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I'm looking for the correct way to define a sum of all elements in a set that fullfill a certain condition.

Example: $$ n(A) = \frac{\sum_{F \in A}{ \begin{cases} A_F\text{ is }\mathrm{NULL} & w_F\\ \text{else} & 0 \end{cases}}}{\sum_{F \in A}{w_F}} $$

A is to supposed to be a record with a number of fields $F$. $A_F$ is the value of field F in the record A. $w_F$ is a weight associated with the Field $F$. I want to calculate how much percent of the record is NULL weighted according to $w_F$.

What I want to know is if this below is a valid way to formulate this or if there is a better way.

If someone has some pointers on a more formal / mathematical notation for database/records I would also be very grateful.

Thanks!

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Although I don't think it is quite mainstream, but you could look into Iverson Notation. In short, you have a bracket and inside you write a condition. The bracket value is 1 if the condition holds, otherwise it's 0. Its most popular usage was in Concrete Mathematics (by Knuth et al).

$$ n(A) = \frac{\sum_{F \in A}{w_F[A_F \neq NULL]}}{\sum_{F \in A}{w_F}} $$

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  • $\begingroup$ I like this one. Looks very natural. I think I will go for this notation at least for the moment and hope that it's common enough for my professor to recognize it ;-) $\endgroup$ – aKzenT Jan 15 '12 at 20:06
  • $\begingroup$ @aKzenT Why possibly puzzle your professor? Define the notation you want to use. Footnote stackexchange and Knuth and s/he won't complain. (Just noticed that this is an ancient answer, so my comment isn't much help.) $\endgroup$ – Ethan Bolker Apr 27 '18 at 16:37
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For the sake of clarity, I would describe such an operation with the help of a short sentence. Also, it is quite common to put some summation conditions in the index spot of the sum symbol, as in

$$\sum_{p \text{ prime}} 1/p$$

In my opinion, it is best to describe what you have written above as

(Brief description of what kinds of objects $A$, $w_F$, $A_f$, etc are and a quick name for what $n(A)$ describes)

$$n(A) = \sum_{F \in A, A_F = 0} w_F/\sum_{F \in A}w_F$$

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  • $\begingroup$ I thought about this one, but I would like the condition to be a bit more visible instead of putting it only in the subscript of the sum. The formula is inside a larger text explaining the operation in text also, just like I did above. $\endgroup$ – aKzenT Jan 15 '12 at 20:04
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I know it is not any cool notation, but maybe you can simply define a dummy set, one that satisfies all your conditions, and then, make the sum of wf on each element of this dummy set .

By adding a dummy set B, this part of your formula $$\sum_{F \in A}{ \begin{cases} A_F\text{ is }\mathrm{NULL} & w_F\\ \text{else} & 0 \end{cases}}$$

could be transformed into

$$ \sum_{F \in B}{w_F} $$ Then add at the end of your formula that set B is all the elements f of set A for which Af is null: $$B=\{f \in A | A_f\text{ is }\mathrm{NULL}\} $$

Thus, my final answer to your question would look like this: $$ n(A) = \frac{\sum_{F \in B}{w_F}}{\sum_{F \in A}{w_F}} $$ where

$$B=\{f \in A | A_f\text{ is }\mathrm{NULL}\} $$

**Also, I don't know yet if this notation is allowed: $$ n(A) = \frac{\sum_{F \in \{f \in A | A_f\text{ is }\mathrm{NULL}\}}{w_F}}{\sum_{F \in A}{w_F}} $$ but it is the integration of the B set definition into one big formula. I personnaly think that defining B on a separated formula is clearer, but I guess sometimes we need to use only one formula.

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  • $\begingroup$ Welcome to stackexchange. I'm glad you want to help by providing answers, but working a five year old question with an accepted answer isn't likely to be useful. You'd be better off spending your time on new questions that have no good answers. $\endgroup$ – Ethan Bolker Apr 27 '18 at 16:41
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    $\begingroup$ Thank you! As a Google wanderer myself, I believe that any solution may help many of my fellow wanderers. I know this question is old, but I try to make internet a better place everyday. I may be not advanced enough in mathematics to formulate excellent answers to many specific questions, but I will sure be glad to help if I find the opportunity. $\endgroup$ – V. Brunelle Apr 27 '18 at 16:51

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