I've come across the requirement to collect the percentiles from a list a few times:
Within what percentile is a certain number?
What is the nth percentile in a list?
I have written these methods to solve the issue:
/for 1:
percentileWithinThreshold:{[threshold;list] (100 * count where list <= threshold) % count list};
/for 2:
thresholdForPercentile:{[percentile;list] (asc list)[-1 + "j"$((percentile % 100) * count list)]};
They work well for both use cases, but I was thinking this is a too common use case, so probably Q offers already something out of the box that does the same. Any idea if there already exists something else?
'100 xrank' generates percentiles.
q) 100 xrank 1 2 3 4
q) 0 25 50 75
Solution for your second requirement:
q) f:{ y (100 xrank y:asc y) bin x}
Also, note that your second function result will not be always same as xrank. Reason for that is 'xrank' uses floor for fractional index output which is the normal scenario with calculating percentiles and your function round up the value and subtracts -1 which ensures that output will always be lesser-equal to input percentile. For example:
q) thresholdForPercentile[63;til 21] / output 12
q) f[63;til 21] / output 13
For first requirement, there is no inbuilt function. However you could improve your function if you keep your input list sorted because in that case you could use 'bin' function which runs faster on big lists.
q) percentileWithinThreshold:{[threshold;list] (100 * 1+list bin threshold) % count list};
Remember that 'bin' will throw type error if one argument is of float type and other is an integer. So make sure to cast them correctly inside the function.
qtln:{[x;y;z]cf:(0 1;1%2 2;0 0;1 1;1%3 3;3%8 8) z-4;n:count y:asc y;?[hf<1;first y;last y]^y[hf-1]+(h-hf)*y[hf]-y -1+hf:floor h:cf[0]+x*n+1f-sum cf}
qtl:qtln[;;8];
Basically I'm looking for a formula to see how many times Yes was used vs. No.
I have something like this:
(({Command.result} ="Yes") / {Command.result})*100
Which makes sense in my head, but I keep getting:
A number, or currency amount is required.
Your current formula attempts to divide a boolean type through by a string. You can only perform division with numbers.
Instead, create two formulas as individual counts of Yes or No:
#YesCount:
If ({Command.result} = "Yes") Then 1 Else 0
#NoCount:
If ({Command.result} = "No") Then 1 Else 0
For the percentage, create two more formulas:
#YesPercent:
100 / Count ({Command.result}) * Sum ({#YesCount})
#NoPercent:
100 / Count ({Command.result}) * Sum ({#NoCount})
I have MDX (similar to one questioned and answered here):
(
[PX Market].[PX MARKET NAME].&[Elbas],
[Measures].[PX QUANTITY]
)
It works for me (it filters measures to value "Elbas" only). But I need another filtering - to have only values which are < or > than 0. There shoud be some condition similar to "[Measures].[PX QUANTITY] < 0". But I do not know how to implement it.
Thank for any of your advice.
Ondra
Table looks similar like this:
PX_MARKET_NAME; PX_QUANTITY
Elbas; 5
Elbas; -3
Elspot; 4
In result I need only 2nd value (-3). Which belongs to Elbas and is smaller then 0.
So far I tried this, but it is now working :(
FILTER
(
[PX Market].[PX MARKET NAME].&[Elbas],
[Measures].[PX PURCHASE]
) < 0
Try that:
IIF(
([PX Market].[PX MARKET NAME].&[Elbas],[Measures].[PX QUANTITY]) < 0,
([PX Market].[PX MARKET NAME].&[Elbas],[Measures].[PX QUANTITY]),
NULL
)
So I've spent hours trying to work out exactly how this code produces prime numbers.
lazy val ps: Stream[Int] = 2 #:: Stream.from(3).filter(i =>
ps.takeWhile{j => j * j <= i}.forall{ k => i % k > 0});
I've used a number of printlns etc, but nothings making it clearer.
This is what I think the code does:
/**
* [2,3]
*
* takeWhile 2*2 <= 3
* takeWhile 2*2 <= 4 found match
* (4 % [2,3] > 1) return false.
* takeWhile 2*2 <= 5 found match
* (5 % [2,3] > 1) return true
* Add 5 to the list
* takeWhile 2*2 <= 6 found match
* (6 % [2,3,5] > 1) return false
* takeWhile 2*2 <= 7
* (7 % [2,3,5] > 1) return true
* Add 7 to the list
*/
But If I change j*j in the list to be 2*2 which I assumed would work exactly the same, it causes a stackoverflow error.
I'm obviously missing something fundamental here, and could really use someone explaining this to me like I was a five year old.
Any help would be greatly appreciated.
I'm not sure that seeking a procedural/imperative explanation is the best way to gain understanding here. Streams come from functional programming and they're best understood from that perspective. The key aspects of the definition you've given are:
It's lazy. Other than the first element in the stream, nothing is computed until you ask for it. If you never ask for the 5th prime, it will never be computed.
It's recursive. The list of prime numbers is defined in terms of itself.
It's infinite. Streams have the interesting property (because they're lazy) that they can represent a sequence with an infinite number of elements. Stream.from(3) is an example of this: it represents the list [3, 4, 5, ...].
Let's see if we can understand why your definition computes the sequence of prime numbers.
The definition starts out with 2 #:: .... This just says that the first number in the sequence is 2 - simple enough so far.
The next part defines the rest of the prime numbers. We can start with all the counting numbers starting at 3 (Stream.from(3)), but we obviously need to filter a bunch of these numbers out (i.e., all the composites). So let's consider each number i. If i is not a multiple of a lesser prime number, then i is prime. That is, i is prime if, for all primes k less than i, i % k > 0. In Scala, we could express this as
nums.filter(i => ps.takeWhile(k => k < i).forall(k => i % k > 0))
However, it isn't actually necessary to check all lesser prime numbers -- we really only need to check the prime numbers whose square is less than or equal to i (this is a fact from number theory*). So we could instead write
nums.filter(i => ps.takeWhile(k => k * k <= i).forall(k => i % k > 0))
So we've derived your definition.
Now, if you happened to try the first definition (with k < i), you would have found that it didn't work. Why not? It has to do with the fact that this is a recursive definition.
Suppose we're trying to decide what comes after 2 in the sequence. The definition tells us to first determine whether 3 belongs. To do so, we consider the list of primes up to the first one greater than or equal to 3 (takeWhile(k => k < i)). The first prime is 2, which is less than 3 -- so far so good. But we don't yet know the second prime, so we need to compute it. Fine, so we need to first see whether 3 belongs ... BOOM!
* It's pretty easy to see that if a number n is composite then the square of one of its factors must be less than or equal to n. If n is composite, then by definition n == a * b, where 1 < a <= b < n (we can guarantee a <= b just by labeling the two factors appropriately). From a <= b it follows that a^2 <= a * b, so it follows that a^2 <= n.
Your explanations are mostly correct, you made only two mistakes:
takeWhile doesn't include the last checked element:
scala> List(1,2,3).takeWhile(_<2)
res1: List[Int] = List(1)
You assume that ps always contains only a two and a three but because Stream is lazy it is possible to add new elements to it. In fact each time a new prime is found it is added to ps and in the next step takeWhile will consider this new added element. Here, it is important to remember that the tail of a Stream is computed only when it is needed, thus takeWhile can't see it before forall is evaluated to true.
Keep these two things in mind and you should came up with this:
ps = [2]
i = 3
takeWhile
2*2 <= 3 -> false
forall on []
-> true
ps = [2,3]
i = 4
takeWhile
2*2 <= 4 -> true
3*3 <= 4 -> false
forall on [2]
4%2 > 0 -> false
ps = [2,3]
i = 5
takeWhile
2*2 <= 5 -> true
3*3 <= 5 -> false
forall on [2]
5%2 > 0 -> true
ps = [2,3,5]
i = 6
...
While these steps describe the behavior of the code, it is not fully correct because not only adding elements to the Stream is lazy but every operation on it. This means that when you call xs.takeWhile(f) not all values until the point when f is false are computed at once - they are computed when forall wants to see them (because it is the only function here that needs to look at all elements before it definitely can result to true, for false it can abort earlier). Here the computation order when laziness is considered everywhere (example only looking at 9):
ps = [2,3,5,7]
i = 9
takeWhile on 2
2*2 <= 9 -> true
forall on 2
9%2 > 0 -> true
takeWhile on 3
3*3 <= 9 -> true
forall on 3
9%3 > 0 -> false
ps = [2,3,5,7]
i = 10
...
Because forall is aborted when it evaluates to false, takeWhile doesn't calculate the remaining possible elements.
That code is easier (for me, at least) to read with some variables renamed suggestively, as
lazy val ps: Stream[Int] = 2 #:: Stream.from(3).filter(i =>
ps.takeWhile{p => p * p <= i}.forall{ p => i % p > 0});
This reads left-to-right quite naturally, as
primes are 2, and those numbers i from 3 up, that all of the primes p whose square does not exceed the i, do not divide i evenly (i.e. without some non-zero remainder).
In a true recursive fashion, to understand this definition as defining the ever increasing stream of primes, we assume that it is so, and from that assumption we see that no contradiction arises, i.e. the truth of the definition holds.
The only potential problem after that, is the timing of accessing the stream ps as it is being defined. As the first step, imagine we just have another stream of primes provided to us from somewhere, magically. Then, after seeing the truth of the definition, check that the timing of the access is okay, i.e. we never try to access the areas of ps before they are defined; that would make the definition stuck, unproductive.
I remember reading somewhere (don't recall where) something like the following -- a conversation between a student and a wizard,
student: which numbers are prime?
wizard: well, do you know what number is the first prime?
s: yes, it's 2.
w: okay (quickly writes down 2 on a piece of paper). And what about the next one?
s: well, next candidate is 3. we need to check whether it is divided by any prime whose square does not exceed it, but I don't yet know what the primes are!
w: don't worry, I'l give them to you. It's a magic I know; I'm a wizard after all.
s: okay, so what is the first prime number?
w: (glances over the piece of paper) 2.
s: great, so its square is already greater than 3... HEY, you've cheated! .....
Here's a pseudocode1 translation of your code, read partially right-to-left, with some variables again renamed for clarity (using p for "prime"):
ps = 2 : filter (\i-> all (\p->rem i p > 0) (takeWhile (\p->p^2 <= i) ps)) [3..]
which is also
ps = 2 : [i | i <- [3..], and [rem i p > 0 | p <- takeWhile (\p->p^2 <= i) ps]]
which is a bit more visually apparent, using list comprehensions. and checks that all entries in a list of Booleans are True (read | as "for", <- as "drawn from", , as "such that" and (\p-> ...) as "lambda of p").
So you see, ps is a lazy list of 2, and then of numbers i drawn from a stream [3,4,5,...] such that for all p drawn from ps such that p^2 <= i, it is true that i % p > 0. Which is actually an optimal trial division algorithm. :)
There's a subtlety here of course: the list ps is open-ended. We use it as it is being "fleshed-out" (that of course, because it is lazy). When ps are taken from ps, it could potentially be a case that we run past its end, in which case we'd have a non-terminating calculation on our hands (a "black hole"). It just so happens :) (and needs to ⁄ can be proved mathematically) that this is impossible with the above definition. So 2 is put into ps unconditionally, so there's something in it to begin with.
But if we try to "simplify",
bad = 2 : [i | i <- [3..], and [rem i p > 0 | p <- takeWhile (\p->p < i) bad]]
it stops working after producing just one number, 2: when considering 3 as the candidate, takeWhile (\p->p < 3) bad demands the next number in bad after 2, but there aren't yet any more numbers there. It "jumps ahead of itself".
This is "fixed" with
bad = 2 : [i | i <- [3..], and [rem i p > 0 | p <- [2..(i-1)] ]]
but that is a much much slower trial division algorithm, very far from the optimal one.
--
1 (Haskell actually, it's just easier for me that way :) )
First of all, excuse me if I do any mistakes, but English is not a language I use very often.
I have a data frame with numbers. A small part of the data frame is this:
nominal 2
2
2
2
ordinal
2
1
1
2
So, I want to use the gower distance function on these numbers.
Here ( http://rgm2.lab.nig.ac.jp/RGM2/R_man-2.9.0/library/StatMatch/man/gower.dist.html ) says that in order to use gower.dist, all nominal variables must be of class "factor" and all ordinal variables of class "ordered".
By default, all the columns are of class "integer" and mode "numeric". In order to change the class of the columns, i use these commands:
DF=read.table("clipboard",header=TRUE,sep="\t")
# I select all the cells and I copy them to the clipboard.
#Then R, with this command, reads the data from there.
MyHeader=names(DF) # I save the headers of the data frame to a temp matrix
for (i in 1:length(DF)) {
if (MyHeader[[i]]=="nominal") DF[[i]]=as.factor(DF[[i]])
}
for (i in 1:length(DF)) {
if (MyHeader[[i]]=="ordinal") DF[[i]]=as.ordered(DF[[i]])
}
The first for/if loop changes the class from integer to factor, which is what I want, but the second changes the class of ordinal variables to: "ordered" "factor".
I need to change all the columns with the header "ordinal" to "ordered", as the gower.dist function says.
Thanks in advance,
B.T.
What you are doing is fine --- if perhaps a little inelegantly.
With your ordered factor, you have something like:
> foo <- as.ordered(1:10)
> foo
[1] 1 2 3 4 5 6 7 8 9 10
Levels: 1 < 2 < 3 < 4 < 5 < 6 < 7 < 8 < 9 < 10
> class(foo)
[1] "ordered" "factor"
Notice that it has two classes, indicating that it is an ordered factor and that is is a factor:
> is.ordered(as.ordered(1:10))
[1] TRUE
> is.factor(as.ordered(1:10))
[1] TRUE
In some senses, you might like to think that foo is an ordered factor but also inherits from the factor class too. Alternatively, if there isn't a specific method that handles ordered factors, but there is a method for factors, R will use the factor method. As far as R is concerned, an ordered factor is an object with classes "ordered" and "factor". This is what your function for Gower's distance will require.
You could easily do this with:
DF$nominal <- as.factor(DF$nominal)
DF$ordinal <- as.ordered(DF$ordinal)
which gives you a dataframe with the correct structure. If you work with data frames, please stay away from [[]] unless you know very well what you're doing. Take Dirks advice, and check Owen's R Guide as well. You definitely need it.
If i do the conversion as I showed above, gower.dist() works perfectly fine. On a sidenote, the gowers distance can easily be calculated using the daisy() function as well:
DF <- data.frame(
ordinal= c(1,2,3,1,2,1),
nominal= c(2,2,2,2,2,2)
)
DF$nominal <- as.factor(DF$nominal)
DF$ordinal <- as.ordered(DF$ordinal)
library(cluster)
daisy(DF,metric="gower")
library(StatMatch)
gower.dist(DF)