I wrote a Secret Santa program (ala Ruby Quiz...ish), but occasionally when the program runs, I get an error.
Stats: If there's 10 names in the pot, the error comes up about 5% of the time. If there's 100 names in the pot, it's less than 1%. This is on a trial of 1000 times in bash. I've determined that the gift arrays are coming up nil at some point, but I'm not sure why or how to avoid it.
Providing code...
0.upto($lname.length-1).each do |i|
j = rand($giftlname.length) # should be less each time.
while $giftlname[j] == $lname[i] # redo random if it picks same person
if $lname[i] == $lname.last # if random gives same output again, means person is left with himself; needs to switch with someone
$giftfname[j], $fname[i] = $giftfname[i], $fname[j]
$giftlname[j], $lname[i] = $giftlname[i], $lname[j]
$giftemail[j], $email[i] = $giftemail[i], $email[j]
else
j = rand($giftlname.length)
end
end
$santas.push('Santa ' + $fname[i] + ' ' + $lname[i] + ' sends gift to ' + $giftfname[j] + ' ' + $giftlname[j] + ' at ' + '<' + $giftemail[j] + '>.') #Error here, something is sometimes nil
$giftfname.delete_at(j)
$giftlname.delete_at(j)
$giftemail.delete_at(j)
end
Thanks SO!
I think your problem is right here:
$giftfname[j], $fname[i] = $giftfname[i], $fname[j]
Your i values range between zero to the last index in $fname (inclusive) and, presumably, your $giftfname starts off as a clone of $fname (or at least another array with the same length). But, as you spin through the each, you're shrinking $giftfname so $giftfname[i] will be nil and the swap operation above will put nil into $giftfname[j] (which is supposed to be a useful entry of $giftfname). Similar issues apply to $giftlname and $giftemail.
I'd recommend using one array with three element objects (first name, last name, email) instead of your three parallel arrays. There's also a shuffle method on Array that might be of use to you:
Start with an array of people.
Make copy of that array.
Shuffle the copy until it is different at every index from that original array.
Then zip the together to get your final list of giver/receiver pairs.
Figured it out and used the retry statement. the if statement now looks like this (all other variables have been edited to be non-global as well)
if lname[i] == lname.last
santas = Array.new
giftfname = fname.clone
giftlname = lname.clone
giftemail = email.clone
retry
That, aside from a few other edits, created the solution I needed without breaking apart the code too much again. Will definitely try out mu's solution as well, but I'm just glad I have this running error-free for now.
Related
I am building a tool to help me reverse engineer database files. I am targeting my tool towards fixed record length flat files.
What I know:
1) Each record has an index(ID).
2) Each record is separated by a delimiter.
3) Each record is fixed width.
4) Each column in each record is separated by at least one x00 byte.
5) The file header is at the beginning (I say this because the header does not contain the delimiter..)
Delimiters I have found in other files are: ( xFAxFA, xFExFE, xFDxFD ) But this is kind of irrelevant considering that I may use the tool on a different database in the future. So I will need something that will be able to pick out a 'pattern' despite how many bytes it is made of. Probably no more than 6 bytes? It would probably eat up too much data if it was more. But, my experience doing this is limited.
So I guess my question is, how would I find UNKNOWN delimiters in a large file? I feel that given, 'what I know' I should be able to program something, I just dont know where to begin...
# Really loose pseudo code
def begin_some_how
# THIS IS THE PART I NEED HELP WITH...
# find all non-zero non-ascii sets of 2 or more bytes that repeat more than twice.
end
def check_possible_record_lengths
possible_delimiter = begin_some_how
# test if any of the above are always the same number of bytes apart from each other(except one instance, the header...)
possible_records = file.split(possible_delimiter)
rec_length_count = possible_records.map{ |record| record.length}.uniq.count
if rec_length_count == 2 # The header will most likely not be the same size.
puts "Success! We found the fixed record delimiter: #{possible_delimiter}
else
puts "Wrong delimiter found"
end
end
possible = [",", "."]
result = [0, ""]
possible.each do |delimiter|
sizes = file.split( delimiter ).map{ |record| record.size }
next if sizes.size < 2
average = 0.0 + sizes.inject{|sum,x| sum + x }
average /= sizes.size #This should be the record length if this is the right delimiter
deviation = 0.0 + sizes.inject{|sum,x| sum + (x-average)**2 }
matching_value = average / (deviation**2)
if matching_value > result[0] then
result[0] = matching_value
result[1] = delimiter
end
end
Take advantage of the fact that the records have constant size. Take every possible delimiter and check how much each record deviates from the usual record length. If the header is small enough compared rest of the file this should work.
My CSV contains about 60 million rows. The 10th column contains some alphanumeric entries, some of which repeat, that I want to convert into integers with a one-to-one mapping. That is, I don't want the same entry in Original.csv to have multiple corresponding integer values in Processed.csv. So, initially, I wrote the following code:
require 'csv'
udids = []
CSV.open('Original.csv', "wb") do |csv|
CSV.foreach('Processed.csv', :headers=>true) do |row|
unless udids.include?(row[9])
udids << row[9]
end
udid = udids.index(row[9]) + 1
array = [udid]
csv<<array
end
end
But, the program was taking a lot of time, which I soon realized was because it had to check all the previous rows to make sure only the new values get assigned a new integer value, and the existing ones are not assigned any new value.
So, I thought of hashing them, because when exploring the web about this issue, I learnt that hashing is faster than sequential comparing, somehow (I have not read the details about the how, but anyway...) So, I wrote the following code to hash them:
arrayUDID=[]
arrayUser=[]
arrayHash=[]
array1=[]
f = File.open("Original.csv", "r")
f.each_line { |line|
row = line.split(",");
arrayUDID<<row[9]
arrayUser<<row[9]
}
arrayUser = arrayUser.uniq
arrayHash = []
for i in 0..arrayUser.size-1
arrayHash<<arrayUser[i]
arrayHash<<i
end
hash = Hash[arrayHash.each_slice(2).to_a]
array1=hash.values_at *arrayUDID
logfile = File.new("Processed.csv","w")
for i in 0..array1.size-1
logfile.print("#{array1[i]}\n")
end
logfile.close
But here again, I observed that the program was taking a lot of time, which I realized must be due to the hash array (or hash table) running out of memory.
So, can you kindly suggest any method that will work for my huge file in a reasonable amount of time? By reasonable amount, I mean within 10 hours, because I realize that it's going to take some hours at least as it took about 5 hours to extract that dataset from an even bigger dataset. So, with my aforementioned codes, it was not getting finished even after 2 days of running the programs. So, if you can suggest a method which can do the task by leaving the computer on overnight, that would be great. Thanks.
I think this should work:
udids = {}
unique_count = 1
output_csv = CSV.open("Processed.csv", "w")
CSV.foreach("Original.csv").with_index do |row, i|
output_csv << row and next if i == 0 # skip first row (header info)
val = row[9]
if udids[val.to_sym]
row[9] = udids[val.to_sym]
else
udids[val.to_sym] = unique_count
row[9] = unique_count
unique_count += 1
end
output_csv << row
end
output_csv.close
The performance depends heavily on how many duplicates there are (the more the better), but basically it keeps track of each value as a key in a hash, and checks to see if it has encountered that value yet or not. If so, it uses the corresponding value, and if not it increments a counter, stores that count as the new value for that key and continues.
I was able to process a 10 million line test CSV file in about 3 minutes.
it run corectly but it should have around 500 matches but it only has around 50 and I dont know why!
This is a probelm for my comsci class that I am having isues with
we had to make a function that checks a list for duplication I got that part but then we had to apply it to the birthday paradox( more info here http://en.wikipedia.org/wiki/Birthday_problem) thats where I am runing into problem because my teacher said that the total number of times should be around 500 or 50% but for me its only going around 50-70 times or 5%
duplicateNumber=0
import random
def has_duplicates(listToCheck):
for i in listToCheck:
x=listToCheck.index(i)
del listToCheck[x]
if i in listToCheck:
return True
else:
return False
listA=[1,2,3,4]
listB=[1,2,3,1]
#print has_duplicates(listA)
#print has_duplicates(listB)
for i in range(0,1000):
birthdayList=[]
for i in range(0,23):
birthday=random.randint(1,365)
birthdayList.append(birthday)
x= has_duplicates(birthdayList)
if x==True:
duplicateNumber+=1
else:
pass
print "after 1000 simulations with 23 students there were", duplicateNumber,"simulations with atleast one match. The approximate probibilatiy is", round(((duplicateNumber/1000)*100),3),"%"
This code gave me a result in line with what you were expecting:
import random
duplicateNumber=0
def has_duplicates(listToCheck):
number_set = set(listToCheck)
if len(number_set) is not len(listToCheck):
return True
else:
return False
for i in range(0,1000):
birthdayList=[]
for j in range(0,23):
birthday=random.randint(1,365)
birthdayList.append(birthday)
x = has_duplicates(birthdayList)
if x==True:
duplicateNumber+=1
print "after 1000 simulations with 23 students there were", duplicateNumber,"simulations with atleast one match. The approximate probibilatiy is", round(((duplicateNumber/1000.0)*100),3),"%"
The first change I made was tidying up the indices you were using in those nested for loops. You'll see I changed the second one to j, as they were previously bot i.
The big one, though, was to the has_duplicates function. The basic principle here is that creating a set out of the incoming list gets the unique values in the list. By comparing the number of items in the number_set to the number in listToCheck we can judge whether there are any duplicates or not.
Here is what you are looking for. As this is not standard practice (to just throw code at a new user), I apologize if this offends any other users. However, I believe showing the OP a correct way to write a program should be could all do us a favor if said user keeps the lack of documentation further on in his career.
Thus, please take a careful look at the code, and fill in the blanks. Look up the python doumentation (as dry as it is), and try to understand the things that you don't get right away. Even if you understand something just by the name, it would still be wise to see what is actually happening when some built-in method is being used.
Last, but not least, take a look at this code, and take a look at your code. Note the differences, and keep trying to write your code from scratch (without looking at mine), and if it messes up, see where you went wrong, and start over. This sort of practice is key if you wish to succeed later on in programming!
def same_birthdays():
import random
'''
This is a program that does ________. It is really important
that we tell readers of this code what it does, so that the
reader doesn't have to piece all of the puzzles together,
while the key is right there, in the mind of the programmer.
'''
count = 0
#Count is going to store the number of times that we have the same birthdays
timesToRun = 1000 #timesToRun should probably be in a parameter
#timesToRun is clearly defined in its name as well. Further elaboration
#on its purpose is not necessary.
for i in range(0,timesToRun):
birthdayList = []
for j in range(0,23):
random_birthday = random.randint(1,365)
birthdayList.append(random_birthday)
birthdayList = sorted(birthdayList) #sorting for easier matching
#If we really want to, we could provide a check in the above nester
#for loop to check right away if there is a duplicate.
#But again, we are here
for j in range(0, len(birthdayList)-1):
if (birthdayList[j] == birthdayList[j+1]):
count+=1
break #leaving this nested for-loop
return count
If you wish to find the percent, then get rid of the above return statement and add:
return (count/timesToRun)
Here's a solution that doesn't use set(). It also takes a different approach with the array so that each index represents a day of the year. I also removed the hasDuplicate() function.
import random
sim_total=0
birthdayList=[]
#initialize an array of 0's representing each calendar day
for i in range(365):
birthdayList.append(0)
for i in range(0,1000):
first_dup=True
for n in range(365):
birthdayList[n]=0
for b in range(0, 23):
r = random.randint(0,364)
birthdayList[r]+=1
if (birthdayList[r] > 1) and (first_dup==True):
sim_total+=1
first_dup=False
avg = float(sim_total) / 1000 * 100
print "after 1000 simulations with 23 students there were", sim_total,"simulations with atleast one duplicate. The approximate problibility is", round(avg,3),"%"
For a project that I am working on for school, one of the parts of the project asks us to take a collection of all the Federalist papers and run it through a program that essentially splits up the text and writes new files (per different Federalist paper).
The logic I decided to go with is to run a search, and every time the search is positive for "Federalist No." it would save into a new file everything until the next "Federalist No".
This is the algorithm that I have so far:
file_name = "Federalist"
section_number = "1"
new_text = File.open(file_name + section_number, 'w')
i = 0
n= 1
while i < l.length
if (l[i]!= "federalist") and (l[i+1]!= "No")
new_text.puts l[i]
i = i + i
else
new_text.close
section_number = (section_number.to_i +1).to_s
new_text = File.open(file_name + section_number, "w")
new_text.puts(l[i])
new_text.puts(l[i+1])
i=i+2
end
end
After debugging the code as much as I could (I am a beginner at Ruby), the problem that I run into now is that because the while function always holds true, it never proceeds to the else command.
In terms of going about this in a different way, my TA suggested the following:
Put the entire text in one string by looping through the array(l) and adding each line to the one big string each time.
Split the string using the split method and the key word "FEDERALIST No." This will create an array with each element being one section of the text:
arrayName = bigString.split("FEDERALIST No.")
You can then loop through this new array to create files for each element using a similar method you use in your program.
But as simple as it may sound, I'm having an extremely difficult time putting even that code together.
i = i + i
i starts at 0, and 0 gets added to it, which gives 0, which will always be less than l, whatever that value is/means.
Since this is a school assignment, I hesitate to give you a straight-up answer. That's really not what SO is for, and I'm glad that you haven't solicited a full solution either.
So I'll direct you to some useful methods in Ruby instead that could help.
In Array: .join, .each or .map
In String: .split
Fyi, your TA's suggestion is far simpler than the algorithm you've decided to embark on... although technically, it is not wrong. Merely more complex.
I have a function that takes a variable amount of ints as arguments.
thisFunction(1,1,1,2,2,2,2,3,4,4,7,4,2)
this function was given in a framework and I'd rather not change the code of the function or the .lua it is from. So I want a function that repeats a number for me a certain amount of times so this is less repetitive. Something that could work like this and achieve what was done above
thisFunction(repeatNum(1,3),repeatNum(2,4),3,repeatNum(4,2),7,4,2)
is this possible in Lua? I'm even comfortable with something like this:
thisFunction(repeatNum(1,3,2,4,3,1,4,2,7,1,4,1,2,1))
I think you're stuck with something along the lines of your second proposed solution, i.e.
thisFunction(repeatNum(1,3,2,4,3,1,4,2,7,1,4,1,2,1))
because if you use a function that returns multiple values in the middle of a list, it's adjusted so that it only returns one value. However, at the end of a list, the function does not have its return values adjusted.
You can code repeatNum as follows. It's not optimized and there's no error-checking. This works in Lua 5.1. If you're using 5.2, you'll need to make adjustments.
function repeatNum(...)
local results = {}
local n = #{...}
for i = 1,n,2 do
local val = select(i, ...)
local reps = select(i+1, ...)
for j = 1,reps do
table.insert(results, val)
end
end
return unpack(results)
end
I don't have 5.2 installed on this computer, but I believe the only change you need is to replace unpack with table.unpack.
I realise this question has been answered, but I wondered from a readability point of view if using tables to mark the repeats would be clearer, of course it's probably far less efficient.
function repeatnum(...)
local i = 0
local t = {...}
local tblO = {}
for j,v in ipairs(t) do
if type(v) == 'table' then
for k = 1,v[2] do
i = i + 1
tblO[i] = v[1]
end
else
i = i + 1
tblO[i] = v
end
end
return unpack(tblO)
end
print(repeatnum({1,3},{2,4},3,{4,2},7,4,2))