average

This will seem rudimentary but I can't find a concise example online that matches up. I have three fields; m1 , m2 , and m3 . I need to create a column or field that is the average of them three. The calculated field would be titled employment. Would the following code be suffice? ALTER TABLE dbo.tablename ADD Employment AS Select ((m1+m2+m3)/3) Sample data m1 20 20 30 m2 15 17 25 m3 60 77 13 desired result. Name m1 m2 m3 Employment Auto body 20 20 30 23 Auto Parts 15 17 25 19 Auto Sales 60 77 13 50 You are very close, it's called Computed Column https://technet.microsoft.com/en-us/library

I'm trying to implement system-wide login throttling and I need to calculate the daily average number of failed login attempts from the last 3 months. I'm currently inserting a record on every login fail, each with a timestamp. How can I do this in MySQL? Thanks in advance for your help SELECT AVG(cnt) FROM (SELECT COUNT(*) AS cnt FROM mytable WHERE `date` BETWEEN DATE_SUB(NOW(), INTERVAL 3 MONTH) AND NOW() GROUP BY DATE(`date`)) x Assuming you have a table mytable with field date of type date , datetime or timestamp 来源： https://stackoverflow.com/questions/5036176/moving-average-mysql

I've searched around SO and can't seem to find a question with an answer that works fine for me. I have a table with almost 2 million rows in, and each row has a MySQL Date formatted field. I'd like to work out (in seconds) how often a row was inserted, so work out the average difference between the dates of all the rows with a SQL query. Any ideas? -- EDIT -- Here's what my table looks like id, name, date (datetime), age, gender If you want to know how often (on average) a row was inserted, I don't think you need to calculate all the differences. You only need to sum up the differences

suppose I want to calculate average value of a data-set such as class Averager { float total; size_t count; float addData (float value) { this->total += value; return this->total / ++this->count; } } sooner or later the total or count value will overflow, so I make it doesn't remember the total value by : class Averager { float currentAverage; size_t count; float addData (float value) { this->currentAverage = (this->currentAverage*count + value) / ++count; return this->currentAverage; } } it seems they will overflow longer, but the multiplication between average and count lead to overflow

pretty basic question, but was wondering: What is the 'proper' way to average every 2 rows together in pandas Dataframe, and thus end up with only half the number of rows? Note that this is different than the rolling_mean since it reduces the number of entries. A fast way to do it: >>> s = pd.Series(range(10)) >>> s 0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 >>> ((s + s.shift(-1)) / 2)[::2] 0 0.5 2 2.5 4 4.5 6 6.5 8 8.5 The "proper way" I guess would be something like: >> a = s.index.values >>> idx = np.array([a, a]).T.flatten()[:len(a)] >>> idx [0 0 1 1 2 2 3 3 4 4] >>> s.groupby(idx).mean() 0 0

Is possible to calculate average of three encrypted integer? No constrain on the method of encrypting. The point of this is just to hide the three numbers and find average. What you seem to be looking for is called Homomorphic Encryption : an encryption scheme which allows you to perform operations on encrypted data, with the encrypted result as the outcome. Such a scheme would allow you to give encrypted data to a 3rd party, which could then do computations on it for you without knowing what they were computing. In your case, you need two operations: addition and division. Until recently,

Using any tools which you would expect to find on a nix system (in fact, if you want, msdos is also fine too), what is the easiest/fastest way to calculate the mean of a set of numbers, assuming you have them one per line in a stream or file? Awk awk '{total += $1; count++ } END {print total/count}' awk ' { n += $1 }; END { print n / NR }' This accumulates the sum in n , then divides by the number of items ( NR = Number of Records). Works for integers or reals. Using Num-Utils for UNIX: average 1 2 3 4 5 6 7 8 9 perl -e 'while (<>) { $sum += $_; $count++ } print $sum / $count, "\n"'; In