在R中，如何提取所有文本直到左括号？

Question

在数据框的“获胜者”列中，我想删除所有从左括号开始的文本。

搜索stackoverflow.com，我发现this response，并且在代码中应用了它的更严格的解决方案，但是它不起作用。我的代码未更改我的输入。

我将很乐意为您解决此问题。

输入：

Year    Lg  Winner                  Team
1956    NL  Don Newcombe (1 | MVP)  Brooklyn (1)
1957    NL  Warren Spahn (1 | HOF | ASG)    Milwaukee (1)
1958    AL  Bob Turley (1 | ASG)    New York (1)

这是我希望输出显示的样子：

Year    Lg  Winner                  Team
1956    NL  Don Newcombe            Brooklyn (1)
1957    NL  Warren Spahn            Milwaukee (1)
1958    AL  Bob Turley              New York (1)

dput（dfx）：

structure(list(Year = 1956:1958, Lg = structure(c(2L, 2L, 1L), .Label = c("AL", 
"NL"), class = "factor"), Winner = structure(c(2L, 3L, 1L), .Label = c("Bob Turley (1 | ASG)", 
"Don Newcombe (1 | MVP)", "Warren Spahn (1 | HOF | ASG)"
), class = "factor"), Team = structure(1:3, .Label = c("Brooklyn (1)", 
"Milwaukee (1)", "New York (1)"), class = "factor")), class = "data.frame", row.names = c(NA, 
-3L))

代码：

library(stringr)
dfnoparens <- dfx
str_replace(dfnoparens$Winner, " \\(.*\\)", "")
head(dfnoparens)

Answer 1

带有问题中的测试数据（仅相关列）。

x <- c('Don Newcombe (1 | MVP)', 'Warren Spahn (1 | HOF | ASG)', 'Bob Turley (1 | ASG)')

使用regexpr/regmatches。

m <- regexpr('^[^\\(]*', x)
y <- regmatches(x, m)
y
#[1] "Don Newcombe " "Warren Spahn " "Bob Turley "

此输出字符串在左括号之前仍然有空白，如果需要，请立即将其删除。

trimws(y)
#[1] "Don Newcombe" "Warren Spahn" "Bob Turley"

Answer 2

df <- structure(list(Year = 1956:1958, 
                     Lg = structure(c(2L, 2L, 1L), .Label = c("AL", "NL"), class = "factor"), 
                     Winner = structure(c(2L, 3L, 1L), 
                                        .Label = c("Bob Turley (1 | ASG)", "Don Newcombe (1 | MVP)", 
                                                   "Warren Spahn (1 | HOF | ASG)"), class = "factor"),
                     Team = structure(1:3, .Label = c("Brooklyn (1)", "Milwaukee (1)", "New York (1)"), 
                                      class = "factor")), class = "data.frame", row.names = c(NA,-3L))

这里是strsplit解决方案。

df$Winner <- unlist(lapply(strsplit(as.character(df$Winner)," (",fixed=TRUE), `[[`, 1))
df
  Year Lg       Winner          Team
1 1956 NL Don Newcombe  Brooklyn (1)
2 1957 NL Warren Spahn Milwaukee (1)
3 1958 AL   Bob Turley  New York (1)

Answer 3

我们可以将trimws与whitespace一起使用

 trimws(x, whitespace = "\\s*\\(.*")
 #[1] "Don Newcombe" "Warren Spahn" "Bob Turley"  

数据

x <- c('Don Newcombe (1 | MVP)', 'Warren Spahn (1 | HOF | ASG)', 'Bob Turley (1 | ASG)')

Answer 4

使用str_extract库中的stringr：

df$Winner <- str_extract(df$Winner, ".*(?=\\s\\(\\d)")

此解决方案在(?=...)中使用正向超前；可以将前瞻表述为“匹配在空白（.*）之前出现的任何内容（\\s），后跟一个圆括号（\\(）和一个数字（\\d）”。

结果：

df
  Year Lg       Winner          Team
1 1956 NL Don Newcombe  Brooklyn (1)
2 1957 NL Warren Spahn Milwaukee (1)
3 1958 AL   Bob Turley  New York (1)