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For a data quality check I need to collect data in a specific interval.
Some tables are huge in size.
Is there any hack to do this without affecting the performance?
Like select 100 rows randomly.
How random do you need? The classic way to do this is with "WHERE RANDOM() < .001". If you need it to give you a repeatable "random" set then you can add a seed. The issue is that your tables are huge and this means reading (scanning) every row from disk just to throw most of them away and since table scan can take a significant time this isn't what you want to do.
So you may want to take advantage of Redshift "limited table scan" capabilities as part of your "random" sampling. (The fastest data to read from disk is the data you don't read from disk.) The issue here is that this solution will depend on your table sort keys and ordering which will push the solution into even "more pseudo" random territory (less of a true random sampling). In many cases this isn't a big deal but if the statistics really matter then this may not work for you.
This is done by sampling "blocks", not rows, based on the sort key(s). This sampling of blocks can be done randomly and each block of data will represent about 250K rows (based on sort key data type, compression etc. and COULD range anywhere from <100K rows to 2M rows). Doing this process will take a little inspection of STV_BLOCKLIST. The storage quanta for Redshift is the 1MB block and each and every block's metadata in the system can be referenced in STV_BLOCKLIST. This system table contains min and max values for each block. First find all the blocks for the sort key for the table in question. Next pick a random sample of these blocks (and if you are still dealing with a lot of data make sure that this sampling picks an even number from across all the slices to avoid execution skew).
Now the trick is to translate these min a max metadata values into a WHERE clause the performs the desired sampling. These min and max values are BIGINTs and are hashed from the data in the sort key column. This hash is data type dependent. If the data type is BIGINT then the has is quite simple - if the data type is timestamp then it is a bit more complex. But the ordering will be preserved across the hashing function for the data type involved. Reverse engineering this hash isn't hard - just perform a few experiments - but I can help if you tell me the type involved as I've done this for just about every data type at this point.
You can even do a random sampling rows on top of this random sampling of blocks. Or if you want you can just pick some narrow ranges of the sort key value and then randomly sample row and avoid all this reverse engineering business. The idea is to use Redshift "reduced scan" capability to greatly reduce the amount of data read from disk. To do this you need to be metadata aware in your choice of sampling windows which often means a sort key where clause. This is all about understanding how the database engine works and using its capabilities to your advantage.
I understand that this answer is based on some unstated information so please reach out in a comment if something isn't clear.
currently I'm experimenting a bit with RRDTool. I'm aware that the accuracy gets lower the longer the time periods are selected. But I thought I could bypass this with my datasource settings.
For example temperature and humidity from my house, resoultion 1h:
And now with the resolution of 1d:
As you could see, there is a great difference for the max. value of the blue line.
I created my datasources and archives with this values:
"rrdtool create temp.rrd --step 30",
"DS:temp:GAUGE:60:U:U",
"DS:humidity:GAUGE:60:U:U",
"RRA:AVERAGE:0.5:1:1051200",
"RRA:MAX:0.5:1:1051200",
"RRA:MIN:0.5:1:1051200",
I thought that 1051200 (1 year = 31536000 / 30 s (resoulution) = 1051200) is correct for saving every value for a year and that there should be no need for interpolating.
Is it possible to get the exact values displayed even if the resolution changes (for example the max humidity (Luftfeuchtigkeit) at 99.9%)?
Here are my values for image creation:
"--start" => "-1h", (-1d etc-)
"--title" => "Haustemperatur",
"--vertical-label" => "°C / % RLF",
"--width" => 800,
"--height" => 600,
"--lower-limit" => "-5",
"DEF:temperatur=$rrdFile:temperatur:LAST",
"DEF:humidity=$rrdFile:humidity:LAST",
"LINE1:temperatur#33CC33:Temperatur",
"GPRINT:temperatur:LAST:\t\tAktuell\: %4.2lf °C",
"GPRINT:temperatur:AVERAGE:Schnitt\: %4.2lf °C",
"GPRINT:temperatur:MAX:Maximum\: %4.2lf °C\j",
"LINE1:humidity#0000FF:Relative Luftfeuchtigkeit",
"GPRINT:humidity:LAST:Aktuell\: %4.2lf %%",
"GPRINT:humidity:AVERAGE:Schnitt\: %4.2lf %%",
"GPRINT:humidity:MAX:Maximum\: %4.2lf %%\j",
Thanks for your help and any suggestions.
P.S. I'm using a library to generate the graphs and the database, please do not be surprised about possible syntax errors.
Your problem is that you are causing the values to be rolled-up on the fly at graph time, but have not correctly specified which rollup function to use. Your second graph is showing the MAXIMUM of the LAST in the interval, not the true Maximum.
There are a few issues to explain with this configuration:
Firstly, your RRD is defined using 3 RRAs with 1cdp=1pdp and different consolidation functions (AVG, MIN, MAX). This means they are functionally identical, but they do not save you any time at graphing as they have not done any pre-rollup for you! You should definitely consider having just one of these (probably AVG) and adding others at lower resolution to help speed up graphing when you have a bigger time window.
Secondly, you need to specify the on-the-fly rollup function. When graphing, RRDTool will work out the best RRA to use based on your DEF lines, and will perform any additional consolidation required on the fly. This can take a long time if the only available RRA is too high-granularity.
Your graph request uses DEF:temperatur=$rrdFile:temperatur:LAST but you do not actually have a LAST type RRA, so RRDTool will grab the last average. Your RRA data points are at 30s interval, but your second graph has (approx) 5min per pixel, meaning that RRDTool needs to grab the 10 entries from the RRA, and print the last. Looking at the data in the top graph, it seems that the last in that interval was the 66 value, though previous ones were 100.
So you have a choice. Do you want the graph to show the average for the time period, the maximum, or both? Do you want the figures at the bottom to show the maximum of the average, or the maximum of everything?
For example
"DEF:temperatur=$rrdFile:temperatur:AVERAGE",
"DEF:humidity=$rrdFile:humidity:AVERAGE",
"DEF:temperaturmax=$rrdFile:temperatur:MAX;reduce=MAX",
"DEF:humiditymax=$rrdFile:humidity:MAX;reduce=MAX",
"LINE1:temperatur#33CC33:Temperatur",
"LINE1:temperaturmax#66EE66:Maximum Temperatur",
"GPRINT:temperatur:LAST:\t\tAktuell\: %4.2lf °C",
"GPRINT:temperatur:AVERAGE:Schnitt\: %4.2lf °C",
"GPRINT:temperaturmax:MAX:Maximum\: %4.2lf °C\j",
"LINE1:humidity#0000FF:Relative Luftfeuchtigkeit",
"LINE1:humiditymax#3333FF:Maximum Luftfeuchtigkeit",
"GPRINT:humidity:LAST:Aktuell\: %4.2lf %%",
"GPRINT:humidity:AVERAGE:Schnitt\: %4.2lf %%",
"GPRINT:humiditymax:MAX:Maximum\: %4.2lf %%\j",
In this case, we define a separate DEF for the maximum data set, so that we can always obtain the highest value even after consolidation. This is also used in the GPRINT so that we get the MAX of the MAX rather than the MAX of the AVERAGE. The Maximum line is now drawn separately to the average line, so that we can see the effect of any rollup of data - the lines will be together at high-resolution but get further apart as the time window widens and resolution decreases.
TheDEF is set to force any rollup function used for the maxima to be MAX rather than AVG, so we can be sure to get the maximum rather than average of maxima.
We are also using AVERAGE rather than LAST in order to get more meaningful data after rollup. Note that we could also use a separate DEF for the LAST as well if we wanted to though it is of less usefulness.
Note that, if you ever expect to be generating graphs over more than a few days, you should definitely consider adding some lower-resolution RRAs for AVERAGE and MAX or else the graphs will generate very slowly. RRDTool is designed with the intention that data will be rolled up over time, rather than (as in a traditional database) every sample kept as-is. So, unless you really need to have 30s resolution data kept for an entire year, you may prefer to keep this high resolution data for only a week, and then have separate RRAs that roll up to 1 hour resolution and keep for longer. Many people keep the 30s for 2 days, then 30min-summary for 2 weeks, 2h-summary for 2 months, and then 1day-summary for 2 years.
For more information, see the RRDTool manual pages.
I have a table with properties as ReadingTime, Frequency and I would like to insert 3 values in between those records where the time difference is greater than 12 hours. I could determine the time difference using the "Time Difference" node available but could not insert rows as per the requirement. Is there any way to attain this in knime ?
In case you are using Time Generator in a chunk loop (with the lagged column and Use second column option on the Time Difference node), you can generate as many nodes as you want (I assume you already use some switches/if nodes).
I've been doing some work for my exams in a few days and I'm going through some past papers but unfortunately there are no corresponding answers. I've answered the question and I was wondering if someone could tell me if I am correct.
My question is
(c) A transactional dataset, T, is given below:
t1: Milk, Chicken, Beer
t2: Chicken, Cheese
t3: Cheese, Boots
t4: Cheese, Chicken, Beer,
t5: Chicken, Beer, Clothes, Cheese, Milk
t6: Clothes, Beer, Milk
t7: Beer, Milk, Clothes
Assume that minimum support is 0.5 (minsup = 0.5).
(i) Find all frequent itemsets.
Here is how I worked it out:
Item : Amount
Milk : 4
Chicken : 4
Beer : 5
Cheese : 4
Boots : 1
Clothes : 3
Now because the minsup is 0.5 you eliminate boots and clothes and make a combo of the remaining giving:
{items} : Amount
{Milk, Chicken} : 2
{Milk, Beer} : 4
{Milk, Cheese} : 1
{Chicken, Beer} : 3
{Chicken, Cheese} : 3
{Beer, Cheese} : 2
Which leaves milk and beer as the only frequent item set then as it is the only one above the minsup?
I agree you should go for the Apriori Algorithm.
The Apriori algorithm is based on the idea that for a pair o items to be frequent, each individual item should also be frequent.
If the hamburguer-ketchup pair is frequent, the hamburger itself must also appear frequently in the baskets. The same can be said about the ketchup.
So for the algorithm, it is established a "threshold X" to define what is or it is not frequent. If an item appears more than X times, it is considered frequent.
The first step of the algorithm is to pass for each item in each basket, and calculate their frequency (count how many time it appears).
This can be done with a hash of size N, where the position y of the hash, refers to the frequency of Y.
If item y has a frequency greater than X, it is said to be frequent.
In the second step of the algorithm, we iterate through the items again, computing the frequency of pairs in the baskets. The catch is that
we compute only for items that are individually frequent. So if item y and item z are frequent on itselves,
we then compute the frequency of the pair. This condition greatly reduces the pairs to compute, and the amount of memory taken.
Once this is calculated, the frequencies greater than the threshold are said frequent itemset.
(http://girlincomputerscience.blogspot.com.br/2013/01/frequent-itemset-problem-for-mapreduce.html)
There are two ways to solve the problem:
using Apriori algorithm
Using FP counting
Assuming that you are using Apriori, the answer you got is correct.
The algorithm is simple:
First you count frequent 1-item sets and exclude the item-sets below minimum support.
Then count frequent 2-item sets by combining frequent items from previous iteration and exclude the item-sets below support threshold.
The algorithm can go on until no item-sets are greater than threshold.
In the problem given to you, you only get 1 set of 2 items greater than threshold so you can't move further.
There is a solved example of further steps on Wikipedia here.
You can refer "Data Mining Concepts and Techniques" by Han and Kamber for more examples.
OK to start, you must first understand, data mining (sometimes called data or knowledge discovery) is the process of analyzing data from different perspectives and summarizing it into useful information - information that can be used to increase revenue, cuts costs, or both. Data mining software is one of a number of analytical tools for analyzing data. It allows users to analyze data from many different dimensions or angles, categorize it, and summarize the relationships identified. Technically, data mining is the process of finding correlations or patterns among dozens of fields in large relational databases.
Now, the amount of raw data stored in corporate databases is exploding. From trillions of point-of-sale transactions and credit card purchases to pixel-by-pixel images of galaxies, databases are now measured in gigabytes and terabytes. (One terabyte = one trillion bytes. A terabyte is equivalent to about 2 million books!) For instance, every day, Wal-Mart uploads 20 million point-of-sale transactions to an A&T massively parallel system with 483 processors running a centralized database. Raw data by itself, however, does not provide much information. In today's fiercely competitive business environment, companies need to rapidly turn these terabytes of raw data into significant insights into their customers and markets to guide their marketing, investment, and management strategies.
Now you must understand that association rule mining is an important model in data mining. Its mining algorithms discover all item associations (or rules) in the data that satisfy the user-specified minimum support (minsup) and minimum confidence (minconf) constraints. Minsup controls the minimum number of data cases that a rule must cover. Minconf controls the predictive strength of the rule. Since only one minsup is used for the whole database, the model implicitly assumes that all items in the data are of the same nature and/or have similar frequencies in the data. This is, however, seldom the case in real- life applications. In many applications, some items appear very frequently in the data, while others rarely appear. If minsup is set too high, those rules that involve rare items will not be found. To find rules that involve both frequent and rare items, minsup has to be set very low. This may cause combinatorial explosion because those frequent items will be associated with one another in all possible ways. This dilemma is called the rare item problem. This paper proposes a novel technique to solve this problem. The technique allows the user to specify multiple minimum supports to reflect the natures of the items and their varied frequencies in the database. In rule mining, different rules may need to satisfy different minimum supports depending on what items are in the rules.
Given a set of transactions T (the database), the problem of mining association rules is to discover all association rules that have support and confidence greater than the user-specified minimum support (called minsup) and minimum confidence (called minconf).
I hope that once you understand the very basics of data mining that the answer to this question shall become apparent.
Input: total cost.
Output: all the combinations of levels that give the desired cost.
Every level of each stack costs a different amount (level 1 in stack 1 doesn't cost the same as level 1 in stack 2). I have a function that converts the level to the actual cost based on the base cost (level 1), which I entered manually (hard coded).
I need to find the combination of levels that give me the inputed cost. I realize there are more than one possible solutions, but I only need a way to iterate trough every possibility.
Here is what I need:
input = 224, this is one of the solutions:
I'm making a simple program that needs to select levels of different stacks and then calculate the cost, and I need to know every possible cost that exists... Each level of each stack costs a different amount of money, but that is not the problem, the problem is how to select one level for each stack.
I probably explained that very vaguely, so here's a picture (you'll have to excuse my poor drawing skills):
So, all stacks have the level 0, and level 0 always costs 0 money.
Additional info:
I have an array called "maxLevels", length of that array is the number of stacks, and each element is the number of the highest level in that stack (for example, maxLevels[0] == 2).
You can iterate from the 1st level because the level 0 doesn't matter at all.
The selected levels should be saved in an array (name: "currentLevels) that is similar to maxLevels (same length) but, instead of containing the maximum level of a stack, it contains the selected level of a stack (for example: currentLevels[3] == 2).
I'm programming in C++, but pseudocode is fine as well.
This isn't homework, I'm doing it for fun (it's basically for a game).
I'm not sure I understand the question, but here's how to churn through all the possible combinations of selecting one item from each stack (in this case 3*1*2*3*1 = 18 possibilities):
void visit_each_combination(size_t *maxLevels, size_t num_of_stacks, Visitor &visitor, std::vector<size_t> &choices_so_far) {
if (num_of_stacks == 0) {
visitor.visit(choices_so_far);
} else {
for (size_t pos = 0; pos <= maxLevels[0]; ++pos) {
choices_so_far.push_back(pos);
visit_each_combination(maxLevels+1, num_of_stacks-1, visitor, choices_so_far);
choices_so_far.pop_back();
}
}
}
You can replace visitor.visit with whatever you want to do with each combination, to make the code more specific. I've used a vector choices_so_far instead of your array currentLevels, but it could just as well work with an array.
This is very simple, if I've understood it correctly. The minimum cost is 0, and the maximum cost is just the sum of the heights of the stacks. To achieve any specific cost between these limits, you can start from the left, selecting the maximum level for each stack until your target is achieved, and then select level 0 for the remaining stacks. (You may have to adjust the last non-zero stack if you overshoot the target.)
I solved it, I think. #Steve Jessop gave me the idea to use recursion.
Algorithm:
circ(currentStack)
{
for (i = 0; i <= allStacks[currentStack]; i ++)
if (currentStack == lastStack && i == allStacks[currentStack])
return 0;
else if (currentStack != lastStack)
circ(++ currentStack);
}