Question

说我有这个数据：

df <- data.frame(
   text = c("Treatment1: This text is","on two lines","","Treatment2:This text","has","three lines","","Treatment3: This has one")
                  )
df
                      text
1 Treatment1: This text is
2             on two lines
3                         
4     Treatment2:This text
5                      has
6              three lines
7                         
8 Treatment3: This has one

我如何解析这个文本，以便所有“处理”都在他们自己的行上，同一行下面的所有文本都是？

例如，这是所需的输出：

text
1 Treatment1: This text is on two lines
2 Treatment2: This text has three lines                
3 Treatment3: This has one

有人可以推荐一种方法吗？

Answer 1

可能类似以下内容首先是dput格式的数据，这是在帖子中共享数据集的最佳格式。

df <-
structure(list(text = c("Treatment1: This text is", "on two lines", 
"", "Treatment2:This text", "has", "three lines", "", "Treatment3: This has one"
)), .Names = "text", class = "data.frame", row.names = c("1", 
"2", "3", "4", "5", "6", "7", "8"))

现在是base R代码。

fact <- cumsum(grepl("treatment", df$text, , ignore.case = TRUE))
result <- do.call(rbind, lapply(split(df, fact), function(x)
                     trimws(paste(x$text, collapse = " "))))
result <- as.data.frame(result)
names(result) <- "text"
result
#                                   text
#1 Treatment1: This text is on two lines
#2  Treatment2:This text has three lines
#3              Treatment3: This has one

修改
正如Rich Scriven在评论中指出的那样，tapply可以大大简化上面的代码。（我没有看到，有时候我太复杂了。）

result2 <- data.frame( text = tapply(df$text, fact, function(x) trimws(paste(x, collapse = " "))) ) all.equal(result, result2) #[1] "Component “text”: 'current' is not a factor"

Answer 2

x <- gsub("\\s+Treatment", "*BREAK*Treatment",
          as.character(paste(df[[1]], collapse = " ")))
data.frame(text = unlist(strsplit(x, "\\*BREAK\\*")))

根据条件将多行字符串折叠为一行。

2 个答案: