Okay, from what I understand, an integer that is a fraction will be rounded one way or the other so that if a formula comes up with say 5/6 - it will automatically round it to 1. I have a calculation:
xyz = ((1300 - [abc intValue])/6) + 100;
xyz is defined as an NSInteger, abc is an NSString that is chosen via a UIPicker. I want the calculation (1300 - [abc intValue]) to add 1 to 100 for each 6 units below 1300. For example, 1255 should result in xyz having a value of 100 and 1254 should result in a value of 101.
Now, I understand that my formula above is wrong because of the rounding principles, but I am getting some CRAZY results from the program itself. When I punched in 1259 - I got 106. When I punched in 1255 - I got 107. Why would it behave that way?
Your understanding is wrong. Integer division truncates:
5 / 6 == 0
(1300 - 1259) / 6 == 41 / 6 == 6
(1300 - 1255) / 6 == 45 / 6 = 7
You can use:
xyz = ((1300.0 - [abc intValue])/6) + 100;
and make xyz a NSDouble. That will ensure it does floating-point division.
You may also be confusing numbers and time. 1255 is 45 below 1300, not 5 below :-)
Related
Im trying to implement a custom min max scaler in kdb+/q. I have taken note of the implementation located in the ml package however I'm looking to be able to scale data between a custom range i.e. 0 and 255. What would be an efficient implementation of min max scaling in kdb+/q?
Thanks
Looking at the link to github on the page you referenced it looks like you may be able to define a function like so:
minmax255:{[sf;x]sf*(x-mnx)%max[x]-mnx:min x}[255]
Where sf is your scaling factor (here given by 255).
q)minmax255 til 10
0 28.33333 56.66667 85 113.3333 141.6667 170 198.3333 226.6667 255
If you don't like decimals you could round to the nearest whole number like:
q)minmax255round:{[sf;x]floor 0.5+sf*(x-mnx)%max[x]-mnx:min x}[255]
q)minmax255round til 10
0 28 57 85 113 142 170 198 227 255
(logic here is if I have a number like 1.7, add .5, and floor I'll wind up with 2, whereas if I had a number like 1.2, add .5, and floor I'll end up with 1)
If you don't want to start at 0 you could use | which takes the max of it's left and right arguments
q)minmax255roundlb:{[sf;lb;x]lb|floor sf*(x-mnx)%max[x]-mnx:min x}[255;10]
q)minmax255roundlb til 10
10 28 56 85 113 141 170 198 226 255
Where I'm using lb to mean 'lower bound'
If you want to apply this to a table you could use
q)show testtab:([]a:til 10;b:til 10)
a b
---
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
q)update minmax255 a from testtab
a b
----------
0 0
28.33333 1
56.66667 2
85 3
113.3333 4
141.6667 5
170 6
198.3333 7
226.6667 8
255 9
The following will work nicely
minmaxCustom:{[l;u;x]l + (u - l) * (x-mnx)%max[x]-mnx:min x}
As petty as it sounds, it is my strong recommendation that you do not follow through with Shehir94 solution for a custom minimum value. Applying a maximum to get a starting range, it will mess with the original distribution. A custom minmax scaling should be a simple linear transformation on a standard 0-1 minmax transformation.
X' = a + bX
For example, to get a custom scaling of 10-255, that would be a b=245 and a=10, we would expect the new mean to follow this formula and the standard deviation to only be a Multiplicative, but applying lower bound messes with this, for example.
q)dummyData:10000?100.0
q)stats:{`transform`minVal`maxVal`avgVal`stdDev!(x;min y;max y; avg y; dev y)}
q)minmax255roundlb:{[sf;lb;x]lb|sf*(x-mnx)%max[x]-mnx:min x}[255;10]
q)minmaxCustom:{[l;u;x]l + (u - l) * (x-mnx)%max[x]-mnx:min x}
q)res:stats'[`orig`lb`linear;(dummyData;minmax255roundlb dummyData;minmaxCustom[10;255;dummyData])]
q)res
transform minVal maxVal avgVal stdDev
-----------------------------------------------
orig 0.02741043 99.98293 50.21896 28.92852
lb 10 255 128.2518 73.45999
linear 10 255 133.024 70.9064
// The transformed average should roughly be
q)10 + ((255-10)%100)*49.97936
132.4494
// The transformed std devaition should roughly be
q)2.45*28.92852
70.87487
To answer the comment, this could be applied over a large number of coluwould be applied to a table in the following manner
q)n:10000
q)tab:([]sym:n?`3;col1:n?100.0)
q)multiColApply:{[tab;scaler;colList]flip ft,((),colList)!((),scaler each (ft:flip tab)[colList])}
q)multiColApply[tab;minmaxCustom[10;20];`col1`col2]
sym col1 col2 col3
------------------------------
cag 13.78461 10.60606 392.7524
goo 15.26201 16.76768 517.0911
eoh 14.05111 19.59596 515.9796
kbc 13.37695 19.49495 406.6642
mdc 10.65973 12.52525 178.0839
odn 16.24697 17.37374 301.7723
ioj 15.08372 15.05051 785.033
mbc 16.7268 20 534.7096
bhj 12.95134 18.38384 711.1716
gnf 19.36005 15.35354 411.597
gnd 13.21948 18.08081 493.1835
khi 12.11997 17.27273 578.5203
If I have two lists:
a:1 2 3 4;
b:10 20 30 40;
I want to sum the product of the two lists within a window of 2. So the result set should be:
10 50 130 250
For example, to get the result of 130 it would be (2*20)+(3*30) = 130
sums 2 mavg '(a*b)
seems to get me part way there, but the window of 2 isn't being applied. I've tried experimenting with sum, sums, sum each, wavg, mavg, etc. and I am completely stuck. Could anyone help? Thanks!
This line should work for you:
2 msum a*b
as demonstrated here:
q)a:1 2 3 4
q)b:10 20 30 40
q)2 msum a*b
10 50 130 250
For more information about the keyword msum, you could check out the Kx Reference page:
https://code.kx.com/wiki/Reference/msum
Hope that helps!
Alternatively you could use the adverb each prior:
q)+':[a*b]
However this will only work with a window size of 2 and if your data contains null values this needs to be padded with 0:
q)+':[0^a*b2]
On a positive note it is faster than using msum in this situation.
q)\ts:1000000 +':[0^a*b2]
940 1264
q)\ts:1000000 2 msum a*b2
1556 1104
Here is my problem:
I have two tables:
q)t1:([]sym:1 5;x: 90 90)
q)t2:([]sym: 2 3 4 6 7 8; y: 100 200 300 400 500 600)
If I do aj[`sym;t2;t1], all the 6 columns in the result table will contain x with value 90.
But what I want is value 90 in column x only in row with sym 2 and 6, i.e the first time that sym in table t2 appear before table t1.
In other words, I want the result table to be like this:
q)([]sym:2 3 4 6 7 8; y: 100 200 300 400 500 600; x:90 0N 0N 90 0N 0N)
sym y x
----------
2 100 90
3 200
4 300
6 400 90
7 500
8 600
Could anyone tell me how I can achieve this? Thank you so much!
Not sure if aj can be used in this sense. This might give you what you need:
q)t2 lj 1!update sym:{x x binr y}[t2.sym;sym] from t1
sym y x
----------
2 100 90
3 200
4 300
6 400 90
7 500
8 600
Uses binr to find the next value greater than the value in t1 then joins only on that.
EDIT: note also that binr is >= ..... If you need strictly greater than you could use:
q)t2 lj 1!update sym:{x 1+x bin y}[t2.sym;sym] from t1
sym y x
----------
2 100 90
3 200
5 300
6 400 90
7 500
8 600
You can do aj to get the index where nearest smaller number of x will fit in, then a vector condition to get x when that index has got incremented, i.e.
select sym, y, x:?[c>prev c;x;0n] from aj[`sym; t2; update c:i from t1]
Once again, sorry if this has been asked before and if its too specific but I'm very stuck and can't quite find a solution.
I have a matrix of say 3 members of a structure called 2, 4 and 16 (in column 1) that have values along their relative distance e.g. member 2 has values at the start, 0m, then at 0.5m then the end of its length 1.5m, where member 4 starts at 0m etc. So that my matrix looks like this:
2 0 125
2 0.5 25
2 1.5 365
4 0 25
4 0.6 57
16 0 354
16 0.2 95
16 0.8 2
and I want to create a matrix that has the overall distance along all the members 2, 4 and 16 combined:
2 0 125
2 0.5 25
2 1.5 365
4 1.5 25
4 2.1 57
16 2.1 354
16 2.3 95
16 3.1 2
is there any way to do this in matlab? Like possibly locating the first zero and adding the value above it to all the rest of the values below then find the next zero value and so on?
Please tell me if this isn't clear, I realise it's a bit confusing but not too sure how to explain it better!
I came up with the following:
idx = find(diff(M(:,1)));
v = zeros(size(M,1),1);
v(idx+1) = M(idx,2);
M(:,2) = M(:,2) + cumsum(v);
The result:
M =
2 0 125
2 0.5 25
2 1.5 365
4 1.5 25
4 2.1 57
16 2.1 354
16 2.3 95
16 2.9 2
Note the last value in the second column disagrees with what you described (2.9 vs 3.1). Either you had a typo, or I'm still not getting it...
data = [2 0 125;
2 0.5 25;
2 1.5 365;
4 0 25;
4 0.6 57;
16 0 354;
16 0.2 95;
16 0.8 2];
idx0 = find(data(:,2)==0);
idx0 = idx0(2:end); %ignore first zero of first member, doesn't need an offset
offset = data(idx0-1,2);
N = size(data,1);
for ii=1:numel(idx0)
idxs = 1:N>=idx0(ii);
data(idxs,2) = data(idxs,2) + offset(ii);
end
I'm working on a program that converts between number bases. For example Octal is 8, decimal is 10. Letters A to Z could be considered as base 26.
I want to convert a number like "A" into 0, Z into 25, "AA" into 27 and "BA" into 53.
Before I start coding I'm doing it on paper so I understand the process. To start out I'm trying to convert 533 to base 26.
What algorithm is best for doing this?
You need to assign a "digit" to each letter, like:
A = 0 N = 13
B = 1 O = 14
C = 2 P = 15
D = 3 Q = 16
E = 4 R = 17
F = 5 S = 18
G = 6 T = 19
H = 7 U = 20
I = 8 V = 21
J = 9 W = 22
K = 10 X = 23
L = 11 Y = 24
M = 12 Z = 25
Then, your {20,13} becomes UN.
Converting back is UN -> {20,13} -> (20 * 26 + 13) -> 52.
By way of further example, let's try the number 10163, just plucked out of the air at random.
Divide that by 26 until you get a number less than 26 (i.e., twice), and you get 15 with a fractional part of 0.03402366.
Multiply that by 26 and you get 0 with a fractional part of 0.88461516.
Multiply that by 26 and you get 23 (actually 22.99999416 on my calculator but, since the initial division was only two steps, we stop here - the very slight inaccuracy is due to the fact that the floating point numbers are being rounded).
So the "digits" are {15,0,23} which is the "number" PAX. Wow, what a coincidence?
To convert PAX back into decimal, its
P * 262 + A * 261 + X * 260
or
(15 * 676) + (0 * 26) + 23
= 10140 + 0 + 23
= 10163
Let's take a step back for a second, and look at decimal.
What does a number like "147" mean? Or rather, what do the characters '1', '4' and '7', when arranged like that, indicate?
There are ten digits in decimal, and after that, we add another digit to the left of the first, and so on as our number increases. So after "9" = 9*1, we get "10" = 1*10 + 0*1. So "147" is 1*10^2 + 4*10 + 7*1 = 147. Similarly, we can go backwards - 147/10^2 = 1, which maps to the character '1'. (147 % 10^2) / 10 = 4, which maps to the character '4'. And 147 % 10 = 7, which maps to the character '7'.
This works works for any base N - if we get the number 0, that maps to the first character in our set. The number 1 maps to the second character, and so on until the number N-1 maps to the last character in our set of digits.
You convert 20 and 13 to the symbols that represent 20 and 13 in your base 26 notation. It sounds like you are using the letters of the alphabet so, that would be UN (where A is 0 and Z is 25).
What language are you writing this in? If you're doing this in Perl you can use the CPAN module Math::Fleximal that I wrote many years ago while I was bored. If you're using a language with infinite precision integers, then life becomes much easier. All you have to do is take characters, convert them into an array of integers, then do the calculation to turn that into a number.