Question

我正在尝试计算以 AK 开头且在交易中也包含 AK 的交易数量，但不要以 AK结尾

示例：

排除的：示例：AK-＆gt; se（中间没有AK）

EXCLUDE ：AK-＆gt; gg-＆gt; se-＆gt; ll：在交易中不包含 AK

包含： 的 例如： AK-＆GT; SE-＆GT; AK-＆GT; GG

示例数据：

f<- data.frame(
id=c("A","A","A","A","C","C","D","D","E"),
Mode=c("AK->se","se->AK->gg, bishan->K","AK->se","se->gr->gg, bishan->AK","AK->se","se->gr->gg, bishan->AK","AK->se","se->gr->gg, bishan->AK","se->AK->df, hg->pp->sk")
)

我需要处理大量数据，因此优化至关重要。

提前致谢。

被修改

f<- data.frame(
id=c("A","A","A","A","C","C","D","D","E"),
Mode=c("AK->se","se->AK->gg, bishan->K","AK->se","se->gr->gg, bishan->AK","AK->se","AK->AK->gg, bishan->AK","AK->se->Ak->gg","se->gr->gg, bishan->AK","AK->AK->df, hg->pp->sk")
)

Answer 1

使用正则表达式

f<- data.frame(
  id=c("A","A","A","A","C","C","D","D","E"),
  Mode=c("AK->se","se->AK->gg, bishan->K","AK->se","se->gr->gg, bishan->AK","AK->se","se->gr->gg, bishan->AK","AK->se->AK->gg","se->gr->gg, bishan->AK","se->AK->df, hg->pp->sk")
)

selection = grepl(pattern="^AK->.*AK->",x=f$Mode,perl=TRUE)
f$Mode[selection]
f$id[selection]

使用lapply（如果有很多字符串，可能会慢一点）

f<- data.frame(
  id=c("A","A","A","A","C","C","D","D","E"),
  Mode=c("AK->se","se->AK->gg, bishan->K","AK->se","se->gr->gg, bishan->AK","AK->se","se->gr->gg, bishan->AK","AK->se->AK->gg","se->gr->gg, bishan->AK","se->AK->df, hg->pp->sk")
)

selection = sapply(strsplit(x=f$Mode,split="->"),FUN=function(x) (x[1]=="AK")&(x[length(x)]!="AK")&(sum(x=="AK")>1))
f$Mode[selection]
f$id[selection]

检测R中的部分字符串匹配

1 个答案: