want to assign a letter or number based on calculated value (much like a
grading scale.) If calc value is:
..0 to .25= I
..26 to .50= H
..
..
..
final one 5.1 =A

can you do as an array beside doing the If and < formula
thx

Use the =VLOOKUP() function
--
Gary''s Student - gsnu200909


"Reno" wrote:

want to assign a letter or number based on calculated value (much like a
grading scale.) If calc value is:
.0 to .25= I
.26 to .50= H
.
.
.
final one 5.1 =A

can you do as an array beside doing the If and < formula
thx

You can use index match as follows...
=IF(D2A21, C21, INDEX(C1:C21, MATCH(D2, A1:A21,1)))

Where your data looks like this and your formula is in D2
A B C D
0 0.25 A T
0.26 0.5 B 4.95
0.51 0.75 C
0.76 1 D
1.01 1.25 E
1.26 1.5 F
1.51 1.75 G
1.76 2 H
2.01 2.25 I
2.26 2.5 J
2.51 2.75 K
2.76 3 L
3.01 3.25 M
3.26 3.5 N
3.51 3.75 O
3.76 4 P
4.01 4.25 Q
4.26 4.5 R
4.51 4.75 S
4.76 5 T
5.01 U

--
HTH...

Jim Thomlinson


"Reno" wrote:

want to assign a letter or number based on calculated value (much like a
grading scale.) If calc value is:
.0 to .25= I
.26 to .50= H
.
.
.
final one 5.1 =A

can you do as an array beside doing the If and < formula
thx

Sorry Formula is in D1 and data entry to look up is in D2.
--
HTH...

Jim Thomlinson


"Jim Thomlinson" wrote:

You can use index match as follows...
=IF(D2A21, C21, INDEX(C1:C21, MATCH(D2, A1:A21,1)))

Where your data looks like this and your formula is in D2
A B C D
0 0.25 A T
0.26 0.5 B 4.95
0.51 0.75 C
0.76 1 D
1.01 1.25 E
1.26 1.5 F
1.51 1.75 G
1.76 2 H
2.01 2.25 I
2.26 2.5 J
2.51 2.75 K
2.76 3 L
3.01 3.25 M
3.26 3.5 N
3.51 3.75 O
3.76 4 P
4.01 4.25 Q
4.26 4.5 R
4.51 4.75 S
4.76 5 T
5.01 U

--
HTH...

Jim Thomlinson


"Reno" wrote:

want to assign a letter or number based on calculated value (much like a
grading scale.) If calc value is:
.0 to .25= I
.26 to .50= H
.
.
.
final one 5.1 =A

can you do as an array beside doing the If and < formula
thx

On Wed, 20 Jan 2010 13:29:01 -0800, Reno
wrote:

want to assign a letter or number based on calculated value (much like a
grading scale.) If calc value is:
.0 to .25= I
.26 to .50= H
.
.
.
final one 5.1 =A

can you do as an array beside doing the If and < formula
thx


If your calculated value is in cell A1, try this formula in the cell
wher you want your "grade":

=LOOKUP(A1,{0,0.26,0.51,1.01,1.51,2.01,3.01,4.01,5 .11},{"I","H","G","F","E","D","C","B","A"})

Change the values to get the correct limits between the different
grades.

Hope this helps / Lars-Åke

If your ranges were all of equal length, you could do it with a simple
formula, but I'm guessing your ranges are not of equal size because you
can't start the way you show (starting at I,H and end up at A) given 0 to
0.25 is I and 0.26 to 0.50 is H. Can you show us all of your actual ranges
and their letter assignments?

--
Rick (MVP - Excel)


"Reno" wrote in message
...
want to assign a letter or number based on calculated value (much like a
grading scale.) If calc value is:
.0 to .25= I
.26 to .50= H
.
.
.
final one 5.1 =A

can you do as an array beside doing the If and < formula
thx

C# 4.0: Provides many new, useful and powerful features. One of those is "dynamic" type. "You can have a number of references with search keyword [Dynamic Type: C# 4.0 ].

I love to use it. I was facing a particular situation while executing sql statement (particulary returning a scalar value) and then returning the result.

I thought of using dynamic type: "A function made to execute the scalar sql statment and return the result as type dynamic." By this i would be able to get any type of data without any explicit conversion.

So, i built one function to accept randomSql, execute it(Execute NonScalar), return the result.

public dynamic FunctionToReturnScalarValue(string aRandomSql)
{
///////lines for database connectivity and executing the
///////random sql

dynamic result= commandObject.ExecuteScalar();

return result;

}

http://www.mindfiresolutions.com/Be-...0-beta-906.php




Reno wrote:

Assigning letter for numerical value
20-Jan-10

want to assign a letter or number based on calculated value (much like a
grading scale.) If calc value is:
..0 to .25= I
..26 to .50= H
..
..
..
final one 5.1 =A

can you do as an array beside doing the If and < formula
thx

Previous Posts In This Thread:

On Wednesday, January 20, 2010 4:29 PM
Reno wrote:

Assigning letter for numerical value
want to assign a letter or number based on calculated value (much like a
grading scale.) If calc value is:
..0 to .25= I
..26 to .50= H
..
..
..
final one 5.1 =A

can you do as an array beside doing the If and < formula
thx

On Wednesday, January 20, 2010 5:06 PM
Gary''s Student wrote:

Use the =VLOOKUP() function--Gary''s Student - gsnu200909"Reno" wrote:
Use the =VLOOKUP() function
--
Gary''s Student - gsnu200909


"Reno" wrote:

On Wednesday, January 20, 2010 5:12 PM
Jim Thomlinson wrote:

You can use index match as follows...
You can use index match as follows...
=IF(D2A21, C21, INDEX(C1:C21, MATCH(D2, A1:A21,1)))

Where your data looks like this and your formula is in D2
A B C D
0 0.25 A T
0.26 0.5 B 4.95
0.51 0.75 C
0.76 1 D
1.01 1.25 E
1.26 1.5 F
1.51 1.75 G
1.76 2 H
2.01 2.25 I
2.26 2.5 J
2.51 2.75 K
2.76 3 L
3.01 3.25 M
3.26 3.5 N
3.51 3.75 O
3.76 4 P
4.01 4.25 Q
4.26 4.5 R
4.51 4.75 S
4.76 5 T
5.01 U

--
HTH...

Jim Thomlinson


"Reno" wrote:

On Wednesday, January 20, 2010 5:14 PM
Jim Thomlinson wrote:

Sorry Formula is in D1 and data entry to look up is in D2.--HTH...
Sorry Formula is in D1 and data entry to look up is in D2.
--
HTH...

Jim Thomlinson


"Jim Thomlinson" wrote:

On Wednesday, January 20, 2010 5:20 PM
Lars-?ke Aspelin wrote:

If your calculated value is in cell A1, try this formula in the cellwher you
If your calculated value is in cell A1, try this formula in the cell
wher you want your "grade":

=LOOKUP(A1,{0,0.26,0.51,1.01,1.51,2.01,3.01,4.01,5 .11},{"I","H","G","F","E","D","C","B","A"})

Change the values to get the correct limits between the different
grades.

Hope this helps / Lars-?ke

On Thursday, January 21, 2010 3:54 AM
Rick Rothstein wrote:

If your ranges were all of equal length, you could do it with a simpleformula,
If your ranges were all of equal length, you could do it with a simple
formula, but I am guessing your ranges are not of equal size because you
cannot start the way you show (starting at I,H and end up at A) given 0 to
0.25 is I and 0.26 to 0.50 is H. Can you show us all of your actual ranges
and their letter assignments?

--
Rick (MVP - Excel)


Submitted via EggHeadCafe - Software Developer Portal of Choice
Free Online Courses Available for Eggheadcafe.com Users
http://www.eggheadcafe.com/tutorials...ses-avail.aspx

C# 4.0: Provides many new, useful and powerful features. One of those is "dynamic" type. "You can have a number of references with search keyword [Dynamic Type: C# 4.0 ].

I love to use it. I was facing a particular situation while executing sql statement (particulary returning a scalar value) and then returning the result.

I thought of using dynamic type: "A function made to execute the scalar sql statment and return the result as type dynamic." By this i would be able to get any type of data without any explicit conversion.

So, i built one function to accept randomSql, execute it(Execute NonScalar), return the result.

public dynamic FunctionToReturnScalarValue(string aRandomSql)
{
///////lines for database connectivity and executing the
///////random sql

dynamic result= commandObject.ExecuteScalar();

return result;

}



Reno wrote:

Assigning letter for numerical value
20-Jan-10

want to assign a letter or number based on calculated value (much like a
grading scale.) If calc value is:
..0 to .25= I
..26 to .50= H
..
..
..
final one 5.1 =A

can you do as an array beside doing the If and < formula
thx

Previous Posts In This Thread:

On Wednesday, January 20, 2010 4:29 PM
Reno wrote:

Assigning letter for numerical value
want to assign a letter or number based on calculated value (much like a
grading scale.) If calc value is:
..0 to .25= I
..26 to .50= H
..
..
..
final one 5.1 =A

can you do as an array beside doing the If and < formula
thx

On Wednesday, January 20, 2010 5:06 PM
Gary''s Student wrote:

Use the =VLOOKUP() function--Gary''s Student - gsnu200909"Reno" wrote:
Use the =VLOOKUP() function
--
Gary''s Student - gsnu200909


"Reno" wrote:

On Wednesday, January 20, 2010 5:12 PM
Jim Thomlinson wrote:

You can use index match as follows...
You can use index match as follows...
=IF(D2A21, C21, INDEX(C1:C21, MATCH(D2, A1:A21,1)))

Where your data looks like this and your formula is in D2
A B C D
0 0.25 A T
0.26 0.5 B 4.95
0.51 0.75 C
0.76 1 D
1.01 1.25 E
1.26 1.5 F
1.51 1.75 G
1.76 2 H
2.01 2.25 I
2.26 2.5 J
2.51 2.75 K
2.76 3 L
3.01 3.25 M
3.26 3.5 N
3.51 3.75 O
3.76 4 P
4.01 4.25 Q
4.26 4.5 R
4.51 4.75 S
4.76 5 T
5.01 U

--
HTH...

Jim Thomlinson


"Reno" wrote:

On Wednesday, January 20, 2010 5:14 PM
Jim Thomlinson wrote:

Sorry Formula is in D1 and data entry to look up is in D2.--HTH...
Sorry Formula is in D1 and data entry to look up is in D2.
--
HTH...

Jim Thomlinson


"Jim Thomlinson" wrote:

On Wednesday, January 20, 2010 5:20 PM
Lars-?ke Aspelin wrote:

If your calculated value is in cell A1, try this formula in the cellwher you
If your calculated value is in cell A1, try this formula in the cell
wher you want your "grade":

=LOOKUP(A1,{0,0.26,0.51,1.01,1.51,2.01,3.01,4.01,5 .11},{"I","H","G","F","E","D","C","B","A"})

Change the values to get the correct limits between the different
grades.

Hope this helps / Lars-?ke

On Thursday, January 21, 2010 3:54 AM
Rick Rothstein wrote:

If your ranges were all of equal length, you could do it with a simpleformula,
If your ranges were all of equal length, you could do it with a simple
formula, but I am guessing your ranges are not of equal size because you
cannot start the way you show (starting at I,H and end up at A) given 0 to
0.25 is I and 0.26 to 0.50 is H. Can you show us all of your actual ranges
and their letter assignments?

--
Rick (MVP - Excel)

On Thursday, April 29, 2010 1:42 AM
eliza sahoo wrote:

Be Careful while dealing with dynamic type and System.DBNull in C# 4.0 (beta)
C# 4.0: Provides many new, useful and powerful features. One of those is "dynamic" type. "You can have a number of references with search keyword [Dynamic Type: C# 4.0 ].

I love to use it. I was facing a particular situation while executing sql statement (particulary returning a scalar value) and then returning the result.

I thought of using dynamic type: "A function made to execute the scalar sql statment and return the result as type dynamic." By this i would be able to get any type of data without any explicit conversion.

So, i built one function to accept randomSql, execute it(Execute NonScalar), return the result.

public dynamic FunctionToReturnScalarValue(string aRandomSql)
{
///////lines for database connectivity and executing the
///////random sql

dynamic result= commandObject.ExecuteScalar();

return result;

}

http://www.mindfiresolutions.com/Be-...0-beta-906.php


Submitted via EggHeadCafe - Software Developer Portal of Choice
WPF Reflection Effect
http://www.eggheadcafe.com/tutorials...on-effect.aspx

C# 4.0: Provides many new, useful and powerful features. One of those is "dynamic" type. "You can have a number of references with search keyword [Dynamic Type: C# 4.0 ].

I love to use it. I was facing a particular situation while executing sql statement (particulary returning a scalar value) and then returning the result.

I thought of using dynamic type: "A function made to execute the scalar sql statment and return the result as type dynamic." By this i would be able to get any type of data without any explicit conversion.

So, i built one function to accept randomSql, execute it(Execute NonScalar), return the result.

public dynamic FunctionToReturnScalarValue(string aRandomSql)
{
///////lines for database connectivity and executing the
///////random sql

dynamic result= commandObject.ExecuteScalar();

return result;

}




Reno wrote:

Assigning letter for numerical value
20-Jan-10

want to assign a letter or number based on calculated value (much like a
grading scale.) If calc value is:
..0 to .25= I
..26 to .50= H
..
..
..
final one 5.1 =A

can you do as an array beside doing the If and < formula
thx

Previous Posts In This Thread:

On Wednesday, January 20, 2010 4:29 PM
Reno wrote:

Assigning letter for numerical value
want to assign a letter or number based on calculated value (much like a
grading scale.) If calc value is:
..0 to .25= I
..26 to .50= H
..
..
..
final one 5.1 =A

can you do as an array beside doing the If and < formula
thx

On Wednesday, January 20, 2010 5:06 PM
Gary''s Student wrote:

Use the =VLOOKUP() function--Gary''s Student - gsnu200909"Reno" wrote:
Use the =VLOOKUP() function
--
Gary''s Student - gsnu200909


"Reno" wrote:

On Wednesday, January 20, 2010 5:12 PM
Jim Thomlinson wrote:

You can use index match as follows...
You can use index match as follows...
=IF(D2A21, C21, INDEX(C1:C21, MATCH(D2, A1:A21,1)))

Where your data looks like this and your formula is in D2
A B C D
0 0.25 A T
0.26 0.5 B 4.95
0.51 0.75 C
0.76 1 D
1.01 1.25 E
1.26 1.5 F
1.51 1.75 G
1.76 2 H
2.01 2.25 I
2.26 2.5 J
2.51 2.75 K
2.76 3 L
3.01 3.25 M
3.26 3.5 N
3.51 3.75 O
3.76 4 P
4.01 4.25 Q
4.26 4.5 R
4.51 4.75 S
4.76 5 T
5.01 U

--
HTH...

Jim Thomlinson


"Reno" wrote:

On Wednesday, January 20, 2010 5:14 PM
Jim Thomlinson wrote:

Sorry Formula is in D1 and data entry to look up is in D2.--HTH...
Sorry Formula is in D1 and data entry to look up is in D2.
--
HTH...

Jim Thomlinson


"Jim Thomlinson" wrote:

On Wednesday, January 20, 2010 5:20 PM
Lars-?ke Aspelin wrote:

If your calculated value is in cell A1, try this formula in the cellwher you
If your calculated value is in cell A1, try this formula in the cell
wher you want your "grade":

=LOOKUP(A1,{0,0.26,0.51,1.01,1.51,2.01,3.01,4.01,5 .11},{"I","H","G","F","E","D","C","B","A"})

Change the values to get the correct limits between the different
grades.

Hope this helps / Lars-?ke

On Thursday, January 21, 2010 3:54 AM
Rick Rothstein wrote:

If your ranges were all of equal length, you could do it with a simpleformula,
If your ranges were all of equal length, you could do it with a simple
formula, but I am guessing your ranges are not of equal size because you
cannot start the way you show (starting at I,H and end up at A) given 0 to
0.25 is I and 0.26 to 0.50 is H. Can you show us all of your actual ranges
and their letter assignments?

--
Rick (MVP - Excel)

On Thursday, April 29, 2010 1:42 AM
eliza sahoo wrote:

Be Careful while dealing with dynamic type and System.DBNull in C# 4.0 (beta)
C# 4.0: Provides many new, useful and powerful features. One of those is "dynamic" type. "You can have a number of references with search keyword [Dynamic Type: C# 4.0 ].

I love to use it. I was facing a particular situation while executing sql statement (particulary returning a scalar value) and then returning the result.

I thought of using dynamic type: "A function made to execute the scalar sql statment and return the result as type dynamic." By this i would be able to get any type of data without any explicit conversion.

So, i built one function to accept randomSql, execute it(Execute NonScalar), return the result.

public dynamic FunctionToReturnScalarValue(string aRandomSql)
{
///////lines for database connectivity and executing the
///////random sql

dynamic result= commandObject.ExecuteScalar();

return result;

}

http://www.mindfiresolutions.com/Be-...0-beta-906.php

On Thursday, April 29, 2010 1:43 AM
eliza sahoo wrote:

Be Careful while dealing with dynamic type and System.DBNull in C# 4.0 (beta)
C# 4.0: Provides many new, useful and powerful features. One of those is "dynamic" type. "You can have a number of references with search keyword [Dynamic Type: C# 4.0 ].

I love to use it. I was facing a particular situation while executing sql statement (particulary returning a scalar value) and then returning the result.

I thought of using dynamic type: "A function made to execute the scalar sql statment and return the result as type dynamic." By this i would be able to get any type of data without any explicit conversion.

So, i built one function to accept randomSql, execute it(Execute NonScalar), return the result.

public dynamic FunctionToReturnScalarValue(string aRandomSql)
{
///////lines for database connectivity and executing the
///////random sql

dynamic result= commandObject.ExecuteScalar();

return result;

}


Submitted via EggHeadCafe - Software Developer Portal of Choice
Get Silverlight 4 Installed: Tips and Tricks
http://www.eggheadcafe.com/tutorials...4-install.aspx