I'm trying to create a number that's inside of a range with scalacheck generators like:
Gen.chooseNum(min, max)
But it defaults to the increment value of 1 but I wanted it to increment with 0.5.
What is the solution for this?
You can generate integers that are multiples of 5 and then divide them by 10.
Gen.choose(min * 10, max * 10)
.map(_ / 5 * 5)
.map(_.toDouble / 10)
Related
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];
I've searched for this but i didn't find anything, i hope this is not a doubled question.
I'm doing a formula in TSQL like this:
#Temp = SQRT((((#Base1 - 1) * (#StDev1 * #StDev1))
+ ((#AvgBase - 1) * (#AvgStDev * #AvgStDev)))
* ((1 / #Base1) + (1 / #AvgBase))
/ (#Base1 + #AvgBase - 2))
But it always returns me a 0.
#Base1 and #AvgBase are int, the rest of parameters are float, but i've also tried with decimal(15,15).
I tried also changing the self multiplication with the function POWER() but the only problem i can't solve is this part: (1 / #Base1) + (1 / #AvgBase), because #Base1 and #AvgBase are so big, and the result of the calc is 0,0001... and some more numbers. How can i force the engine to not round the result to 0? Thanks
EDIT: I solved it changing the #AvgBase and the #Base1 to float type. I guess that the result 1/#param, with #param -> int gives you the rounded result and when you go for casting it or whatever, you are working on a rounded result anyway.
have you tried to create a #INVBase1 = (1/#Base1) ? will this also be rounded to 0? what happens when you play around with the data format of this new variable?
alternatively have you tried
/ ((#Base1) + (#AvgBase))
instead of
* ((1 / #Base1) + (1 / #AvgBase))
I am trying to understand why does the following:
say 4 x 2.5 * 2;
produce
88
instead of
44444
as the following:
say 4 x (2.5 * 2);
The way I tried to explain the second one is that since operator x expects a number as right argument 2.5 * 2 gets computed into number 5 and then 4 is treated as string which yields "44444". But I can't explain the 1st one! Changing it to print does not matter, either.
print 4 x 2.5 * 2 . "\n"
also gives
88
I am not about to write such code, of course. I'm just trying to understand the behavior.
The repetition operator x and the multiplication * have the same precedence. Both are also left-associative. Therefore, the correct parenthezisation for
4 x 2.5 * 2
is
(4 x 2.5) * 2
Because the 2nd argument for x must be an integer, this is equivalent to
(4 x 2) * 2
Which does "44" * 2. There we have our 88.
If in doubt, use explicit parens to make your intentions clear to the compiler:
say 4 x (2.5 * 2);
I have been trying this for a while but thus far haven't had any luck.
What is the easiest way to retrieve a random number between two very precise numbers on iOS?
For example, I want a random number between 41.37783830549337 and 41.377730629131634, how would I accomplish this?
Thank you so much in advance!
Edit: I tried this:
double min = 41.37783830549337;
double max = 41.377730629131634;
double test = ((double)rand() / RAND_MAX) * (max - min) + min;
NSLog(#"Min:%lf, max:%lf, result:%lf",min,max,test);
But the results weren't quite as precise as I was hoping, and ended up like this::
Min:41.377838, max:41.377731, result:41.377838
You can normalise the output of rand to any range you want:
((double)rand() / RAND_MAX) * (max - min) + min
[Note: This is pure C, I'm assuming it works equivalently in Obj-C.]
[Note 2: Replace double with the data-type of your choice as appropriate.]
[Note 3: Replace rand with the random-number source of your choice as appropriate.]
I am new to iPhone programming. I have 10 number say (1,2,3,4,5,6,7,8,9,10). I want to choose randomly 1 number from the above 10 numbers. How can I choose a random number from a set of numbers?
If you simply want a value between 1 and 10, you can use the standard C rand() method. This returns an integer between zero and RAND_MAX.
To get a value between 0 and 9 you can use the % operator. So to get a value between 1 and 10 you can use:
rand()%10 + 1
If you don't want the same series of pseudo random numbers each time, you'll need to use srand to seed the random number generator. A good value to seed it with would be the current time.
If you're asking about choosing a number from a list of arbitrary (and possibly non consecutive) numbers, you could use the following.
int numbers[] = {2,3,5,7,11,13,17,19,23,29};
int randomChoice = numbers[rand()%10];
To generate a random number you should use random() function. But if you call it twice it gives you two equal answers. Before calling random(), call srand(time()) to get fresh new random number. if you want to use for(int i = 0; ...) to create numbers,
use srand(time() + i).
Something like this:
- (IBAction)generate:(id)sender
{
// Generate a number between 1 and 10 inclusive
int generated;
generated = (random() % 10) + 1;
}