r语言 - 将"firstname lastname"更改为"lastname, firstname"

Question

我有一个名称列表，需要将其从"Firstname Lastname"转换为"Lastname，Firstname"。

Barack Obama
Donald J. Trump
J. Edgar Hoover
Beyonce Knowles-Carter
Sting

我用G.Grothendieck对"；姓、名"->quot；名字姓氏"；在序列化的字符串中到达gsub("([^ ]*) ([^ ]*)", "\2, \1", str)，这给了我-

Obama, Barack
J., DonaldTrump, 
Edgar, J.Hoover, 
Knowles-Carter, Beyonce
Sting

我想要什么-

Obama, Barack
Trump, Donald J. 
Hoover, J. Edgar
Knowles-Carter, Beyonce
Sting

我想要一个正则表达式的答案。

Answer 1

有一个名为person的深奥函数，它是为保存名称而设计的，一个为您进行解析的转换函数as.person，以及一个稍后使用它的format方法(创造性地使用大括号参数(。它甚至适用于复杂的姓氏(如范尼斯特鲁伊(，但单一姓氏的结果并不令人满意。不过，它可以用一个快速结束的sub来修复。

x <- c("Barack Obama","Donald J. Trump","J. Edgar Hoover","Beyonce Knowles-Carter","Sting", "Ruud van Nistelrooy", "John von Neumann")
y <- as.person(x)
format(y, include=c("family","given"), braces=list(family=c("",",")))
[1] "Obama, Barack"           "Trump, Donald J."       
[3] "Hoover, J. Edgar"        "Knowles-Carter, Beyonce"
[5] "Sting,"                  "van Nistelrooy, Ruud"   
[7] "von Neumann, John"
## fix for single names - curse you Sting!
sub(",$", "", format(y, include=c("family","given"), braces=list(family=c("",","))))
[1] "Obama, Barack"           "Trump, Donald J."       
[3] "Hoover, J. Edgar"        "Knowles-Carter, Beyonce"
[5] "Sting"                   "van Nistelrooy, Ruud"   
[7] "von Neumann, John" 

Answer 2

使用

gsub("(.*[^van])\s(.*)", "\2, \1", people)

正则表达式：

(.*[^van])                                         \s                      (.*)
Any ammount of characters exluding "van"...  the last white space...  The last name containing any character.

数据：

people <- c("Barack Obama",
"Donald J. Trump",
"J. Edgar Hoover",
"Beyonce Knowles-Carter",
"Sting",
"Ruud van Nistelrooy",
"Xi Jinping",
"Hans Zimvanmer")

结果：

[1] "Obama, Barack"           "Trump, Donald J."        "Hoover, J. Edgar"       
[4] "Knowles-Carter, Beyonce" "Sting"                   "van Nistelrooy, Ruud"   
[7] "Jinping, Xi"             "Zimvanmer, Hans"