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如何在列表中拆分并获得不同的单词?

[英]How can i split and get distinct words in a list?

My sample data coloumn, which come from an CSV file is 我的样本数据来自CSV文件,

|----Category------------|

 SHOES
 SHOES~SHOCKS
 SHOES~SHOCKS~ULTRA SOCKS

I would love to split the specific column and get the distinct values in a list like 我想拆分特定的列,并在列表中获取不同的值,例如

SHOES
SHOCKS
ULTRA SOCKS

I tried the following, but it does not work as expected. 我尝试了以下操作,但是它没有按预期工作。

var test = from c in products select c.Category.Split('~').Distinct().ToList();

It actually returns the following. 它实际上返回以下内容。

在此处输入图片说明

Any thoughts please? 有什么想法吗? Thank you. 谢谢。

I would use SelectMany to "flatten" the list before removing duplicates: 在删除重复项之前,我将使用SelectMany来“拉平”列表:

products.SelectMany(c => c.Category.Split('~'))
        .Distinct()

您可以使用SelectMany展平集合:

products.SelectMany(p => p.Category.Split('~')).Distinct().ToList();

You were close, you just needed to flatten out your collection to pull the individual items of each grouping via a SelectMany() call : 亲密无间,您只需要展平集合即可通过SelectMany()调用提取每个分组的各个项目:

// The SelectMany will map the results of each of your Split() calls
// into a single collection (instead of multiple)
var test = products.SelectMany(p => p.Category.Split('~'))
                   .Distinct()
                   .ToList();

You can see a complete working example demonstrated here and seen below : 您可以在这里和下面看到一个完整的工作示例

// Example input
var input = new string[] { "SHOES","SHOES~SHOCKS","SHOES~SHOCKS~ULTRA SOCKS" };
// Get your results (yields ["SHOES","SHOCKS","ULTRA SOCKS"])
var output = input.SelectMany(p => p.Split('~'))
                  .Distinct()
                  .ToList();

通过使用SelectMany()将此list of strings列表list of strings合并到一个列表中,然后向列表中添加另一个Distinct。

var test = from c in products select c.Category.Split('~').Distinct().ToList().SelectMany(x => x).Distinct().ToList();

Here's how you'd do it in query syntax. 这是您在查询语法中的处理方式。

var test = (from p in products 
            from item in p.Category.Split('~')
            select item).Distinct().ToList();

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