block: 2, Reaction Time: 220 ms"))

df$y <- gsub("Reaction Time:", "", 
         gsub("Block:", "",
              gsub("Position:", "",
                   gsub("Characters:", "",
                        gsub("Subject:" ,"",df$x)))))

df$x <- NULL
df <- df %>% separate(y, into = c("Subject", "Character", "Position", "Block", "Reaction"),
            sep = "[,]")

df
   Subject Character Position Block Reaction
1  MS1989        G3       TL     2   957 ms
2  MS1989        7A       TR     2   220 ms

Answer 2

将数据复制到文本文件＆＃34; tidy.txt＆＃34;：

library(tidyverse)

untidy  <- read_csv("untidy.txt", col_names=FALSE)
tidier <- untidy %>% 
          separate_rows(col=1:ncol(untidy), sep=":") %>% 
          distinct()
colnames(tidier)  <- as.character(tidier[1,])
tidy <- tidier[-1,]

然后：

$input = '[image]http://picture.com/1.jpg[/image]';
$output_pf = '<img src="%s">';

if(preg_match('#\[image\](.+)\[/image\]#', $input, $matches)) {
    $output = sprintf($output_pf, $matches[1]);
}

print $output;

如何在r中格式化此数据？

2 个答案: