I want to simulate dice roll functionality. However, I don't get what I expect. I want to get a Dice with value ranging from 1 to 6 inclusively (dice).
I tried to find it in Eiffel Documentation, but it is very hard to do.
The following code prints values for 10 consecutive rolls:
local
r: RANDOM
do
across
1 |..| 10 as i
from
create r.set_seed (...) -- ... is the initial "seed"
r.start
loop
io.put_integer (r.item \\ 6 + 1)
io.put_new_line
r.forth
end
end
Related
so this is what I'm trying to do, and I'm not sure how cause I'm new to python. I've searched for a few options and I'm not sure why this doesn't work.
So I have 6 different nodes, in maya, called aiSwitch. I need to generate random different numbers from 0 to 6 and input that value in the aiSiwtch*.index.
In short the result should be
aiSwitch1.index = (random number from 0 to 5)
aiSwitch2.index = (another random number from 0 to 5 different than the one before)
And so on unil aiSwitch6.index
I tried the following:
import maya.cmds as mc
import random
allswtich = mc.ls('aiSwitch*')
for i in allswitch:
print i
S = range(0,6)
print S
shuffle = random.sample(S, len(S))
print shuffle
for w in shuffle:
print w
mc.setAttr(i + '.index', w)
This is the result I get from the prints:
aiSwitch1 <-- from print i
[0,1,2,3,4,5] <--- from print S
[2,3,5,4,0,1] <--- from print Shuffle (random.sample results)
2
3
5
4
0
1 <--- from print w, every separated item in the random.sample list.
Now, this happens for every aiSwitch, cause it's in a loop of course. And the random numbers are always a different list cause it happens every time the loop runs.
So where is the problem then?
aiSwitch1.index = 1
And all the other aiSwitch*.index always take only the last item in the list but the time I get to do the setAttr. It seems to be that w is retaining the last value of the for loop. I don't quite understand how to
Get a random value from 0 to 5
Input that value in aiSwitch1.index
Get another random value from 0 to 6 different to the one before
Input that value in aiSwitch2.index
Repeat until aiSwitch5.index.
I did get it to work with the following form:
allSwitch = mc.ls('aiSwitch')
for i in allSwitch:
mc.setAttr(i + '.index', random.uniform(0,5))
This gave a random number from 0 to 5 to all aiSwitch*.index, but some of them repeat. I think this works cause the value is being generated every time the loop runs, hence setting the attribute with a random number. But the numbers repeat and I was trying to avoid that. I also tried a shuffle but failed to get any values from it.
My main mistake seems to be that I'm generating a list and sampling it, but I'm failing to assign every different item from that list to different aiSwitch*.index nodes. And I'm running out of ideas for this.
Any clues would be greatly appreciated.
Thanks.
Jonathan.
Here is a somewhat Pythonic way: shuffle the list of indices, then iterate over it using zip (which is useful for iterating over structures in parallel, which is what you need to do here):
import random
index = list(range(6))
random.shuffle(index)
allSwitch = mc.ls('aiSwitch*')
for i,j in zip(allSwitch,index):
mc.setAttr(i + '.index', j)
my problem is the following. I have a BIG file with many rows containing ordered numbers (repetitions are possible)
1
1.5
3
3.5
6
6
...
1504054
1504056
I would like to print all the pair of row numbers such that their difference is smaller than a given threshold thr. Let us say for instance thr=2.01, I want
0 1
0 2
1 2
1 3
2 3
4 5
...
N-1 N
I wrote a thing in python but the file is huge and I think I need a smart way to do this in bash.
Actually, in the complete data structure there exists also a second column containing a string:
1 s0
1.5 s1
3 s2
3.5 s3
6 s4
6 s5
...
1504054 sN-1
1504056 sN
and, if easy to do, I would like to write in each row the pair of linked strings, possibly separated by "|":
s0|s1
s0|s2
s1|s2
s1|s3
s2|s3
s4|s5
...
sN-1|sN
Thanks for your help, I am not too familiar with bash
In any language you can white a program implementing this pseudo code:
while read line:
row = line.split(sep)
new_kept_rows = []
for kr in kept_rows :
if abs(kr[0], row[0])<=thr:
print "".join(kr[1:]) "|" "".join(row[1:])
new_kept_rows.append(kr)
kept_rows = new_kept_rows
This program only keep the few lines which could match the condition. All other are freed from memory. So the memory footprint should remain small even for big files.
I would use awk language because I'm comfortable with. But python would fit too (the pseudo code I give is very close to be python).
A quick and hopefully easy question:
I need the means to select from among a given set of hash keys at random. The perl6.org documentation on neither rand nor Hash offer many hints.
my %a = 1,2,3,4,5,6;
Given the above,
%a.keys;
returns (5 1 3) and if I simply try
%a.rand;
I get a pseudorandom float rather than any one key.
I eventually cobbled together %a.keys[Int(%a.elems.rand)], but was hoping for something simpler.
Use pick or roll, eg
%a.keys.pick
As always, Christoph's answer is correct (he knows Perl 6 well). However, I thought I'd elaborate a little since pick and roll can easily be confused at first.
If you only want one random item, then pick and roll seem identical and can be used interchangeably because they both default to returning one item from the original list or array:
my $rand-keyA = %a.keys.pick;
my $rand-keyB = %a.keys.roll;
However, think of them this way:
pick means "Since there are only N things in this container, I can only pick up to N things at once."
roll means "I have an N-sided dice that I can roll as many times as I want."
my %a = 1,2,3,4,5,6; # { 1 => 2, 3 => 4, 5 => 6 }
# (i.e. keys are 1, 3, and 5)
say %a.keys.pick(2); # (5 3)
say %a.keys.pick(2); # (3 1)
say %a.keys.pick(5); # (3 5 1) no more, because there were only three to pick from
say %a.keys.pick(10); # (3 1 5)
say %a.keys.roll(5); # (1 5 1 5 3) as many "rolls" as you request
say %a.keys.roll(10); # (5 5 1 1 5 5 3 1 3 1)
pick(*) is an easy way to create a randomly reordered list from an array without having to know how many elements it has:
my #array = <foo bar baz foobar>;
#array.pick(*); # (bar foobar baz foo)
roll(*) is an easy way to create an infinite list whose elements are chosen randomly from the original array:
my #rolls = #array.roll(*);
say #rolls[0]; # foobar
say #rolls[10]; # bar
say #rolls[351]; # baz
say #rolls[19123]; # foobar
say #rolls[1000000]; # bar
say #rolls[1000001]; # bar
say #rolls[1000002]; # foo
I am using numerical recipes scheme to generate random numbers (ran3, page 7 in this PDF file). I didn't notice anything strange but this time, I got a negative numbers at the "warm up" stage which are larger than MBIG. The code look as if this shouldn't happen. I can easily fix this with changing the if statement to be a while statement at the line that says if(mk.lt.MZ)mk=mk+MBIG but I want to know what are the consequences.
Edit:here is the function
FUNCTION ran3a(idum)
INTEGER idum
INTEGER MBIG,MSEED,MZ
C REAL MBIG,MSEED,MZ
REAL ran3a,FAC
PARAMETER (MBIG=1000000000,MSEED=161803398,MZ=0,FAC=1./MBIG)
C PARAMETER (MBIG=4000000.,MSEED=1618033.,MZ=0.,FAC=1./MBIG)
INTEGER i,iff,ii,inext,inextp,k
INTEGER mj,mk,ma(55)
C REAL mj,mk,ma(55)
SAVE iff,inext,inextp,ma
DATA iff /0/
if(idum.lt.0.or.iff.eq.0)then
iff=1
mj=MSEED-iabs(idum)
mj=mod(mj,MBIG)
ma(55)=mj
mk=1
do 11 i=1,54
ii=mod(21*i,55)
ma(ii)=mk
mk=mj-mk
if(mk.lt.MZ)mk=mk+MBIG
mj=ma(ii)
11 continue
do 13 k=1,4
do 12 i=1,55
ma(i)=ma(i)-ma(1+mod(i+30,55))
if(ma(i).lt.MZ)ma(i)=ma(i)+MBIG
12 continue
13 continue
inext=0
inextp=31
idum=1
endif
inext=inext+1
if(inext.eq.56)inext=1
inextp=inextp+1
if(inextp.eq.56)inextp=1
mj=ma(inext)-ma(inextp)
if(mj.lt.MZ)mj=mj+MBIG
ma(inext)=mj
ran3a=mj*FAC
return
END
I was getting Seg Faults (using gfortran 4.8) because the function was trying to change the input value idum from the negative number to 1. There is no reason for that line (nor anything with iff), so I deleted it and printed out the array ma at several different places and found no negative numbers in the array.
One possibility, though, is if iabs(idum) is larger than MSEED, you might have a problem with the line mj=MSEED - iabs(idum). You should protect from this by using mj=abs(MSEED-abs(idum)) like the book has written.
Had a look at the pdf. What you need to do is
1) Seed it: value = ran3(-1)
2) Use it: value = ran3(0)
I'm attempting to solve http://projecteuler.net/problem=1.
I want to create a method which takes in an integer and then creates an array of all the integers preceding it and the integer itself as values within the array.
Below is what I have so far. Code doesn't work.
def make_array(num)
numbers = Array.new num
count = 1
numbers.each do |number|
numbers << number = count
count = count + 1
end
return numbers
end
make_array(10)
(1..num).to_a is all you need to do in Ruby.
1..num will create a Range object with start at 1 and end at whatever value num is. Range objects have to_a method to blow them up into real Arrays by enumerating each element within the range.
For most purposes, you won't actually need the Array - Range will work fine. That includes iteration (which is what I assume you want, given the problem you're working on).
That said, knowing how to create such an Array "by hand" is valuable learning experience, so you might want to keep working on it a bit. Hint: you want to start with an empty array ([]) instead with Array.new num, then iterate something num.times, and add numbers into the Array. If you already start with an Array of size num, and then push num elements into it, you'll end up with twice num elements. If, as is your case, you're adding elements while you're iterating the array, the loop never exits, because for each element you process, you add another one. It's like chasing a metal ball with the repulsing side of a magnet.
To answer the Euler Question:
(1 ... 1000).to_a.select{|x| x%3==0 || x%5==0}.reduce(:+) # => 233168
Sometimes a one-liner is more readable than more detailed code i think.
Assuming you are learning Ruby by examples on ProjectEuler, i'll explain what the line does:
(1 ... 1000).to_a
will create an array with the numbers one to 999. Euler-Question wants numbers below 1000. Using three dots in a Range will create it without the boundary-value itself.
.select{|x| x%3==0 || x%5==0}
chooses only elements which are divideable by 3 or 5, and therefore multiples of 3 or 5. The other values are discarded. The result of this operation is a new Array with only multiples of 3 or 5.
.reduce(:+)
Finally this operation will sum up all the numbers in the array (or reduce it to) a single number: The sum you need for the solution.
What i want to illustrate: many methods you would write by hand everyday are already integrated in ruby, since it is a language from programmers for programmers. be pragmatic ;)