I'm working on a codewars kata that uses the Luhn Algorithm
to validate credit card numbers. Already have my answer, which uses string conversion to split the digits and then reconvert to an integer and then use the rest of the algorithm. All well and good until I test it on an octal number.
The code starts as:
def validate(n)
n.to_s.split("") #n = 7867 gives me ["7","8","6","7"], which is fine
n.to_s.split("") #n = 0776 gives me ["5","1","0], when I need ["0","7","7","6"]
n.to_s.split("") #n = 0100 gives me ["6", "4"] when I need ["0","1","0","0"]
#other code here
end
where the method should be called on whatever n is plugged in.
How do I prevent an octal, binary, or hexidecimal, etc number from converting like that? Is there a way to keep the digits as is so I can use them?
Your problem is that you are obviously storing the number as a number, in n; or else you would not need to say n.to_s.
There must be a place where those numbers enter your program; a user prompt, or by reading a file or whatever. You likely have something like n = user_input.to_i somewhere, forcing ruby to convert the string to a number. Make sure that you treat n as string everywhere, and leading zeroes will stick around.
If you are using numeric literals in your program, you can explicitly say that they are decimal:
n = 0d0100 # or "0100".to_i
n.to_s.rjust(4,"0").split("")
#=> ["0", "1", "0", "0"]
Also, use rjust to make them 4 digits long in string form as credit/debit cards have four 4-digit numbers
To split credit card into an array of integer, you could do something like below:
cc = "4111 1111 1111 1111".gsub(/ /,"").split("").map(&:to_i)
#=> [4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
Related
I'm working on a Ruby challenge to take any non-negative integer as an argument and return it with its digits in descending order. Essentially, rearrange the digits to create the highest possible number.
for example:
Input: 145263 Output: 654321
Input: 123456789 Output: 987654321
Currently this is what my solution looks like:
def descending_order(n)
# take any non-negative integer as an argument
# return it with digits in descending order
n.sqrt(1){ |digits| digits.sort_by.reverse }
end
However, it keeps throwing an error message saying:
`descending_order': undefined method `sqrt' for 0:Integer (NoMethodError)
def descending_order(n)
n.to_s.split(//).sort.reverse.join.to_i
end
This creates an array of strings, each string being a single digit as text. We can use the normal sort here, because we can reasonably assume that the collating sequence of the digits in an encoding obeys the same order than their numeric counterpart. In particular, we know that i.e. '4' < '8'.
I think the easiest to understand answer is:
n = 145263
n.digits.sort.reverse.join.to_i
n.digits => [3, 6, 2, 5, 4, 1]
n.digits.sort => [1, 2, 3, 4, 5, 6]
n.digits.sort.reverse => [6, 5, 4, 3, 2, 1]
n.digits.sort.reverse.join => "654321"
n.digits.sort.reverse.join.to_i => 654321
Summary of methods
digits => Returns the digits of ints place-value representation with radix base (default: 10). The digits are returned as an array with the least significant digit as the first array element.
sort => Returns a new array created by sorting self.
reverse => Returns a new array containing selfs elements in reverse order.
join => Returns a string created by converting each element of the array to a string, separated by the given separator. If the separator is nil, it uses current $,. If both the separator and $, are nil, it uses an empty string.
to_i => Returns the result of interpreting leading characters in str as an integer base base (between 2 and 36).
I like Tom Lord's answer in his comment. It's a bit cryptic at first, but Stefan's explanation is excellent.
I'm converting arrays consisting of four byte values to 32bit numbers by executing the following code:
a = [0, 16, 82, 0]
i = a.map { |e| "%02x" % e }.join.to_i(16)
# => 1069568
It works as intended, but I wonder if there's a more elegant way to perform this task. Maybe not utilizing strings.
Using pack and unpack1:
a = [0, 16, 82, 0]
a.pack('C4').unpack1('L>')
#=> 1069568
C4 means 8-bit unsigned (4 times) and L> means 32-bit unsigned (big endian).
However, pack returns a binary string, so this is not string-free.
If you have one byte, that would be the result as is. If you add one byte to the right side, that would make the original result move two positions to the left (which means multiplying by 0x100, or 16 ** 2 = 256), and add the new byte. You can repeat this as many times as there are bytes.
a.inject{|acc, byte| acc * 0x100 + byte}
# => 1069568
I tried SecureRandom.random_number(9**6) but it sometimes returns 5 and sometimes 6 numbers. I'd want it to be a length of 6 consistently. I would also prefer it in the format like SecureRandom.random_number(9**6) without using syntax like 6.times.map so that it's easier to be stubbed in my controller test.
You can do it with math:
(SecureRandom.random_number(9e5) + 1e5).to_i
Then verify:
100000.times.map do
(SecureRandom.random_number(9e5) + 1e5).to_i
end.map { |v| v.to_s.length }.uniq
# => [6]
This produces values in the range 100000..999999:
10000000.times.map do
(SecureRandom.random_number(9e5) + 1e5).to_i
end.minmax
# => [100000, 999999]
If you need this in a more concise format, just roll it into a method:
def six_digit_rand
(SecureRandom.random_number(9e5) + 1e5).to_i
end
To generate a random, 6-digit string:
# This generates a 6-digit string, where the
# minimum possible value is "000000", and the
# maximum possible value is "999999"
SecureRandom.random_number(10**6).to_s.rjust(6, '0')
Here's more detail of what's happening, shown by breaking the single line into multiple lines with explaining variables:
# Calculate the upper bound for the random number generator
# upper_bound = 1,000,000
upper_bound = 10**6
# n will be an integer with a minimum possible value of 0,
# and a maximum possible value of 999,999
n = SecureRandom.random_number(upper_bound)
# Convert the integer n to a string
# unpadded_str will be "0" if n == 0
# unpadded_str will be "999999" if n == 999999
unpadded_str = n.to_s
# Pad the string with leading zeroes if it is less than
# 6 digits long.
# "0" would be padded to "000000"
# "123" would be padded to "000123"
# "999999" would not be padded, and remains unchanged as "999999"
padded_str = unpadded_str.rjust(6, '0')
Docs to Ruby SecureRand, lot of cool tricks here.
Specific to this question I would say: (SecureRandom.random_number * 1000000).to_i
Docs: random_number(n=0)
If 0 is given or an argument is not given, ::random_number returns a float: 0.0 <= ::random_number < 1.0.
Then multiply by 6 decimal places (* 1000000) and truncate the decimals (.to_i)
If letters are okay, I prefer .hex:
SecureRandom.hex(3) #=> "e15b05"
Docs:
hex(n=nil)
::hex generates a random hexadecimal string.
The argument n specifies the length, in bytes, of the random number to
be generated. The length of the resulting hexadecimal string is twice
n.
If n is not specified or is nil, 16 is assumed. It may be larger in
future.
The result may contain 0-9 and a-f.
Other options:
SecureRandom.uuid #=> "3f780c86-6897-457e-9d0b-ef3963fbc0a8"
SecureRandom.urlsafe_base64 #=> "UZLdOkzop70Ddx-IJR0ABg"
For Rails apps creating a barcode or uid with an object you can do something like this in the object model file:
before_create :generate_barcode
def generate_barcode
begin
return if self.barcode.present?
self.barcode = SecureRandom.hex.upcase
end while self.class.exists?(barcode: barcode)
end
SecureRandom.random_number(n) gives a random value between 0 to n. You can achieve it using rand function.
2.3.1 :025 > rand(10**5..10**6-1)
=> 742840
rand(a..b) gives a random number between a and b. Here, you always get a 6 digit random number between 10^5 and 10^6-1.
This is what I have so far:
myArray.map!{ rand(max) }
Obviously, however, sometimes the numbers in the list are not unique. How can I make sure my list only contains unique numbers without having to create a bigger list from which I then just pick the n unique numbers?
Edit:
I'd really like to see this done w/o loop - if at all possible.
(0..50).to_a.sort{ rand() - 0.5 }[0..x]
(0..50).to_a can be replaced with any array.
0 is "minvalue", 50 is "max value"
x is "how many values i want out"
of course, its impossible for x to be permitted to be greater than max-min :)
In expansion of how this works
(0..5).to_a ==> [0,1,2,3,4,5]
[0,1,2,3,4,5].sort{ -1 } ==> [0, 1, 2, 4, 3, 5] # constant
[0,1,2,3,4,5].sort{ 1 } ==> [5, 3, 0, 4, 2, 1] # constant
[0,1,2,3,4,5].sort{ rand() - 0.5 } ==> [1, 5, 0, 3, 4, 2 ] # random
[1, 5, 0, 3, 4, 2 ][ 0..2 ] ==> [1, 5, 0 ]
Footnotes:
It is worth mentioning that at the time this question was originally answered, September 2008, that Array#shuffle was either not available or not already known to me, hence the approximation in Array#sort
And there's a barrage of suggested edits to this as a result.
So:
.sort{ rand() - 0.5 }
Can be better, and shorter expressed on modern ruby implementations using
.shuffle
Additionally,
[0..x]
Can be more obviously written with Array#take as:
.take(x)
Thus, the easiest way to produce a sequence of random numbers on a modern ruby is:
(0..50).to_a.shuffle.take(x)
This uses Set:
require 'set'
def rand_n(n, max)
randoms = Set.new
loop do
randoms << rand(max)
return randoms.to_a if randoms.size >= n
end
end
Ruby 1.9 offers the Array#sample method which returns an element, or elements randomly selected from an Array. The results of #sample won't include the same Array element twice.
(1..999).to_a.sample 5 # => [389, 30, 326, 946, 746]
When compared to the to_a.sort_by approach, the sample method appears to be significantly faster. In a simple scenario I compared sort_by to sample, and got the following results.
require 'benchmark'
range = 0...1000000
how_many = 5
Benchmark.realtime do
range.to_a.sample(how_many)
end
=> 0.081083
Benchmark.realtime do
(range).sort_by{rand}[0...how_many]
end
=> 2.907445
Just to give you an idea about speed, I ran four versions of this:
Using Sets, like Ryan's suggestion.
Using an Array slightly larger than necessary, then doing uniq! at the end.
Using a Hash, like Kyle suggested.
Creating an Array of the required size, then sorting it randomly, like Kent's suggestion (but without the extraneous "- 0.5", which does nothing).
They're all fast at small scales, so I had them each create a list of 1,000,000 numbers. Here are the times, in seconds:
Sets: 628
Array + uniq: 629
Hash: 645
fixed Array + sort: 8
And no, that last one is not a typo. So if you care about speed, and it's OK for the numbers to be integers from 0 to whatever, then my exact code was:
a = (0...1000000).sort_by{rand}
Yes, it's possible to do this without a loop and without keeping track of which numbers have been chosen. It's called a Linear Feedback Shift Register: Create Random Number Sequence with No Repeats
[*1..99].sample(4) #=> [64, 99, 29, 49]
According to Array#sample docs,
The elements are chosen by using random and unique indices
If you need SecureRandom (which uses computer noise instead of pseudorandom numbers):
require 'securerandom'
[*1..99].sample(4, random: SecureRandom) #=> [2, 75, 95, 37]
How about a play on this? Unique random numbers without needing to use Set or Hash.
x = 0
(1..100).map{|iter| x += rand(100)}.shuffle
You could use a hash to track the random numbers you've used so far:
seen = {}
max = 100
(1..10).map { |n|
x = rand(max)
while (seen[x])
x = rand(max)
end
x
}
Rather than add the items to a list/array, add them to a Set.
If you have a finite list of possible random numbers (i.e. 1 to 100), then Kent's solution is good.
Otherwise there is no other good way to do it without looping. The problem is you MUST do a loop if you get a duplicate. My solution should be efficient and the looping should not be too much more than the size of your array (i.e. if you want 20 unique random numbers, it might take 25 iterations on average.) Though the number of iterations gets worse the more numbers you need and the smaller max is. Here is my above code modified to show how many iterations are needed for the given input:
require 'set'
def rand_n(n, max)
randoms = Set.new
i = 0
loop do
randoms << rand(max)
break if randoms.size > n
i += 1
end
puts "Took #{i} iterations for #{n} random numbers to a max of #{max}"
return randoms.to_a
end
I could write this code to LOOK more like Array.map if you want :)
Based on Kent Fredric's solution above, this is what I ended up using:
def n_unique_rand(number_to_generate, rand_upper_limit)
return (0..rand_upper_limit - 1).sort_by{rand}[0..number_to_generate - 1]
end
Thanks Kent.
No loops with this method
Array.new(size) { rand(max) }
require 'benchmark'
max = 1000000
size = 5
Benchmark.realtime do
Array.new(size) { rand(max) }
end
=> 1.9114e-05
Here is one solution:
Suppose you want these random numbers to be between r_min and r_max. For each element in your list, generate a random number r, and make list[i]=list[i-1]+r. This would give you random numbers which are monotonically increasing, guaranteeing uniqueness provided that
r+list[i-1] does not over flow
r > 0
For the first element, you would use r_min instead of list[i-1]. Once you are done, you can shuffle the list so the elements are not so obviously in order.
The only problem with this method is when you go over r_max and still have more elements to generate. In this case, you can reset r_min and r_max to 2 adjacent element you have already computed, and simply repeat the process. This effectively runs the same algorithm over an interval where there are no numbers already used. You can keep doing this until you have the list populated.
As far as it is nice to know in advance the maxium value, you can do this way:
class NoLoopRand
def initialize(max)
#deck = (0..max).to_a
end
def getrnd
return #deck.delete_at(rand(#deck.length - 1))
end
end
and you can obtain random data in this way:
aRndNum = NoLoopRand.new(10)
puts aRndNum.getrnd
you'll obtain nil when all the values will be exausted from the deck.
Method 1
Using Kent's approach, it is possible to generate an array of arbitrary length keeping all values in a limited range:
# Generates a random array of length n.
#
# #param n length of the desired array
# #param lower minimum number in the array
# #param upper maximum number in the array
def ary_rand(n, lower, upper)
values_set = (lower..upper).to_a
repetition = n/(upper-lower+1) + 1
(values_set*repetition).sample n
end
Method 2
Another, possibly more efficient, method modified from same Kent's another answer:
def ary_rand2(n, lower, upper)
v = (lower..upper).to_a
(0...n).map{ v[rand(v.length)] }
end
Output
puts (ary_rand 5, 0, 9).to_s # [0, 8, 2, 5, 6] expected
puts (ary_rand 5, 0, 9).to_s # [7, 8, 2, 4, 3] different result for same params
puts (ary_rand 5, 0, 1).to_s # [0, 0, 1, 0, 1] repeated values from limited range
puts (ary_rand 5, 9, 0).to_s # [] no such range :)
This is what I have so far:
myArray.map!{ rand(max) }
Obviously, however, sometimes the numbers in the list are not unique. How can I make sure my list only contains unique numbers without having to create a bigger list from which I then just pick the n unique numbers?
Edit:
I'd really like to see this done w/o loop - if at all possible.
(0..50).to_a.sort{ rand() - 0.5 }[0..x]
(0..50).to_a can be replaced with any array.
0 is "minvalue", 50 is "max value"
x is "how many values i want out"
of course, its impossible for x to be permitted to be greater than max-min :)
In expansion of how this works
(0..5).to_a ==> [0,1,2,3,4,5]
[0,1,2,3,4,5].sort{ -1 } ==> [0, 1, 2, 4, 3, 5] # constant
[0,1,2,3,4,5].sort{ 1 } ==> [5, 3, 0, 4, 2, 1] # constant
[0,1,2,3,4,5].sort{ rand() - 0.5 } ==> [1, 5, 0, 3, 4, 2 ] # random
[1, 5, 0, 3, 4, 2 ][ 0..2 ] ==> [1, 5, 0 ]
Footnotes:
It is worth mentioning that at the time this question was originally answered, September 2008, that Array#shuffle was either not available or not already known to me, hence the approximation in Array#sort
And there's a barrage of suggested edits to this as a result.
So:
.sort{ rand() - 0.5 }
Can be better, and shorter expressed on modern ruby implementations using
.shuffle
Additionally,
[0..x]
Can be more obviously written with Array#take as:
.take(x)
Thus, the easiest way to produce a sequence of random numbers on a modern ruby is:
(0..50).to_a.shuffle.take(x)
This uses Set:
require 'set'
def rand_n(n, max)
randoms = Set.new
loop do
randoms << rand(max)
return randoms.to_a if randoms.size >= n
end
end
Ruby 1.9 offers the Array#sample method which returns an element, or elements randomly selected from an Array. The results of #sample won't include the same Array element twice.
(1..999).to_a.sample 5 # => [389, 30, 326, 946, 746]
When compared to the to_a.sort_by approach, the sample method appears to be significantly faster. In a simple scenario I compared sort_by to sample, and got the following results.
require 'benchmark'
range = 0...1000000
how_many = 5
Benchmark.realtime do
range.to_a.sample(how_many)
end
=> 0.081083
Benchmark.realtime do
(range).sort_by{rand}[0...how_many]
end
=> 2.907445
Just to give you an idea about speed, I ran four versions of this:
Using Sets, like Ryan's suggestion.
Using an Array slightly larger than necessary, then doing uniq! at the end.
Using a Hash, like Kyle suggested.
Creating an Array of the required size, then sorting it randomly, like Kent's suggestion (but without the extraneous "- 0.5", which does nothing).
They're all fast at small scales, so I had them each create a list of 1,000,000 numbers. Here are the times, in seconds:
Sets: 628
Array + uniq: 629
Hash: 645
fixed Array + sort: 8
And no, that last one is not a typo. So if you care about speed, and it's OK for the numbers to be integers from 0 to whatever, then my exact code was:
a = (0...1000000).sort_by{rand}
Yes, it's possible to do this without a loop and without keeping track of which numbers have been chosen. It's called a Linear Feedback Shift Register: Create Random Number Sequence with No Repeats
[*1..99].sample(4) #=> [64, 99, 29, 49]
According to Array#sample docs,
The elements are chosen by using random and unique indices
If you need SecureRandom (which uses computer noise instead of pseudorandom numbers):
require 'securerandom'
[*1..99].sample(4, random: SecureRandom) #=> [2, 75, 95, 37]
How about a play on this? Unique random numbers without needing to use Set or Hash.
x = 0
(1..100).map{|iter| x += rand(100)}.shuffle
You could use a hash to track the random numbers you've used so far:
seen = {}
max = 100
(1..10).map { |n|
x = rand(max)
while (seen[x])
x = rand(max)
end
x
}
Rather than add the items to a list/array, add them to a Set.
If you have a finite list of possible random numbers (i.e. 1 to 100), then Kent's solution is good.
Otherwise there is no other good way to do it without looping. The problem is you MUST do a loop if you get a duplicate. My solution should be efficient and the looping should not be too much more than the size of your array (i.e. if you want 20 unique random numbers, it might take 25 iterations on average.) Though the number of iterations gets worse the more numbers you need and the smaller max is. Here is my above code modified to show how many iterations are needed for the given input:
require 'set'
def rand_n(n, max)
randoms = Set.new
i = 0
loop do
randoms << rand(max)
break if randoms.size > n
i += 1
end
puts "Took #{i} iterations for #{n} random numbers to a max of #{max}"
return randoms.to_a
end
I could write this code to LOOK more like Array.map if you want :)
Based on Kent Fredric's solution above, this is what I ended up using:
def n_unique_rand(number_to_generate, rand_upper_limit)
return (0..rand_upper_limit - 1).sort_by{rand}[0..number_to_generate - 1]
end
Thanks Kent.
No loops with this method
Array.new(size) { rand(max) }
require 'benchmark'
max = 1000000
size = 5
Benchmark.realtime do
Array.new(size) { rand(max) }
end
=> 1.9114e-05
Here is one solution:
Suppose you want these random numbers to be between r_min and r_max. For each element in your list, generate a random number r, and make list[i]=list[i-1]+r. This would give you random numbers which are monotonically increasing, guaranteeing uniqueness provided that
r+list[i-1] does not over flow
r > 0
For the first element, you would use r_min instead of list[i-1]. Once you are done, you can shuffle the list so the elements are not so obviously in order.
The only problem with this method is when you go over r_max and still have more elements to generate. In this case, you can reset r_min and r_max to 2 adjacent element you have already computed, and simply repeat the process. This effectively runs the same algorithm over an interval where there are no numbers already used. You can keep doing this until you have the list populated.
As far as it is nice to know in advance the maxium value, you can do this way:
class NoLoopRand
def initialize(max)
#deck = (0..max).to_a
end
def getrnd
return #deck.delete_at(rand(#deck.length - 1))
end
end
and you can obtain random data in this way:
aRndNum = NoLoopRand.new(10)
puts aRndNum.getrnd
you'll obtain nil when all the values will be exausted from the deck.
Method 1
Using Kent's approach, it is possible to generate an array of arbitrary length keeping all values in a limited range:
# Generates a random array of length n.
#
# #param n length of the desired array
# #param lower minimum number in the array
# #param upper maximum number in the array
def ary_rand(n, lower, upper)
values_set = (lower..upper).to_a
repetition = n/(upper-lower+1) + 1
(values_set*repetition).sample n
end
Method 2
Another, possibly more efficient, method modified from same Kent's another answer:
def ary_rand2(n, lower, upper)
v = (lower..upper).to_a
(0...n).map{ v[rand(v.length)] }
end
Output
puts (ary_rand 5, 0, 9).to_s # [0, 8, 2, 5, 6] expected
puts (ary_rand 5, 0, 9).to_s # [7, 8, 2, 4, 3] different result for same params
puts (ary_rand 5, 0, 1).to_s # [0, 0, 1, 0, 1] repeated values from limited range
puts (ary_rand 5, 9, 0).to_s # [] no such range :)