Question

我有一个数据库结构 - 下面的缩写版本

structure(list(sex1 = c("totalmaleglobal", "totalfemaleglobal", 
"totalglobal", "totalfemaleGSK", "totalfemaleglobal", 
"totalfemaleUN")), .Names = "sex1", row.names = c(NA, 6L),
class="data.frame")

我想提取“总计＆＃39;”，“总人数”，“总人数”等字样

怎么做？

我使用以下代码尝试了正则表达式

pattern1=c("total")
pattern2=c("totalmale")
pattern3=c("totalfemale")

daly$sex <- str_extract(daly$sex1,pattern1)
daly$sex <- str_extract(daly$sex1,pattern2)
daly$sex <- str_extract(daly$sex1,pattern3)

但它给了我NA。

Answer 1

也许

library(stringr)
daly$sex <- str_extract(daly$sex1,paste(rev(mget(ls(pattern = "pattern\\d+"))), collapse="|"))
daly
#                sex1         sex
# 1   totalmaleglobal   totalmale
# 2 totalfemaleglobal totalfemale
# 3       totalglobal       total
# 4    totalfemaleGSK totalfemale
# 5 totalfemaleglobal totalfemale
# 6     totalfemaleUN totalfemale

Answer 2

gsub，

的两个步骤

v2 <- gsub(paste(v1, collapse='|'), '', d1$sex1)

gsub(paste(v2, collapse='|'), '', d1$sex1)
#[1] "totalmale"   "totalfemale" "total"       "totalfemale" "totalfemale" "totalfemale"

，其中

v1 <- c('total', 'totalmale', 'totalfemale')

Answer 3

试试这个：

{{1}}

Answer 4

我们还可以对所需模式（sapply向量）执行grepl和s1（在基数R中），如下所示：

x <- sapply(s1,function(x) grepl(x, d1$sex1))
colnames(x)[max.col(x, ties.method = "first")]

# [1] "totalmale" "totalfemale" "total" "totalfemale" "totalfemale" "totalfemale"

其中

s1 <- c("totalmale", "totalfemale", "total")

从字符串中提取匹配的单词

4 个答案: