更換日期的特徵向量，以特定的格式

Question

我給出如下特徵向量：

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

我需要通過它來搜索日期的所有出現，並與格式MONTHNAME d，YYYY替換它們。

我知道一般格式應該是％B％d，％Y，我可能必須使用sub()函數，但我不太確定如何將兩者結合在一起。

當我嘗試像

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

我剛剛得到以下結果

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

可能有人請幫助我弄清楚如何把它一起？

---

我與同伴stackoverflowers的幫助下新的代碼如下：

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，當我只輸入：

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

...它只返回NA。我已將問題範圍縮小到gsub，其中替換值值，「如果NA，則結果中對應於匹配的所有元素將被設置爲NA」。因此，當僅輸入最後一行時缺少第一個日期，它僅返回NA。

我該如何讓它接受一個或兩個日期？

Answer 1

第一方法：

使用基礎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種方法：

使用圖書館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." 
> 

編輯OP之後已經改變的問題一點，解決的辦法是更通用，假設該月將始終處於中間位置，並且分隔符僅限於破折號（ - ）和點（。）：

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。然後轉換爲日期，使用的格式，讓您所需的輸出，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（一個比較難看）從stringr包str_replace_all正則表達式，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."