更换日期的特征向量，以特定的格式

Question

我给出如下特征向量：

"On the evening of 2017-04-23, I was too tired" 
"to complete my homework that was due on 24.04.2017." 

我需要通过它来搜索日期的所有出现，并与格式MONTHNAME d，YYYY替换它们。

我知道一般格式应该是％B％d，％Y，我可能必须使用sub()函数，但我不太确定如何将两者结合在一起。

当我尝试像

sub("[0-9]{2}.[0-9]{2}.[0-9]{4}","%B %d, %Y",x) 

我刚刚得到以下结果

"On the evening of 2001-01-15, I was too tired to complete my homework that was due on %B %d, %Y." 

可能有人请帮助我弄清楚如何把它一起？

---

我与同伴stackoverflowers的帮助下新的代码如下：

streamlineDates(x) 
{ 
#set pattern to dates in form of YYYY-MM-DD or DD.MM.YYYY 
pattern <- "\\d{2,4}[.-]\\d{2}[.-]\\d{2,4}" 

y <- c(x) 

val <- unlist(regmatches(y, gregexpr(pattern, y))) 

val1 <- as.Date(val,format=c("%Y-%m-%d","%d.%m.%Y")) 
val2 <- format(val1,"%B %d, %Y") 

y1 <- list() 
for (i in 1:length(y)){ 
    y1[i] <- gsub(pattern,val2[i],y[i]) 
} 
} 

然而，当我只输入：

x <- "to complete my homework that was due on 24.04.2017." 

...它只返回NA。我已将问题范围缩小到gsub，其中替换值值，“如果NA，则结果中对应于匹配的所有元素将被设置为NA”。因此，当仅输入最后一行时缺少第一个日期，它仅返回NA。

我该如何让它接受一个或两个日期？

Answer 1

第一方法：

使用基础R溶液（不使用任何包）：

pattern <- "\\d{2,4}[.-]\\d{2}[.-]\\d{2,4}" 
rep <- c("On the evening of 2017-04-23, I was too tired","to complete my homework that was due on 24.04.2017.") 


val <- unlist(regmatches(rep, gregexpr(pattern, rep))) 

val1 <- as.Date(val,format=c("%Y-%m-%d","%d.%m.%Y")) 
val2 <- format(val1,"%B %d, %Y") 
val2 
rep1 <- list() 
for (i in 1:length(rep)){ 
rep1[i] <- gsub(pattern,val2[i],rep[i]) 
} 

答案：

do.call("c",rep1) 

> do.call("c",rep1)             
[1] "On the evening of April 23, 2017, I was too tired"  
[2] "to complete my homework that was due on April 24, 2017." 
> 

第2种方法：

使用图书馆stringr

library(stringr) 
rep <- c("On the evening of 2017-04-23, I was too tired","to complete my homework that was due on 24.04.2017.") 
val <- str_extract(rep,"\\d{2,4}[.-]\\d{2}[.-]\\d{2,4}") 
val1 <- as.Date(val,format=c("%Y-%m-%d","%d.%m.%Y")) 
val2 <- format(val1,"%B %d, %Y") 
rep1 <- str_replace_all(rep,"\\d{2,4}[.-]\\d{2}[.-]\\d{2,4}",val2) 
rep1 

答：但它

> rep1 
[1] "On the evening of April 23, 2017, I was too tired"  
[2] "to complete my homework that was due on April 24, 2017." 
> 

编辑OP之后已经改变的问题一点，解决的办法是更通用，假设该月将始终处于中间位置，并且分隔符仅限于破折号（ - ）和点（。）：

pattern <- "\\d{2,4}[.-]\\d{2}[.-]\\d{2,4}" 
rep <- c("On the evening of 2017-04-23, I was too tired","to complete my homework that was due on 24.04.2017.") 


val <- unlist(regmatches(rep, gregexpr(pattern, rep))) 

year <- regmatches(val, gregexpr("\\d{4}", val)) 

month <- regmatches(val, gregexpr("(?<=[.-])\\d{1,2}(?=[.-])", val,perl=T)) 

date <- regmatches(val, gregexpr("(?<=[.-])\\d{2}$|^\\d{2}(?=[.-])", val,perl=T)) 
#Extracting year month and date , assuming month always falls in middle string 

date1 <- paste0(year,"-",month,"-",date) 
date1 <- as.Date(date1,"%Y-%m-%d") 
val2 <- format(date1,"%B %d, %Y") 

rep1 <- list() 
for (i in 1:length(rep)){ 
    rep1[i] <- gsub(pattern,val2[i],rep[i]) 
} 


do.call("c",rep1) 

Answer 2

首先您需要指定日期的所有格式。然后转换为日期，使用的格式，让您所需的输出，即

#Note that I don't use any delimiter in the formatting simply because 
#I will use gsub to replace all except the numbers with '' from the string 
v1 <- c('%Y%m%d', '%d%m%Y') 

format(as.Date(gsub('\\D+', '', x), format = v1), "%B %d, %Y") 
#[1] "April 23, 2017" "April 24, 2017" 

可以使用（一个比较难看）从stringr包str_replace_all正则表达式，即

stringr::str_replace_all(x, '\\d+-\\d+-\\d+|\\d+\\.\\d+\\.\\d+', 
         format(as.Date(gsub('\\D+', '', x), format = v1), "%B %d, %Y")) 

#[1] "On the evening of April 23, 2017, I was too tired"  
#[2] "to complete my homework that was due on April 24, 2017."