我刚刚获得了邮递员的工作，我需要您的帮助才能保留它。我必须先订购大量邮件，然后再出去递送。

街道严格按顺序编号，从街道开始处的1开始，然后继续以不跳过任何数字的方式一直到结尾，左侧为奇数，右侧为偶数。加上房屋是在以后添加的，因此我们可能还会在公民编号后附加字母。

箱子中包含城市的所有邮件，因此我只能选择街道上的邮件。

我需要您的帮助才能更快地订购邮件。我要你写的是编写函数或完整程序：
-街道名称。
-按照我的路线订购的公民编号列表。
-地址列表（代表邮件箱）。

并输出仅包含我街道地址的地址列表，并按照公民编号列表进行排序。

地址的格式为：

 Person Name/n   
 CivicN Street Name

其中CivicN是“/”可以遵循由一个数字和一个大写字母（10 / B）。如果愿意，可以使用小写字母。

如果街道名称重叠，则将其视为不同的街道：

Church Road  != Saint Lorenz Church Road 

为了简单起见，我们忽略了其余地址（假设每个邮件都相同）

信件必须快速送达，以便赢得最短的答案。

例：

版面：

 1 1/B     3 5 7      9 11 13     
 ==============================    
       2 4       4/B 6           

输入：

"Tea Avenue"
["1","1/B","2","4","3","5","7","4/B","6","9","11","13"]
["Mrs. Pie O. Pinky\n6 Tea Avenue","Ms. Kita I. Omeeha\n6 Tea Avenue","Mr. Raile A. Lee\n26 Uea Grove","Odd O. Nic\n76 Mira Road","Mrs. Fuel Tee\n78 Uea Grove","Ny O. Ondip\n55 Uea Grove","Mrs. Black\n67 Uea Grove","Ollie E.\n11 Tea Avenue","Mr. Urna Li\n75 Mira Road","Ms. Polly\n2 Tea Avenue"]

输出：

Ms. Polly
2 Tea Avenue  

Mrs. Pie O. Pinky 
6 Tea Avenue 

Ms. Kita I. Omeeha
6 Tea Avenue 

Ollie E.
11 Tea Avenue

输入：

"Church Road"
["1","3","5","5/B","2","4","7","7/B","6","9","9/B","11","11/B"]
["Billy Ray V.\n5 Church Roadside East","Ms. Mia\n5 Church Road","Mrs. Dadeos\n9/B Church Road","Dr. Ymin U.\n3 Church Road","Atty. Nerou\n3 Church Road","Ollie A. Chaim\n6 Saint Lorenz Church Road","Ms. Rose\n5 Church Road","Alf Taohy\n79 Berry Road","Ms. Ootr E.\n5 Saint Lorenz Church Road","Lol E.\n21 Berry Road","Ms. Norton\n2 Church Road"]

输出：

Dr. Ymin U.
3 Church Road

Atty. Nerou
3 Church Road

Ms. Mia
5 Church Road

Ms. Rose
5 Church Road

Ms. Norton 
2 Church Road" 

Mrs. Dadeos
9/B Church Road

测试生成器：在线尝试！ 更改string_wrapper_left / right和分隔符以修改格式。

规则：
-禁止出现标准漏洞。
-标准输入/输出方法。
-以字节为单位的最短答案获胜。

果冻，13 个字节

,⁵KỴṪ⁼ɗƇ@ɗ€ẎY

一个完整的程序，它接受三个参数-命名街道上的公民编号列表，邮筒中的地址列表以及您的街道名称-它将按照给定的公民编号顺序打印要发送到的地址。

在线尝试！

怎么样？

,⁵KỴṪ⁼ɗƇ@ɗ€ẎY - Main link: civic numbers, addresses
          €   - for each (civic number):
         ɗ    -   last three links as a dyad - i.e. f(civic number, addresses):
,⁵            -     pair (the civic number) with 3rd program input (the street name)
  K           -     join (that) with a space -- i.e. X=civic number+' '+street name
        @     -     with swapped arguments i.e. f(addresses, X):
       Ƈ      -       filter (the addresses) keeping those for which:
      ɗ       -         last three links as a dyad - i.e. f(address, X):
   Ỵ          -           split (address) at newlines
    Ṫ         -           tail (get the second line)
     ⁼        -           equals (X)?
           Ẏ  - tighten (the list of lists to a single list)
            Y - join with newlines

05AB1E，15 13 12 字节

ε²ðýUʒ¶¡Xk]˜

通过移植@JonathanAllan的Jelly答案 -2个字节，因此请确保对他进行投票！
-1个字节感谢@Grimy。

输入顺序如下：[市民编号列表]，街道名称，[地址列表]。

在线尝试。

说明：

ε       # Map over the (implicit) input-list of civic numbers
 ²      #  Push the second street-input
  ðý    #  And join the two values by a space
    U   #  Pop and store this street + num string in variable `X`
 ʒ      #  Filter the (implicit) input-list of addresses by:
  ¶¡    #   Split on newlines
    Xk  #   Get the index of string `X` (street + num) in this list,
        #   which will be either -1 (not found), 0 (first address-line), or 
        #   1 (second address-line), and only 1 is truthy in 05AB1E
]       # After both the inner filter and outer map:
 ˜      # Flatten the array to remove any empty inner lists
        # (after which the result is output implicitly)

JavaScript（Node.js），58字节

(s,n,a)=>n.flatMap(n=>a.filter(x=>x.split`
`[1]==n+' '+s))

在线尝试！

已评论

(s, n, a) =>          // s = street name, n[] = civic numbers, a[] = addresses
  n.flatMap(n =>      // for each civic number n in n[]:
    a.filter(x =>     //   for each address x in a[]:
      x.split`\n`[1]  //     keep it if the 2nd part is equal to
      == n + ' ' + s  //     n + space + expected street name
    )                 //   end of filter()
  )                   // end of flatMap() (empty entries are discarded)

红宝石，65个字节

->s,o,n{n.grep(/\d\S* #{s}$/).sort_by{|i|o.index i[/\d+(\/.)?/]}}

在线尝试！

Perl 5，51个字节

sub{my($r,$a,@B)=@_;map{$n=$_;grep/
$n $r$/,@B}@$a}

在线尝试！

C＃（Visual C＃交互式编译器），63字节

(a,b,c)=>b.SelectMany(x=>c.Where(l=>l.Split('\n')[1]==x+" "+a))

在线尝试！

Python 3，79 85个字节（感谢squid）

d=lambda s,n,a:n and[k for k in a if k.split('\n')[1]==n[0]+' '+s]+d(s,n[1:],a)

旧：

d=lambda s,n,a:[k for k in a if k.split('\n')[1]==n[0]+' '+s]+d(s,n[1:],a)if n else[]

Python 3，65个字节

lambda s,n,m:[a for i in n for a in m if a.endswith('\n%s '%i+s)]

在线尝试！

也适用于Python 2。

对于i公民号码列表中的每个数字n，该函数会遍历邮件列表中的所有地址a，m并保留公民号码i和街道名称组合s匹配的地址a。这将导致排序和筛选列表。

Kotlin，145个字节

fun p(s:String,o:List<String>,m:List<String>):List<String>
=if(o.size<1)List(0){""}
else
m.filter{it.split("\n")[1]==o[0]+" "+s}+p(s,o.drop(1),m)

在线尝试！