I'm fairly new to scala/spark, so forgive me if my question is elementary but I've searched everywhere and can't find the answer.
Problem
I'm trying to boost the confidence scores a bunch of network router observations (observations of probable router types at different network junctions).
I have a type NetblockObservation combines device types seen on a network with an associated netblock and a confidence. The confidence is the confidence that we accurately identified which device the device we saw.
case class NetblockObservation(
device_type: String
ip_start: Long,
ip_end: Long,
confidence: Double
)
If the confidence is above some threshold thresh, then I want that observation to be in the returned dataset. If it's below thresh, it should not be.
In addition if I have two observations with the same device_type and that one contains the other, the containee should have its confidence increased by by the confidence of the container.
Example
Let's say I have 3 Netblock Observations
// 0.0.0.0/28
NetblockObservation(device_type: "x", ip_start: 0, ip_end: 15, confidence_score: .4)
// 0.0.0.0/29
NetblockObservation(device_type: "x", ip_start: 0, ip_end: 7, confidence_score: .4)
// 0.0.0.0/30
NetblockObservation(device_type: "x", ip_start: 0, ip_end: 3, confidence_score: .4)
With a confidence threshold of 1, I would expect to have a single output of NetblockObservation(device_type: "x", ip_start: 0, ip_end: 4, confidence_score: 1.2)
Explanation: I am allowed to add the confidence scores of NetblockObservation's together if it's contained and has the same device_type
I was allowed to add the confidence score of the 0.0.0.0/29 to the confidence of the 0.0.0.0/30 because it's contained within it.
I was not allowed to add the confidence score of 0.0.0.0/30 to the 0.0.0.0/29 because 0.0.0.0/29 is not contained within 0.0.0.0/30.
My (pitiful) Attempt
Failure reason: Too slow / never completed
I attempted to implement this while simultaneously learning scala/spark so I'm not sure if it's the idea or the implementation which is wrong. I think it would eventually work but after an hour, it hadn't completed on a dataset of size 300,000 (small compared to production scale) so I gave up on it.
The idea is to find the largest netblock and separate the data into netblocks which are contained and netblocks which are not contained. The netblocks which are not contained are recursively passed back into the same function. If the largest netblock has a confidence_score of 1, the entire contained dataset is disregarded and the largest is added to return dataset. If the confidence_score is less then 1, then its confidence_score is added to everything in the contained dataset and that group is recursively passed back to the same function. Eventually, you should only be left with the data which has a confidence_score greater then 1. This algorithm also has the issue of not taking device_type into account.
def handleDataset(largestInNetData: Option[NetblockObservation], netData: RDD[NetblockObservation]): RDD[NetblockObservation] = {
if (netData.isEmpty) spark.sparkContext.emptyRDD else largestInNetData match {
case Some(largest) =>
val grouped = netData.groupBy(item =>
if (item.ip_start >= largest.ip_start && item.ip_end <= largest.ip_end) largestInNetData
else None)
def lookup(k: Option[NetblockObservation]) = grouped.filter(_._1 == k).flatMap(_._2)
val nos = handleDataset(None, lookup(None))
// Threshold is assumed to be 1
val next = if (largest.confidence_score >= 1) spark.sparkContext.parallelize(Seq(largest)) else
handleDataset(None, lookup(largestInNetData)
.filter(x => x != largest)
.map(x => x.copy(confidence_score = x.confidence_score + largest.confidence_score)))
nos ++ next
case None =>
val largest = netData.reduce((a: NetblockObservation, b: NetblockObservation) => if ((a.ip_end - a.ip_start) > (b.ip_end - b.ip_start)) a else b)
handleDataset(Option(largest), netData)
}
}
It is a fairly involved bit of code, so here is a general algorithm that I hope will help:
Forget about Spark for a moment and write a Scala function, probably in the companion object for NetblockObservation, that takes a collection of them and returns a subset of that collection that is contained. You should unit test the heck out of this function, and again this is pure Scala.
Moving now to Spark. Do a groupBy on your RDD[NetblockObservation] with device_type as the key producing essentially a map of String to Iterable[NetblockObservation].
Filter out all the entries in the map that have a value of size 1 and have a confidence below thresh.
For the entries that remain, apply your function from the first step to the collections of NetblockObservations with a mapValues.
Do a reduceByKey or similar to simply add up the confidence_scores of the contained values.
Enjoy a refreshing beverage.
Related
The code works absolutely fine for the data set containing 500000+ instances but whenever I reduce the data set to 5000/10000/15000 it throws a key error : word "***" not in vocabulary.Not for every data point but for most them it throws the error.The data set is in excel format. [1]: https://i.stack.imgur.com/YCBiQ.png
I don't know how to fix this problem since i have very little knowledge about it,,I am still learning.Please help me fix this problem!
purchases_train = []
for i in tqdm(customers_train):
temp = train_df[train_df["CustomerID"] == i]["StockCode"].tolist()
purchases_train.append(temp)
purchases_val = []
for i in tqdm(validation_df['CustomerID'].unique()):
temp = validation_df[validation_df["CustomerID"] == i]["StockCode"].tolist()
purchases_val.append(temp)
model = Word2Vec(window = 10, sg = 1, hs = 0,
negative = 10, # for negative sampling
alpha=0.03, min_alpha=0.0007,
seed = 14)
model.build_vocab(purchases_train, progress_per=200)
model.train(purchases_train, total_examples = model.corpus_count,
epochs=10, report_delay=1)
model.save("word2vec_2.model")
model.init_sims(replace=True)
# extract all vectors
X = model[model.wv.vocab]
X.shape
products = train_df[["StockCode", "Description"]]
products.drop_duplicates(inplace=True, subset='StockCode', keep="last")
products_dict=products.groupby('StockCode'['Description'].apply(list).to_dict()
def similar_products(v, n = 6):
ms = model.similar_by_vector(v, topn= n+1)[1:]
new_ms = []
for j in ms:
pair = (products_dict[j[0]][0], j[1])
new_ms.append(pair)
return new_ms
similar_products(model['21883'])
If you get a KeyError saying a word is not in the vocabulary, that's a reliable indicator that the word you're looking-up was not in the training data fed to Word2Vec, or did not appear enough (default min_count=5) times.
So, your error indicates the word-token '21883' did not appear at least 5 times in the texts (purchases_train) supplied to Word2Vec. You should do either or both of:
Ensure all words you're going to look-up appear enough times, either with more training data or a lower min_count. (However, words with only one or a few occurrences tend not to get good vectors & instead just drag the quaality of surrounding-words' vectors down - so keeping this value above 1, or even raising it above the default of 5 to discard more rare words, is a better path whenever you have sufficient data.)
If your later code will be looking up words that might not be present, either check for their presence first (word in model.wv.vocab) or set up a try: ... except: ... to catch & handle the case where they're not present.
I have some extremely old legacy procedural code which takes 10 or so enumerated inputs [ i0, i1, i2, ... i9 ] and generates 170 odd enumerated outputs [ r0, r1, ... r168, r169 ]. By enumerated, I mean that each individual input & output has its own set of distinct value sets e.g. [ red, green, yellow ] or [ yes, no ] etc.
I’m putting together the entire state table using the existing code, and instead of puzzling through them by hand, I was wondering if there was an algorithmic way of determining an appropriate function to get to each result from the 10 inputs. Note, not all input columns may be required to determine an individual output column, i.e. r124 might only be dependent on i5, i6 and i9.
These are not continuous functions, and I expect I might end up with some sort of hashing function approach, but I wondered if anyone knew of a more repeatable process I should be using instead? (If only there was some Karnaugh map like approach for multiple value non-binary functions ;-) )
If you are willing to actually enumerate all possible input/output sequences, here is a theoretical approach to tackle this that should be fairly effective.
First, consider the entropy of the output. Suppose that you have n possible input sequences, and x[i] is the number of ways to get i as an output. Let p[i] = float(x[i])/float(n[i]) and then the entropy is - sum(p[i] * log(p[i]) for i in outputs). (Note, since p[i] < 1 the log(p[i]) is a negative number, and therefore the entropy is positive. Also note, if p[i] = 0 then we assume that p[i] * log(p[i]) is also zero.)
The amount of entropy can be thought of as the amount of information needed to predict the outcome.
Now here is the key question. What variable gives us the most information about the output per information about the input?
If a particular variable v has in[v] possible values, the amount of information in specifying v is log(float(in[v])). I already described how to calculate the entropy of the entire set of outputs. For each possible value of v we can calculate the entropy of the entire set of outputs for that value of v. The amount of information given by knowing v is the entropy of the total set minus the average of the entropies for the individual values of v.
Pick the variable v which gives you the best ratio of information_gained_from_v/information_to_specify_v. Your algorithm will start with a switch on the set of values of that variable.
Then for each value, you repeat this process to get cascading nested if conditions.
This will generally lead to a fairly compact set of cascading nested if conditions that will focus on the input variables that tell you as much as possible, as quickly as possible, with as few branches as you can manage.
Now this assumed that you had a comprehensive enumeration. But what if you don't?
The answer to that is that the analysis that I described can be done for a random sample of your possible set of inputs. So if you run your code with, say, 10,000 random inputs, then you'll come up with fairly good entropies for your first level. Repeat with 10,000 each of your branches on your second level, and the same will happen. Continue as long as it is computationally feasible.
If there are good patterns to find, you will quickly find a lot of patterns of the form, "If you put in this that and the other, here is the output you always get." If there is a reasonably short set of nested ifs that give the right output, you're probably going to find it. After that, you have the question of deciding whether to actually verify by hand that each bucket is reliable, or to trust that if you couldn't find any exceptions with 10,000 random inputs, then there are none to be found.
Tricky approach for the validation. If you can find fuzzing software written for your language, run the fuzzing software with the goal of trying to tease out every possible internal execution path for each bucket you find. If the fuzzing software decides that you can't get different answers than the one you think is best from the above approach, then you can probably trust it.
Algorithm is pretty straightforward. Given possible values for each input we can generate all the input vectors possible. Then per each output we can just eliminate these inputs that do no matter for the output. As the result we for each output we can get a matrix showing output values for all the input combinations excluding the inputs that do not matter for given output.
Sample input format (for code snipped below):
var schema = new ConvertionSchema()
{
InputPossibleValues = new object[][]
{
new object[] { 1, 2, 3, }, // input #0
new object[] { 'a', 'b', 'c' }, // input #1
new object[] { "foo", "bar" }, // input #2
},
Converters = new System.Func<object[], object>[]
{
input => input[0], // output #0
input => (int)input[0] + (int)(char)input[1], // output #1
input => (string)input[2] == "foo" ? 1 : 42, // output #2
input => input[2].ToString() + input[1].ToString(), // output #3
input => (int)input[0] % 2, // output #4
}
};
Sample output:
Leaving the heart of the backward conversion below. Full code in a form of Linqpad snippet is there: http://share.linqpad.net/cknrte.linq.
public void Reverse(ConvertionSchema schema)
{
// generate all possible input vectors and record the resul for each case
// then for each output we could figure out which inputs matters
object[][] inputs = schema.GenerateInputVectors();
// reversal path
for (int outputIdx = 0; outputIdx < schema.OutputsCount; outputIdx++)
{
List<int> inputsThatDoNotMatter = new List<int>();
for (int inputIdx = 0; inputIdx < schema.InputsCount; inputIdx++)
{
// find all groups for input vectors where all other inputs (excluding current) are the same
// if across these groups outputs are exactly the same, then it means that current input
// does not matter for given output
bool inputMatters = inputs.GroupBy(input => ExcudeByIndexes(input, new[] { inputIdx }), input => schema.Convert(input)[outputIdx], ObjectsByValuesComparer.Instance)
.Where(x => x.Distinct().Count() > 1)
.Any();
if (!inputMatters)
{
inputsThatDoNotMatter.Add(inputIdx);
Util.Metatext($"Input #{inputIdx} does not matter for output #{outputIdx}").Dump();
}
}
// mapping table (only inputs that matters)
var mapping = new List<dynamic>();
foreach (var inputGroup in inputs.GroupBy(input => ExcudeByIndexes(input, inputsThatDoNotMatter), ObjectsByValuesComparer.Instance))
{
dynamic record = new ExpandoObject();
object[] sampleInput = inputGroup.First();
object output = schema.Convert(sampleInput)[outputIdx];
for (int inputIdx = 0; inputIdx < schema.InputsCount; inputIdx++)
{
if (inputsThatDoNotMatter.Contains(inputIdx))
continue;
AddProperty(record, $"Input #{inputIdx}", sampleInput[inputIdx]);
}
AddProperty(record, $"Output #{outputIdx}", output);
mapping.Add(record);
}
// input x, ..., input y, output z form is needed
mapping.Dump();
}
}
public BinomialModelPrediction predictBinomial(RowData data) throws PredictException {
double[] preds = this.preamble(ModelCategory.Binomial, data);
BinomialModelPrediction p = new BinomialModelPrediction();
double d = preds[0];
p.labelIndex = (int)d;
String[] domainValues = this.m.getDomainValues(this.m.getResponseIdx());
p.label = domainValues[p.labelIndex];
p.classProbabilities = new double[this.m.getNumResponseClasses()];
System.arraycopy(preds, 1, p.classProbabilities, 0, p.classProbabilities.length);
if(this.m.calibrateClassProbabilities(preds)) {
p.calibratedClassProbabilities = new double[this.m.getNumResponseClasses()];
System.arraycopy(preds, 1, p.calibratedClassProbabilities, 0, p.calibratedClassProbabilities.length);
}
return p;
}
Eg: classProbabilities =[0.82333,0,276666]
labelIndex = 1
label = true
domainValues = [false,true]
what does this labelIndex signifies and does the class probabilities
order is same as the domain value order ,If order is same then it means that here probability of false is 0.82333 and probability of true is 0.27666 but why is this labelIndex showing as 1 and label as true.
Please help me to figure out this issue.
Like Tom commented, the prediction is not "wrong". You can infer from this that the threshold H2O has chosen is less than 0.27666. You probably have imbalanced training data, otherwise H2O would have not picked a low threshold for classifying a predicted value of 0.27666 as a 1. Does your training set include fewer examples of the positive class than the negative class?
If you don't like that threshold for whatever reason, then you can manually create your own. Just make sure you know how to properly evaluate the effect of using different thresholds on the performance of your model, otherwise I'd recommend just using the default threshold.
The name, "classProbabilities" is a misnomer. These are not actual probabilities, they are predicted values, though people often use the terms interchangeably. Binary classification algorithms produce "predicted values" that look like probabilities when they're between 0 and 1, but unless a calibration process is performed, they are not going to represent the probabilities. Calibration is not necessarily a straight-forward process and there are many techniques. Here's some more info about calibration methods for imbalanced data. In H2O, you can perform calibration using Platt scaling using the calibrate_model option. But this is probably not really necessary to what you're trying to do.
The proper way to use the raw output from a binary classification model is to only look at the predicted value for the positive class (you can simply ignore the predicted value for the negative class). Then you choose a threshold which suits your needs, or you can use the default threshold in H2O, which is chosen to maximize the F1 score. Some other software will use a hardcoded threshold of 0.5, but that will be a terrible choice if you don't have an even number of positive and negative examples in your training data. If you have only a few positive examples in your training data, then the best threshold will be something much lower than 0.5.
This is an algorithmic question about a somewhat complex problem. The foundation is this:
A scheduling system based on available slots and reserved slots. Slots have certain criteria, let's call them tags. A reservation is matched to an available slot by those tags, if the available slot's tag set is a superset of the reserved slot.
As a concrete example, take this scenario:
11:00 12:00 13:00
+--------+
| A, B |
+--------+
+--------+
| C, D |
+--------+
Between the times of 11:00 to 12:30 reservations for the tags A and B can be made, from 12:00 to 13:30 C and D is available, and there's an overlap from about 12:00 to 12:30.
11:00 12:00 13:00
+--------+
| A, B |
+--------+
+--------+
| C, D |
+--------+
xxxxxx
x A x
xxxxxx
Here a reservation for A has been made, so no other reservations for A or B can be made between 11:15-ish and 12:00-ish.
That's the idea in a nutshell. There are no specific limitations for the available slots:
an available slot can contain any number of tags
any number of slots can overlap at any time
slots are of arbitrary length
reservations can contain any number of tags
The only rule that needs to be obeyed in the system is:
when adding a reservation, at least one remaining available slot must match all the tags in the reservation
To clarify: when there are two available slots at the same time with, say, tag A, then two reservations for A can be made at that time, but no more.
I have that working with a modified implementation of an interval tree; as a quick overview:
all available slots are added to the interval tree (duplicates/overlaps are preserved)
all reserved slots are iterated and:
all available slots matching the time of the reservation are queried from the tree
the first of those matching the reservation's tags is sliced and the slice removed from the tree
When that process is finished, what's left are the remaining slices of available slots, and I can query whether a new reservation can be made for a particular time and add it.
Data structures look something like this:
{
type: 'available',
begin: 1497857244,
end: 1497858244,
tags: [{ foo: 'bar' }, { baz: 42 }]
}
{
type: 'reserved',
begin: 1497857345,
end: 1497857210,
tags: [{ foo: 'bar' }]
}
Tags are themselves key-value objects, a list of them is a "tag set". Those could be serialised if it helps; so far I'm using a Python set type which makes comparison easy enough. Slot begin/end times are UNIX time stamps within the tree. I'm not particularly married to these specific data structures and can refactor them if it's useful.
The problem I'm facing is that this doesn't work bug-free; every once in a while a reservation sneaks its way into the system that conflicts with other reservations, and I couldn't yet figure out how that can happen exactly. It's also not very clever when tags overlap in a complex way where the optimal distribution needs to be calculated so all reservations can be fit into the available slots as best as possible; in fact currently it's non-deterministic how reservations are matched to available slots in overlapping scenarios.
What I want to know is: interval trees are mostly great for this purpose, but my current system to add tag set matching as an additional dimension to this is clunky and bolted-on; is there a data structure or algorithm that can handle this in an elegant way?
Actions that must be supported:
Querying the system for available slots that match certain tag sets (taking into account reservations that may reduce availability but are not themselves part of said tag set; e.g. in the example above querying for an availability for B).
Ensuring no reservations can be added to the system which don't have a matching available slot.
Your problem can be solved using constraint programming. In python this can be implemented using the python-constraint library.
First, we need a way to check if two slots are consistent with each other. this is a function that returns true if two slots share a tag and their rimeframes overlap. In python this can be implemented using the following function
def checkNoOverlap(slot1, slot2):
shareTags = False
for tag in slot1['tags']:
if tag in slot2['tags']:
shareTags = True
break
if not shareTags: return True
return not (slot2['begin'] <= slot1['begin'] <= slot2['end'] or
slot2['begin'] <= slot1['end'] <= slot2['end'])
I was not sure whether you wanted the tags to be completely the same (like {foo: bar} equals {foo: bar}) or only the keys (like {foo: bar} equals {foo: qux}), but you can change that in the function above.
Consistency check
We can use the python-constraint module for the two kinds of functionality you requested.
The second functionality is the easiest. To implement this, we can use the function isConsistent(set) which takes a list of slots in the provided data structure as input. The function will then feed all the slots to python-constraint and will check if the list of slots is consistent (no 2 slots that shouldn't overlap, overlap) and return the consistency.
def isConsistent(set):
#initialize python-constraint context
problem = Problem()
#add all slots the context as variables with a singleton domain
for i in range(len(set)):
problem.addVariable(i, [set[i]])
#add a constraint for each possible pair of slots
for i in range(len(set)):
for j in range(len(set)):
#we don't want slots to be checked against themselves
if i == j:
continue
#this constraint uses the checkNoOverlap function
problem.addConstraint(lambda a,b: checkNoOverlap(a, b), (i, j))
# getSolutions returns all the possible combinations of domain elements
# because all domains are singleton, this either returns a list with length 1 (consistent) or 0 (inconsistent)
return not len(problem.getSolutions()) == 0
This function can be called whenever a user wants to add a reservation slot. The input slot can be added to the list of already existing slots and the consistency can be checked. If it is consistent, the new slot an be reserverd. Else, the new slot overlaps and should be rejected.
Finding available slots
This problem is a bit trickier. We can use the same functionality as above with a few significant changes. Instead of adding the new slot together with the existing slot, we now want to add all possible slots to the already existing slots. We can then check the consistency of all those possible slots with the reserved slots and ask the constraint system for the combinations that are consistent.
Because the number of possible slots would be infinite if we didn't put any restrictions on it, we first need to declare some parameters for the program:
MIN = 149780000 #available time slots can never start earlier then this time
MAX = 149790000 #available time slots can never start later then this time
GRANULARITY = 1*60 #possible time slots are always at least one minut different from each other
We can now continue to the main function. It looks a lot like the consistency check, but instead of the new slot from the user, we now add a variable to discover all available slots.
def availableSlots(tags, set):
#same as above
problem = Problem()
for i in range(len(set)):
problem.addVariable(i, [set[i]])
#add an extra variable for the available slot is added, with a domain of all possible slots
problem.addVariable(len(set), generatePossibleSlots(MIN, MAX, GRANULARITY, tags))
for i in range(len(set) +1):
for j in range(len(set) +1):
if i == j:
continue
problem.addConstraint(lambda a, b: checkNoOverlap(a, b), (i, j))
#extract the available time slots from the solution for clean output
return filterAvailableSlots(problem.getSolutions())
I use some helper functions to keep the code cleaner. They are included here.
def filterAvailableSlots(possibleCombinations):
result = []
for slots in possibleCombinations:
for key, slot in slots.items():
if slot['type'] == 'available':
result.append(slot)
return result
def generatePossibleSlots(min, max, granularity, tags):
possibilities = []
for i in range(min, max - 1, granularity):
for j in range(i + 1, max, granularity):
possibleSlot = {
'type': 'available',
'begin': i,
'end': j,
'tags': tags
}
possibilities.append(possibleSlot)
return tuple(possibilities)
You can now use the function getAvailableSlots(tags, set) with the tags for which you want the available slots and a set of already reserved slots. Note that this function really return all the consistent possible slots, so no effort is done to find the one of maximum lenght or for other optimalizations.
Hope this helps! (I got it to work as you described in my pycharm)
Here's a solution, I'll include all the code below.
1. Create a table of slots, and a table of reservations
2. Create a matrix of reservations x slots
which is populated by true or false values based on whether that reservation-slot combination are possible
3. Figure out the best mapping that allows for the most Reservation-Slot Combinations
Note: my current solution scales poorly with very large arrays as it involves looping through all possible permutations of a list with size = number of slots. I've posted another question to see if anyone can find a better way of doing this. However, this solution is accurate and can be optimized
Python Code Source
Part 1
from IPython.display import display
import pandas as pd
import datetime
available_data = [
['SlotA', datetime.time(11, 0, 0), datetime.time(12, 30, 0), set(list('ABD'))],
['SlotB',datetime.time(12, 0, 0), datetime.time(13, 30, 0), set(list('C'))],
['SlotC',datetime.time(12, 0, 0), datetime.time(13, 30, 0), set(list('ABCD'))],
['SlotD',datetime.time(12, 0, 0), datetime.time(13, 30, 0), set(list('AD'))],
]
reservation_data = [
['ReservationA', datetime.time(11, 15, 0), datetime.time(12, 15, 0), set(list('AD'))],
['ReservationB', datetime.time(11, 15, 0), datetime.time(12, 15, 0), set(list('A'))],
['ReservationC', datetime.time(12, 0, 0), datetime.time(12, 15, 0), set(list('C'))],
['ReservationD', datetime.time(12, 0, 0), datetime.time(12, 15, 0), set(list('C'))],
['ReservationE', datetime.time(12, 0, 0), datetime.time(12, 15, 0), set(list('D'))]
]
reservations = pd.DataFrame(data=reservation_data, columns=['reservations', 'begin', 'end', 'tags']).set_index('reservations')
slots = pd.DataFrame(data=available_data, columns=['slots', 'begin', 'end', 'tags']).set_index('slots')
display(slots)
display(reservations)
Part 2
def is_possible_combination(r):
return (r['begin'] >= slots['begin']) & (r['end'] <= slots['end']) & (r['tags'] <= slots['tags'])
solution_matrix = reservations.apply(is_possible_combination, axis=1).astype(int)
display(solution_matrix)
Part 3
import numpy as np
from itertools import permutations
# add dummy columns to make the matrix square if it is not
sqr_matrix = solution_matrix
if sqr_matrix.shape[0] > sqr_matrix.shape[1]:
# uhoh, there are more reservations than slots... this can't be good
for i in range(sqr_matrix.shape[0] - sqr_matrix.shape[1]):
sqr_matrix.loc[:,'FakeSlot' + str(i)] = [1] * sqr_matrix.shape[0]
elif sqr_matrix.shape[0] < sqr_matrix.shape[1]:
# there are more slots than customers, why doesn't anyone like us?
for i in range(sqr_matrix.shape[0] - sqr_matrix.shape[1]):
sqr_matrix.loc['FakeCustomer' + str(i)] = [1] * sqr_matrix.shape[1]
# we only want the values now
A = solution_matrix.values.astype(int)
# make an identity matrix (the perfect map)
imatrix = np.diag([1]*A.shape[0])
# randomly swap columns on the identity matrix until they match.
n = A.shape[0]
# this will hold the map that works the best
best_map_so_far = np.zeros([1,1])
for column_order in permutations(range(n)):
# this is an identity matrix with the columns swapped according to the permutation
imatrix = np.zeros(A.shape)
for row, column in enumerate(column_order):
imatrix[row,column] = 1
# is this map better than the previous best?
if sum(sum(imatrix * A)) > sum(sum(best_map_so_far)):
best_map_so_far = imatrix
# could it be? a perfect map??
if sum(sum(imatrix * A)) == n:
break
if sum(sum(imatrix * A)) != n:
print('a perfect map was not found')
output = pd.DataFrame(A*imatrix, columns=solution_matrix.columns, index=solution_matrix.index, dtype=int)
display(output)
The suggested approaches by Arne and tinker were both helpful, but not ultimately sufficient. I came up with a hybrid approach that solves it well enough.
The main problem is that it's a three-dimensional issue, which is difficult to solve in all dimensions at once. It's not just about matching a time overlap or a tag overlap, it's about matching time slices with tag overlaps. It's simple enough to match slots to other slots based on time and even tags, but it's then pretty complicated to match an already matched availability slot to another reservation at another time. Meaning, this scenario in which one availability can cover two reservations at different times:
+---------+
| A, B |
+---------+
xxxxx xxxxx
x A x x A x
xxxxx xxxxx
Trying to fit this into constraint based programming requires an incredibly complex relationship of constraints which is hardly manageable. My solution to this was to simplify the problem…
Removing one dimension
Instead of solving all dimensions at once, it simplifies the problem enormously to largely remove the dimension of time. I did this by using my existing interval tree and slicing it as needed:
def __init__(self, slots):
self.tree = IntervalTree(slots)
def timeslot_is_available(self, start: datetime, end: datetime, attributes: set):
candidate = Slot(start.timestamp(), end.timestamp(), dict(type=SlotType.RESERVED, attributes=attributes))
slots = list(self.tree[start.timestamp():end.timestamp()])
return self.model_is_consistent(slots + [candidate])
To query whether a specific slot is available, I take only the slots relevant at that specific time (self.tree[..:..]), which reduces the complexity of the calculation to a localised subset:
| | +-+ = availability
+-|------|-+ xxx = reservation
| +---|------+
xx|x xxx|x
| xxxx|
| |
Then I confirm the consistency within that narrow slice:
#staticmethod
def model_is_consistent(slots):
def can_handle(r):
return lambda a: r.attributes <= a.attributes and a.contains_interval(r)
av = [s for s in slots if s.type == SlotType.AVAILABLE]
rs = [s for s in slots if s.type == SlotType.RESERVED]
p = Problem()
p.addConstraint(AllDifferentConstraint())
p.addVariables(range(len(rs)), av)
for i, r in enumerate(rs):
p.addConstraint(can_handle(r), (i,))
return p.getSolution() is not None
(I'm omitting some optimisations and other code here.)
This part is the hybrid approach of Arne's and tinker's suggestions. It uses constraint-based programming to find matching slots, using the matrix algorithm suggested by tinker. Basically: if there's any solution to this problem in which all reservations can be assigned to a different available slot, then this time slice is in a consistent state. Since I'm passing in the desired reservation slot, if the model is still consistent including that slot, this means it's safe to reserve that slot.
This is still problematic if there are two short reservations assignable to the same availability within this narrow window, but the chances of that are low and the result is merely a false negative for an availability query; false positives would be more problematic.
Finding available slots
Finding all available slots is a more complex problem, so again some simplification is necessary. First, it's only possible to query the model for availabilities for a particular set of tags (there's no "give me all globally available slots"), and secondly it can only be queried with a particular granularity (desired slot length). This suits me well for my particular use case, in which I just need to offer users a list of slots they can reserve, like 9:15-9:30, 9:30-9:45, etc.. This makes the algorithm very simple by reusing the above code:
def free_slots(self, start: datetime, end: datetime, attributes: set, granularity: timedelta):
slots = []
while start < end:
slot_end = start + granularity
if self.timeslot_is_available(start, slot_end, attributes):
slots.append((start, slot_end))
start += granularity
return slots
In other words, it just goes through all possible slots during the given time interval and literally checks whether that slot is available. It's a bit of a brute-force solution, but works perfectly fine.
Ok, so say you have a really big Range in ruby. I want to find a way to get the max value in the Range.
The Range is exclusive (defined with three dots) meaning that it does not include the end object in it's results. It could be made up of Integer, String, Time, or really any object that responds to #<=> and #succ. (which are the only requirements for the start/end object in Range)
Here's an example of an exclusive range:
past = Time.local(2010, 1, 1, 0, 0, 0)
now = Time.now
range = past...now
range.include?(now) # => false
Now I know I could just do something like this to get the max value:
range.max # => returns 1 second before "now" using Enumerable#max
But this will take a non-trivial amount of time to execute. I also know that I could subtract 1 second from whatever the end object is is. However, the object may be something other than Time, and it may not even support #-. I would prefer to find an efficient general solution, but I am willing to combine special case code with a fallback to a general solution (more on that later).
As mentioned above using Range#last won't work either, because it's an exclusive range and does not include the last value in it's results.
The fastest approach I could think of was this:
max = nil
range.each { |value| max = value }
# max now contains nil if the range is empty, or the max value
This is similar to what Enumerable#max does (which Range inherits), except that it exploits the fact that each value is going to be greater than the previous, so we can skip using #<=> to compare the each value with the previous (the way Range#max does) saving a tiny bit of time.
The other approach I was thinking about was to have special case code for common ruby types like Integer, String, Time, Date, DateTime, and then use the above code as a fallback. It'd be a bit ugly, but probably much more efficient when those object types are encountered because I could use subtraction from Range#last to get the max value without any iterating.
Can anyone think of a more efficient/faster approach than this?
The simplest solution that I can think of, which will work for inclusive as well as exclusive ranges:
range.max
Some other possible solutions:
range.entries.last
range.entries[-1]
These solutions are all O(n), and will be very slow for large ranges. The problem in principle is that range values in Ruby are enumerated using the succ method iteratively on all values, starting at the beginning. The elements do not have to implement a method to return the previous value (i.e. pred).
The fastest method would be to find the predecessor of the last item (an O(1) solution):
range.exclude_end? ? range.last.pred : range.last
This works only for ranges that have elements which implement pred. Later versions of Ruby implement pred for integers. You have to add the method yourself if it does not exist (essentially equivalent to special case code you suggested, but slightly simpler to implement).
Some quick benchmarking shows that this last method is the fastest by many orders of magnitude for large ranges (in this case range = 1...1000000), because it is O(1):
user system total real
r.entries.last 11.760000 0.880000 12.640000 ( 12.963178)
r.entries[-1] 11.650000 0.800000 12.450000 ( 12.627440)
last = nil; r.each { |v| last = v } 20.750000 0.020000 20.770000 ( 20.910416)
r.max 17.590000 0.010000 17.600000 ( 17.633006)
r.exclude_end? ? r.last.pred : r.last 0.000000 0.000000 0.000000 ( 0.000062)
Benchmark code is here.
In the comments it is suggested to use range.last - (range.exclude_end? ? 1 : 0). It does work for dates without additional methods, but will never work for non-numeric ranges. String#- does not exist and makes no sense with integer arguments. String#pred, however, can be implented.
I'm not sure about the speed (and initial tests don't seem incredibly fast), but the following might do what you need:
past = Time.local(2010, 1, 1, 0, 0, 0)
now = Time.now
range = past...now
range.to_a[-1]
Very basic testing (counting in my head) showed that it took about 4 seconds while the method you provided took about 5-6. Hope this helps.
Edit 1: Removed second solution as it was totally wrong.
I can't think there's any way to achieve this that doesn't involve enumerating the range, at least unless as already mentioned, you have other information about how the range will be constructed and therefore can infer the desired value without enumeration. Of all the suggestions, I'd go with #max, since it seems to be most expressive.
require 'benchmark'
N = 20
Benchmark.bm(30) do |r|
past, now = Time.local(2010, 2, 1, 0, 0, 0), Time.now
#range = past...now
r.report("range.max") do
N.times { last_in_range = #range.max }
end
r.report("explicit enumeration") do
N.times { #range.each { |value| last_in_range = value } }
end
r.report("range.entries.last") do
N.times { last_in_range = #range.entries.last }
end
r.report("range.to_a[-1]") do
N.times { last_in_range = #range.to_a[-1] }
end
end
user system total real
range.max 49.406000 1.515000 50.921000 ( 50.985000)
explicit enumeration 52.250000 1.719000 53.969000 ( 54.156000)
range.entries.last 53.422000 4.844000 58.266000 ( 58.390000)
range.to_a[-1] 49.187000 5.234000 54.421000 ( 54.500000)
I notice that the 3rd and 4th option have significantly increased system time. I expect that's related to the explicit creation of an array, which seems like a good reason to avoid them, even if they're not obviously more expensive in elapsed time.