Questions tagged [data-partitioning]

Data partitioning deals with the dividing of a collection of data into smaller collections of data for the purpose of faster processing, easier statistics gathering and smaller memory/persistence footprint.

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SQL Server: Get time duration based on continues values of a column

I Have a table Like This ID VehicleID Time EngineStatus -- ---------- ------------------ ---------------- 0 0001 1/11/2016 00:00 off 0 0002 …
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partitioning data by row number in r

Now I have data dat <- data.frame(y,x1,x2) there are about 200 rows. And I want to partition this data by row number like 1 to 150 and 150 to 200 How can I do this?
Samuel Lee
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I have used: Y=datasample(Data,100,'Replace',false) for separating data for training NN.Data has 150 data samples(150*5)

Please tell me how to extract the remaining 50 data samples from the 'Data' dataset for testing the trained NN. Is there any other way to separate the training and testing data for classification purpose. Please help me.. Thank you
RACHITA PATRO
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Python JSON data parsing

Im trying to get the average of "active" for each place under a specific area. So say the output would be ("Andaman and Nicobar Islands": 10, "Andhra Pradesh": 12) however i get type error string indices must be integers from the…
nataku
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Finding where a given number falls in a partition

Suppose I have a sorted array of integers say partition = [0, 3, 7, 12, 18, 23, 27] and then given a value value = 9 I would like to return the interval on which my value sits. For example bounds = function(partition,…
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How to partition array of integers to even and odd?

I want to partition an array (eg [1,2,3,4,5,6,7,8]), first partition should keep even values, second odd values (example result: [2,4,6,8,1,3,5,7]). I managed to resolve this problem twice with built-in Array.prototype methods. First solution uses…
Marecky
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algorithm translation to php or pseudocode from python for partition of an integer

def partitions(n): # base case of recursion: zero is the sum of the empty list if n == 0: yield [] return # modify partitions of n-1 to form partitions of n for p in partitions(n-1): yield [1] + p if…
Sushant
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