blocks|key|1017292|text|您可以使用数组和struct函数来完成这个任务：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1017293|customer.select($"cid",+array(struct(lit("name")+as+"column_name",+$"name"+as+"column_value"),+struct(lit("age")+as+"column_name",+$"age"+as+"column_value")+))|offset|length|style|CODE|1017294|将使：|1017295|+%7C--+cid:+string+(nullable+=+true)
+%7C--+array(named_struct(column_name,+name+AS+`column_name`,+NamePlaceholder(),+name+AS+`column_value`),+named_struct(column_name,+age+AS+`column_name`,+NamePlaceholder(),+age+AS+`column_value`)):+array+(nullable+=+false)
+%7C++++%7C--+element:+struct+(containsNull+=+false)
+%7C++++%7C++++%7C--+column_name:+string+(nullable+=+false)
+%7C++++%7C++++%7C--+column_value:+string+(nullable+=+true)|code-block|syntax|javascript|1017296|entityMap^0|0|0|4F|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@$D|S|E|T|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|U|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|V|8|@]|9|@]|A|$M|N]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

You can do it using array and struct functions:

<code>customer.select($"cid", array(struct(lit("name") as "column_name", $"name" as "column_value"), struct(lit("age") as "column_name", $"age" as "column_value") ))</code>

will make:

<pre><code> |-- cid: string (nullable = true)
 |-- array(named_struct(column_name, name AS `column_name`, NamePlaceholder(), name AS `column_value`), named_struct(column_name, age AS `column_name`, NamePlaceholder(), age AS `column_value`)): array (nullable = false)
 | |-- element: struct (containsNull = false)
 | | |-- column_name: string (nullable = false)
 | | |-- column_value: string (nullable = true)
</code></pre>

blocks|key|2854858|text|唯一的困难是数组必须包含相同类型的元素。因此，在将所有列放入数组之前，都需要将它们转换为字符串(在您的示例中，age是一个int+)。情况如下：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2854859|val+cols+=+customer.columns.tail
val+result+=+customer.select('cid,
++++array(cols.map(c+=>+struct(lit(c)+as+"name",+col(c)+cast+"string"+as+"value"))+:+_*)+as+"array")

result.show(false)

%2B---%2B-----------------------------------------------------------%2B
%7Ccid%7Carray++++++++++++++++++++++++++++++++++++++++++++++++++++++%7C
%2B---%2B-----------------------------------------------------------%2B
%7CC1+%7C[[name,Jackie+Chan],+[age,50],+[city,Dayton],+[sex,M]]+++++%7C
%7CC2+%7C[[name,Harry+Smith],+[age,30],+[city,Beavercreek],+[sex,M]]%7C
%7CC3+%7C[[name,Ellen+Smith],+[age,28],+[city,Beavercreek],+[sex,F]]%7C
%7CC4+%7C[[name,John+Chan],+[age,26],+[city,Dayton],+[sex,M]]+++++++%7C
%2B---%2B-----------------------------------------------------------%2B

result.printSchema()

root
+%7C--+cid:+string+(nullable+=+true)
+%7C--+array:+array+(nullable+=+false)
+%7C++++%7C--+element:+struct+(containsNull+=+false)
+%7C++++%7C++++%7C--+name:+string+(nullable+=+false)
+%7C++++%7C++++%7C--+value:+string+(nullable+=+true)|code-block|syntax|javascript|2854860|entityMap^0|1J|3|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

The only difficulty is that arrays must contain elements of the same type. Therefore, you need to cast all the columns to strings before putting them in an array (<code>age</code> is an int in your case). Here is how it goes:

<pre class="lang-scala prettyprint-override"><code>val cols = customer.columns.tail
val result = customer.select('cid,
 array(cols.map(c =&gt; struct(lit(c) as "name", col(c) cast "string" as "value")) : _*) as "array")

result.show(false)

+---+-----------------------------------------------------------+
|cid|array |
+---+-----------------------------------------------------------+
|C1 |[[name,Jackie Chan], [age,50], [city,Dayton], [sex,M]] |
|C2 |[[name,Harry Smith], [age,30], [city,Beavercreek], [sex,M]]|
|C3 |[[name,Ellen Smith], [age,28], [city,Beavercreek], [sex,F]]|
|C4 |[[name,John Chan], [age,26], [city,Dayton], [sex,M]] |
+---+-----------------------------------------------------------+

result.printSchema()

root
 |-- cid: string (nullable = true)
 |-- array: array (nullable = false)
 | |-- element: struct (containsNull = false)
 | | |-- name: string (nullable = false)
 | | |-- value: string (nullable = true)
</code></pre>

blocks|key|2854889|text|映射列可能是处理总体问题的更好方法。您可以在同一个映射中保留不同的值类型，而不必将其转换为string。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2854890|df.select('cid',
++++create_map(lit("name"),+col("name"),+lit("age"),+col("age"),
+++++++++++++++lit("city"),+col("city"),+lit("sex"),col("sex")
+++++++++++++++).alias('map_col')
++)|code-block|syntax|javascript|2854891|或者，如果您想要的话，可以将它封装在一个数组中。|2854892|这样，您仍然可以对相关的键或值进行数值或字符串转换。例如：|2854893|df.select('cid',
++++create_map(lit("name"),+col("name"),+lit("age"),+col("age"),
+++++++++++++++lit("city"),+col("city"),+lit("sex"),col("sex")
+++++++++++++++).alias('map_col')
++)
df.select('*',+
++++++map_concat(+col('cid'),+create_map(lit('u_age'),when(col('map_col')['age']+<+18,+True)))
)|2854894|希望这是有意义的，在这里直接输入，所以如果某个地方缺少一个括号，请原谅|2854895|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|U|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Map columns might be a better way to deal with the overall problem. You can keep different value types in the same map, without having to cast it to string. 

<pre><code>df.select('cid',
 create_map(lit("name"), col("name"), lit("age"), col("age"),
 lit("city"), col("city"), lit("sex"),col("sex")
 ).alias('map_col')
 )
</code></pre>

or wrap the map col in an array if you want it 

This way you can still do numerical or string transformations on the relevant key or value. For example:

<pre><code>df.select('cid',
 create_map(lit("name"), col("name"), lit("age"), col("age"),
 lit("city"), col("city"), lit("sex"),col("sex")
 ).alias('map_col')
 )
df.select('*', 
 map_concat( col('cid'), create_map(lit('u_age'),when(col('map_col')['age'] &lt; 18, True)))
)
</code></pre>

Hope that makes sense, typed this straight in here so forgive if there's a bracket missing somewhere

Suppose I have a dataframe like this

<pre><code>val customer = Seq(
 ("C1", "Jackie Chan", 50, "Dayton", "M"),
 ("C2", "Harry Smith", 30, "Beavercreek", "M"),
 ("C3", "Ellen Smith", 28, "Beavercreek", "F"),
 ("C4", "John Chan", 26, "Dayton","M")
 ).toDF("cid","name","age","city","sex")
</code></pre>

How can i get cid values in one column and get the rest of the values in an <code>array &lt; struct &lt; column_name, column_value &gt; &gt;</code> in spark

Converting a dataframe to an array of struct of column names and values

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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假设我有这样的数据val customer = Seq(    ("C1", "Jackie Chan", 50, "Dayton", "M"),    ("C2", "Harry Smith", 30, "Beavercreek", "M"),    ("C3", "Ellen Smith", 28, "Beaver...

问将dataframe转换为列名和值的结构数组
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