how to reshape the columns of a matrix into one row? 
hello, 
please i need help in this:
suppose that we have a (3*4) matrix that have only binary values either 0 or 1, now i need to read this matrix column by column from the top of the column to the bottom. and i concatenate the values of each column with the values of the other columns. so how can i do that? 
for example the matrix a= [
0 1 1 1
1 0 1 0
1 0 0 1
]
the ouput of this process hsould be = 011100110101

regards,
it's not clear what it is you want as an output. if it's a row vector, there's no concatenation involved, you just simply reshape your matrix:
a = [0 1 1 1;
 1 0 1 0;
 1 0 0 1]
 v = reshape(a, 1, []) %reshape a into one row. since matlab is column major, this is done column-wise
i would do it like this:

https://www.advanced-excel.com/columns-to-rows.html finding the correct answer here is important, but perhaps equally important is finding it using the method of matrix reduction, since those are the instructions.the augmented matrix is4 -2 -5 -514 -1 -2 -364 -2 4 -24you'll have to forgive the poor formatting. this answer box doesn't have a proper math symbol editor.using elementary row operations, the reduced row echelon form of the augmented matrix is1 0 0 -60 1 0 60 0 1 3from which we conclude that the system has a unique solution (-6,6,3). Notice that the matrix A is upper-triangle, so its determinant is a  
product   
of diagonal elements:  
  
det(A) = 1*2*3 = 6.  
  
Moreover, det(A) is   
non-zero.  
Therefore the columns of A are linearly independent and   
therefore  
they constitutive a basis for R^3, so R^3 = Col(A).  
  
Hence   
any base of R^3 fits to be a base for Col(A),   
for instance the columns of   
A:  
  
(1,0,0), (2,2,0), (3,3,3)  
  
or   
  
(1,0,0), (0,1,0),   
(0,0,1)  
  
are the base.  
  
The dimension of Col(A) is the dimension of   
R^3 and therefore is equal  
to 3. Hi Valdeth. You should use input and output file streams if you want to write and read from a file. Basically you open the file for writing and once you are done close the file. Later you can open the file for reading. The example below reads numbers from the file and just prints them out. If you want to read the file line by line you should look at the getline() function. I have just used your initial array to come up with a result per array which gives the product of all negative numbers in the array. Note that there **HAS** to be at least 1 negative number per array.



```
const int m=4, n=5;
	int i,j;
	int num;
	int result;
	int A[m][n]={{5,3,-4,8,2},
				 {1,5,2,7,-8},
				 {-3,9,5,2,4},
				 {4,-1,3,6,8}};

	ofstream outfile;
	ifstream infile; 

	outfile.open("c:/matrix/matrix.txt");

	for (i=0;i<m;i++)
	{
		result = 1;

		for (j=0;j<n;j++)
		{
			outfile << setw(3)<< A[i][j];

                            // Multiply negative numbers in each array (has to be at least 1 negative number in each array)
			if (A[i][j] < 0)              				                      result *= A[i][j];  // result = result x A[i][j]
		}
		cout << "Result of array " << i << " is " << result << endl;
		outfile << "\n";
	}

	outfile.close(); // Close for writing

	infile.open("c:/matrix/matrix.txt"); // opens the file
    
	infile >> num;
	while ( !infile.eof() ) 
	{ // keep reading until end-of-file
		cout << "The next number is " << num << endl;
		infile >> num; // sets EOF flag if no value found
	}
	infile.close();  //Close for reading
``` hi [@jitse\_b](/u/jitse_b)

welcome to the community !

yes you can, there are various ways to do so.

the simplest:

you may want to create a table with 3 columns,

key\_column, type text, value “k000” (or whatever you want, it doesn’t matter if you stick to only one comparison)

“material1”, type ref, source tabl… Should be able to do something like this:


UPDATE table\_name  

SET new\_column = (calculated value)  

WHERE new\_column is NULL Assuming you're joining on columns with no duplicates, which is a very common case:

* An inner join of A and B gives the result of A intersect B, i.e. the inner part of a [Venn diagram](http://en.wikipedia.org/wiki/Venn_diagram) intersection.
* An outer join of A and B gives the results of A union B, i.e. the outer parts of a Venn diagram union.

**Examples**

Suppose you have two tables, with a single column each, and data as follows:


```
A    B
-    -
1    3
2    4
3    5
4    6

```
Note that (1,2) are unique to A, (3,4) are common, and (5,6) are unique to B.

**Inner join**

An inner join using either of the equivalent queries gives the intersection of the two tables, i.e. the two rows they have in common.


```
select * from a INNER JOIN b on a.a = b.b;
select a.*, b.*  from a,b where a.a = b.b;

a | b
--+--
3 | 3
4 | 4

```
**Left outer join**

A left outer join will give all rows in A, plus any common rows in B.


```
select * from a LEFT OUTER JOIN b on a.a = b.b;
select a.*, b.*  from a,b where a.a = b.b(+);

a |  b
--+
1 | null
2 | null
3 |    3
4 |    4

```
**Right outer join**

A right outer join will give all rows in B, plus any common rows in A.


```
select * from a RIGHT OUTER JOIN b on a.a = b.b;
select a.*, b.*  from a,b where a.a(+) = b.b;

a    |  b
+
3    |  3
4    |  4
null |  5
null |  6

```
**Full outer join**

A full outer join will give you the union of A and B, i.e. all the rows in A and all the rows in B. If something in A doesn't have a corresponding datum in B, then the B portion is null, and vice versa.


```
select * from a FULL OUTER JOIN b on a.a = b.b;

 a   |  b
+
   1 | null
   2 | null
   3 |    3
   4 |    4
null |    6
null |    5

``` Lets say you want to send a table with a single column of GUIDs.


First we need to create a structure using [SqlMetaData](https://msdn.microsoft.com/en-us/library/microsoft.sqlserver.server.sqlmetadata(v=vs.110).aspx) which represents the schema of the table (columns).



```
var tableSchema = new List<SqlMetaData>(1)
{
  new SqlMetaData("Id", SqlDbType.UniqueIdentifier)
}.ToArray();

```

Next you create a list of records that match the schema using [SqlDataRecord](https://msdn.microsoft.com/en-us/library/microsoft.sqlserver.server.sqldatarecord(v=vs.110).aspx).



```
var tableRow = new SqlDataRecord(tableSchema);
tableRow.SetGuid(0, Guid.NewGuid());
var table = new List<SqlDataRecord>(1)
{
  tableRow
};

```

Then create the [SqlParameter](https://msdn.microsoft.com/en-us/library/system.data.sqlclient.sqlparameter(v=vs.110).aspx):



```
var parameter = new SqlParameter();
parameter.SqlDbType = SqlDbType.Structured;
parameter.ParameterName = "@UserIds";
parameter.Value = table;

var parameters = new SqlParameter[1]
{
  parameter
};

```

Then simply call the stored procedure by using the [Database.SqlQuery](https://msdn.microsoft.com/en-us/library/gg696545(v=vs.113).aspx).



```
IEnumerable<ReturnType> result;
using (var myContext = new DbContext())
{
  result = myContext.Database.SqlQuery<User>("GetUsers", parameters)
    .ToList();         // calls the stored procedure
    // ToListAsync();  // Async
{

```

In SQL Server, create your User-Defined Table Type (I suffix them with TTV, Table Typed Value):



```
CREATE TYPE [dbo].[UniqueidentifiersTTV] AS TABLE(
  [Id] [uniqueidentifier] NOT NULL
)
GO

```

Then specify the type as a parameter (don't forget, Table Type Values have to be readonly!):



```
CREATE PROCEDURE [dbo].[GetUsers] (
  @UserIds [UniqueidentifiersTTV] READONLY
) AS
BEGIN
  SET NOCOUNT ON

  SELECT u.* -- Just an example :P
  FROM [dbo].[Users] u
  INNER JOIN @UserIds ids On u.Id = ids.Id
END

``` You could try using the in line view like this:


Select table1.col1, table1.col2, table1…., table2.col1, table2.col2, table2….  

 from  

(SELECT *  

FROM dbo.Shaft\_Scales  

WHERE dbo.Shaft\_Scales.ID = (SELECT max(ID) from dbo.Shaft\_Scales ID) as table1,  

(SELECT *  

FROM OtherTable  

WHERE dbo.OtherTable.ID = (SELECT max(ID) from dbo.OtherTable ID) as table2


—————- another approach from meandyou 


This works on [at least some] SQL processors


SELECT T1.* , T2.*  

FROM table1 T1, table2 T2  

WHERE …


Do we have to discuss why not to use “SELECT * ” in your SQL?<pre>


Try this


SELECT * FROM T1 as a, T2 as b WHERE a.ID = (SELECT max(ID) from T1)  

and b.ID = (SELECT max(ID) from T2);</pre> Here is a sample of traversing a simple xml structure

 
```
    Dim myxml = <root>
                    <level id="1">
                        <prop n="PROPERTY" v="VALUE"/>
                    </level>
                    <level id="2">
                        <prop n="PROPERTY" v="VALUE"/>
                    </level>
                </root>

    For Each level In myxml.Elements
        Debug.WriteLine("id=" & level.@id.ToString)
        For Each prop In level.Elements
            Debug.WriteLine("n=" & prop.@n.ToString & "  v=" & prop.@v.ToString)
        Next
    Next

```