您在处理一个多到多的关系在这里。结果
如果它是一个数据库，这意味着你会放在3个表：

You're dealing with a many-to-many relationship here.
If it were a database that means you'd put in 3 tables:

1. People
2. Colors
3. Link table between 1 and 2

我建议你要么通过利用数据库​​解决问题或者数据库在Delphi模拟的东西，就像它。

I suggest you either fix the problem by utilizing a database or model the thing in Delphi just like it where a database.

用Delphi结构结果
此外 ShortString短他们已经过时，有超过longstrings零收益停止使用。结果
使用3个表意味着您可以迅速得到人们每种颜色和颜色每人列表。

Using Delphi structures
Furthermore stop using shortstring They are outdated and have zero benefits over longstrings.
Using 3 tables means you can quickly get a list of people per color and colors per person.

下面是它如何工作：

TPerson = record
  name: string;
  other_data....
end;

TPeople = array of TPerson;

TFavColor = record
  name: string;
  other_data....
end;

TFavColors = array of TFavColor;

TPersonColor = record
  PersonIndex: Cardinal;  <<-- index into the TPeople array
  ColorIndex: Cardinal;   <<-- index into the TFavColors array
end;

TPersonColors = array of TPersonColor;

现在你可以遍历所有的TPersonColors阵列来提取数据。

Now you can just loop over the TPersonColors array to extract your data.

使用数据库结果
在SQL因为你的数据编制索引，将会变​​得更快（外键都（应该是）永远索引）。

Using a database
In SQL it would be even faster because your data is indexed (foreign key are (should be) always indexed).

SQL语句中看到所有的人，像蓝色和红色会是什么样子（这里使用MySQL的语法）：

The SQL statement the see all people that like blue and red would look like (using MySQL syntax here):

SELECT p.name  
FROM person p
INNER JOIN personcolor pc ON (pc.person_id = p.id)
INNER JOIN color c1 ON (pc.color_id = c1.id)
INNER JOIN color c2 ON (pc.color_id = c2.id)
WHERE c1.name = 'red' AND c2.name = 'blue'
GROUP BY p.id <<-- eliminate duplicates (not sure it's needed)

用Delphi其琐碎的数据库链接到您的应用程序。结果
所以这是我推荐的路线。

Using Delphi its trivial to link a database to your app.
So that's the route I'd recommend.