家庭邮件合并（代码高尔夫）

Question

前几天我写了一些邮件合并代码，虽然它有效，但我对这些代码很感兴趣。我想看看它在其他语言中是什么样子。

因此，对于输入，例程采用联系人列表，

Jim,Smith,2681 Eagle Peak,,Bellevue,Washington,United States,98004
Erica,Johnson,2681 Eagle Peak,,Bellevue,Washington,United States,98004
Abraham,Johnson,2681 Eagle Peak,,Bellevue,Washington,United States,98004
Marge,Simpson,6388 Lake City Way,,Burnaby,British Columbia,Canada,V5A 3A6
Larry,Lyon,52560 Free Street,,Toronto,Ontario,Canada,M4B 1V7
Ted,Simpson,6388 Lake City Way,,Burnaby,British Columbia,Canada,V5A 3A6
Raoul,Simpson,6388 Lake City Way,,Burnaby,British Columbia,Canada,V5A 3A6

然后它将具有相同地址和姓氏的行合并到一条记录中。假设行未排序）。代码还应该足够灵活，可以按任何顺序提供字段（因此需要将字段索引作为参数）。对于两个人的家庭，它连接两个名字字段。对于三人或三人以上的家庭，名字设置为“the”，姓氏设置为“surname family”。

Erica and Abraham,Johnson,2681 Eagle Peak,,Bellevue,Washington,United States,98004
Larry,Lyon,52560 Free Street,,Toronto,Ontario,Canada,M4B 1V7
The,Simpson Family,6388 Lake City Way,,Burnaby,British Columbia,Canada,V5A 3A6
Jim,Smith,2681 Eagle Peak,,Bellevue,Washington,United States,98004

我的 C# 实现是：

var source = File.ReadAllLines(@"sample.csv").Select(l => l.Split(','));
var merged = HouseholdMerge(source, 0, 1, new[] {1, 2, 3, 4, 5});

public static IEnumerable<string[]> HouseholdMerge(IEnumerable<string[]> data, int fnIndex, int lnIndex, int[] groupIndexes)
{            
    Func<string[], string> groupby = fields => String.Join("", fields.Where((f, i) => groupIndexes.Contains(i)));

    var groups = data.OrderBy(groupby).GroupBy(groupby);

    foreach (var group in groups)
    {
        string[] result = group.First().ToArray();

        if (group.Count() == 2)
        {
            result[fnIndex] += " and " + group.ElementAt(1)[fnIndex];
        }
        else if (group.Count() > 2)
        {
            result[fnIndex] = "The";
            result[lnIndex] += " Family";
        }

        yield return result;
    }            
}

我不喜欢如何执行 groupby 委托。我希望 C# 是否有某种方法将字符串表达式转换为委托。例如 Func groupby = f => “f[2] + f[3] + f[4] + f[5] + f[1]；”我有一种感觉，类似的事情可能可以用 Lisp 或 Python 来完成。我期待看到其他语言更好的实现。

编辑：社区 wiki 复选框去了哪里？请某些模组修复该问题。

Answer 1

Ruby — 181 155

姓名索引位于代码：a 和 b 中。输入数据来自 ARGF。

a,b=0,1
[*
lt;].map{|i|i.strip.split ?,}.group_by{|i|i.rotate(a).drop 1}.map{|i,j|k,l,m=j
k[a]+=' and '+l[a]if l
(k[a]='The';k[b]+=' Family')if m
puts k*','}

Answer 2

Python - 没有打高尔夫球，

如果输入文件的索引不是 0 和 1，我不确定行的顺序应该是什么

import csv
from collections import defaultdict

class HouseHold(list):
    def __init__(self, fn_idx, ln_idx):
        self.fn_idx = fn_idx
        self.ln_idx = ln_idx

    def append(self, item):
        self.item = item
        list.append(self, item[self.fn_idx])

    def get_value(self):
        fn_idx = self.fn_idx
        ln_idx = self.ln_idx
        item = self.item
        addr = [j for i,j in enumerate(item) if i not in (fn_idx, ln_idx)]
        if len(self) < 3:
            fn, ln = " and ".join(self), item[ln_idx]
        else:
            fn, ln = "The", item[ln_idx]+" Family"
        return [fn, ln] + addr

def source(fname):
    with open(fname) as in_file:
        for item in csv.reader(in_file):
            yield item

def household_merge(src, fn_idx, ln_idx, groupby):
    res = defaultdict(lambda:HouseHold(fn_idx, ln_idx))
    for item in src:
        key = tuple(item[x] for x in groupby)
        res[key].append(item)
    return res.values()

data =  household_merge(source("sample.csv"), 0, 1, [1,2,3,4,5,6,7])
with open("result.csv", "w") as out_file:
    csv.writer(out_file).writerows(item.get_value() for item in data)

Answer 3

Python - 178 个字符

import sys
d={}
for x in sys.stdin:F,c,A=x.partition(',');d[A]=d.get(A,[])+[F]
print"".join([" and ".join(v)+c+A,"The"+c+A.replace(c,' Family,',1)][2<len(v)]for A,v in d.items())

输出

Jim,Smith,2681 Eagle Peak,,Bellevue,Washington,United States,98004
The,Simpson Family,6388 Lake City Way,,Burnaby,British Columbia,Canada,V5A 3A6
Larry,Lyon,52560 Free Street,,Toronto,Ontario,Canada,M4B 1V7
Erica and Abraham,Johnson,2681 Eagle Peak,,Bellevue,Washington,United States,98004

Answer 4

Python 2.6.6 - 287 个字符

这假设您可以硬编码文件名（名为 i）。如果您想从命令行获取输入，则会增加约 16 个字符。

from itertools import*
for z,g in groupby(sorted([l.split(',')for l in open('i').readlines()],key=lambda x:x[1:]), lambda x:x[2:]):
 l=list(g);r=len(l);k=','.join(z);o=l[0]
 if r>2:print'The,'+o[1],"Family,"+k,
 elif r>1:print o[0],"and",l[1][0]+","+o[1]+","+k,
 else:print','.join(o),

---

输出

Erica and Abraham,Johnson,2681 Eagle Peak,,Bellevue,Washington,United States,98004
Larry,Lyon,52560 Free Street,,Toronto,Ontario,Canada,M4B 1V7
The,Simpson Family,6388 Lake City Way,,Burnaby,British Columbia,Canada,V5A 3A6
Jim,Smith,2681 Eagle Peak,,Bellevue,Washington,United States,98004

我相信这可以改进，但已经晚了。

Answer 5

Haskell - 341 321

（根据评论进行更改）。

不幸的是，Haskell 没有标准的 split 函数，这使得这个函数变得相当长。

输入到 stdin，输出到 stdout。

import List
import Data.Ord
main=interact$unlines.e.lines
s[]=[]
s(',':x)=s x
s l@(x:y)=let(h,i)=break(==k)l in h:(s i)
t[]=[]
t x=tail x
h=head
m=map
k=','
e l=m(t.(>>=(k:)))$(m c$groupBy g$sortBy(comparing t)$m s l)
c(x:[])=x
c(x:y:[])=(h x++" and "++h y):t x
c x="The":((h$t$h x)++" Family"):(t$t$h x)
g a b=t a==t b

Answer 6

Lua，434字节

x,y=1,2 s,p,r,a=string.gsub,pairs,io.read,{}for j,b,c,d,e,f,g,h,i in r('*a'):gmatch('('..('([^,]*),'):rep(7)..'([^,]*))\n')
do k=s(s(s(j,b,''),c,''),'[,%s]','')for l,m in p(a)do if not m.f and (m[y]:match(c) and m[9]==k) then z=1
if m.d then m[x]="The"m[y]=m[y]..' family'm.f=1 else m[x]=m[x].." and "..b m.d=1 end end end if not z then
a[#a+1]={b,c,d,e,f,g,h,i,k} end z=nil end for k,v in p(a)do v[9]=nil print(table.concat(v,','))end