Linear Interpolation
Jun 7, 2007
my data is like this:
header | data | key
-------------------
500 | 3.2 | 10
500 | 3.4 | 20
500 | 3.6 | 25
500 | 3.7 | 40
501 | 4.1 | 10
501 | 4.2 | 15
501 | 4.4 | 30
501 | 4.6 | 35
and what I want to do is find the median of "data", but keyed off of "key", so if my desired median is 30, I want to take the two records (data, key) nearest to key = 30, and get the average of "data".
...and do this within each "header" value.
actually, to be precise, I want the linear interpolation, so for header = 500, I want to get the (data, key) pairs of (3.6, 25) and (3.7, 40) and return the interpolated "data" value of 3.6333 (as done here (http://en.wikipedia.org/wiki/Linear_interpolation))
so for the above example the query would produce:
header | interp
-----------------
500 | 3.633
501 | 4.4
possible, or am I crazy?
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Jul 23, 2005
How can I order the results of my query in non-linear fasion. I have afield with these values: Reg S, 144A, US and want to order my resultsby US, 144A, Reg S.I would prefer not to create another field in the table if possible.
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Jul 24, 2006
This is a real challenge. I hope someone is smart enough to know howto do this.I have a tableTABLE1[Column 1- 2001][Column 2- 2002][Column 3- 2003][Column 4 - 2004][Column 5 - 2005][Column 6 - 2006][Column 7 - Slope][2001][2002][2003][2004][2005][2006] [Slope][1] [2] [3] [4] [5] [6] [1][1.2] [.9] [4] [5] [5.4] [6.2] [?]Slope is defined as "M" in the equation y=mx+bI need a way a finding the linear equation that best fits the points soI can have SQL calculate the slope.Are there any smart people around that would know how to do this?thanks
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May 26, 2008
Hi all
i wants to generate linear sequence number like 1,2,3,.............1000000
,are there any function like NEWID() ( this return unique guide, i want to get integer)
i want to used this generated number inside the SQL query
thanks
IndikaD (Virtusa Cop)
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Apr 22, 2007
I would like to understand the algorithm that the linear regression method uses to choose the regressors in the model from a list of possible regressors.
I think that it is different from the common methods used in statistics like stepwise, forward or backward.
Laura Lerner
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Jan 22, 2007
We are trying to create a model of linear regression with nested table. We used the create mining model sintax as follow :
create mining model rate_plan3002_nested2
( CUST_cycle LONG KEY,
VOICE_CHARGES double CONTINUOUS predict,
DUR_PARTNER_GRP_1 double regressor CONTINUOUS ,
nested_taarif_time_3002 table
( CUST_cycle long CONTINUOUS,
TARIFF_TIME text key,
TARIFF_VOICE_DUR_ALL double regressor CONTINUOUS
)
) using microsoft_linear_regression
INSERT INTO MINING STRUCTURE [rate_plan3002_nested2_Structure]
(CUST_cycle ,
VOICE_CHARGES ,
DUR_PARTNER_GRP_1 ,
[nested_taarif_time_3002](SKIP,TARIFF_TIME ,TARIFF_VOICE_DUR_ALL)
)
SHAPE {
OPENQUERY([Cell],
'SELECT CUST_cycle ,
VOICE_CHARGES ,
DUR_PARTNER_GRP_1
FROM dbo.panel_anality_3002
order by CUST_cycle ')}
APPEND
({OPENQUERY([Cell],
'select CUST_cycle,
TARIFF_TIME,
CYCLE_DATE
from dbo.nested_taarif_time_3002
order by CUST_cycle,TARIFF_TIME')
}
relate CUST_cycle to CUST_cycle
) as nested_taarif_time_3002
The results we got are a model with intercept only. if we don't use the nested variable (the red line) we get a rigth model . (we had more variable ....)
Is there a way to do this regression correctly?
Thanks,
Dror
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Sep 2, 2007
When using linear regression in the SQL Server 2005 Business IntelIigence Studio I interpet the information below as follow: X has a standard deviation of +- 37.046. Is it possible to obtain the standard deviation of each coefficient in the regression expression?
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Jan 18, 2008
Hi,
I am trying to create a model using microsoft Linear Regression algorithm. But I want to constrain the coefficient of the parameters to non-negative value. There is concept of bound in SAS where we can specify the range of the coefficient. Does any of the SSAS mining algorithms support restricting the coefficient value?
Thanks,
DMN
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Sep 18, 2006
Q1. Model Prediction -- Suppose we already have a trained Microsoft Linear Regression Mining Model, say, target y regressed on two variables:
x1 and x2, where y, x1, x2 are of datatype Float. We try to perform Model Prediction with an Input Table in which some records consist of NULL x2 values. How are the resulting predicted y values calculated?
My guess:
The resulting linear regression formula is in the form:
y = constant + coeff1 * (x1 - avg_x1) + coeff2 * (x2 - avg_x2)
where avg_x1 is the average of x1 in the training set, and avg_x2 is the average of x2 in the training set (Correct?).
I guess that for some variable being NULL in the Input Table, Microsoft Linear Regression just treat it as the average of that variable in the training set.
So for x2 being NULL, the whole term coeff2 * (x2 - avg_x2) just disappear, as it is zero if we substitute x2 with its average value.
Is this correct?
Q2. Model Training -- Using the above example that y regressed on x1 and x2, if we have a train set that, say, consist of 100 records in which
y: no NULL value
x1: no NULL value
x2: 70 records out of 100 records are NULL
Can someone help explain the mathematical procedure or algorithm that produce coeff1 and coeff2?
In particular, how is the information in the "partial records" used in the regression to contribute to coeff1 and the constant, etc ?
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Sep 18, 2006
Q1. Model Prediction -- Suppose we already have a trained Microsoft Linear Regression Mining Model, say, target y regressed on two variables:
x1 and x2, where y, x1, x2 are of datatype Float. We try to perform Model Prediction with an Input Table in which some records consist of NULL x2 values. How are the resulting predicted y values calculated?
My guess:
The resulting linear regression formula is in the form:
y = constant + coeff1 * (x1 - avg_x1) + coeff2 * (x2 - avg_x2)
where avg_x1 is the average of x1 in the training set, and avg_x2 is the average of x2 in the training set (Correct?).
I guess that for some variable being NULL in the Input Table, Microsoft Linear Regression just treat it as the average of that variable in the training set.
So for x2 being NULL, the whole term coeff2 * (x2 - avg_x2) just disappear, as it is zero if we substitute x2 with its average value.
Is this correct?
Q2. Model Training -- Using the above example that y regressed on x1 and x2, if we have a train set that, say, consist of 100 records in which
y: no NULL value
x1: no NULL value
x2: 70 records out of 100 records are NULL
Can soemone help explain the mathematical procedure or algorithm that produce coeff1 and coeff2?
In particular, how is the information in the "partial records" used in the regression to contribute to coeff1 and the constant, etc ?
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Dec 19, 2006
With the number of threads it is difficult to know if this has been posted. If I use the Mining Content Viewer for Linear Regression, under Node Distribution, there are values given for Attribute Name, Attribute Value, Support, Probability, Variance, and Value Type. The output is similar to what Joris supplied in his thread about Predict Probability in Decision Trees. My questions:
1. How should these fields be interpreted?
2. With Linear Regression, is it possible to get the coefficient values and tests of significance (t-tests?), if they are not part of the output I have pointed to?
Thanks for your help with this?
Sam
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Nov 24, 2015
I'm using a bullet chart in a SSRS report and I want to set the Maximum value in the Linear Scale properties to highest value of the following 4 fields. Is there any way to do this?? This will make all charts line up properly.
NC_LAST_YEAR
NC_LINKED
NC_CURRENT
NC_PLAN
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