Additionally, how can I format it as a string padded with zeros?
To generate the number call rand with the result of the expression "10 to the power of 10"
rand(10 ** 10)
To pad the number with zeros you can use the string format operator
'%010d' % rand(10 ** 10)
or the rjust method of string
rand(10 ** 10).to_s.rjust(10,'0')
I would like to contribute probably a simplest solution I know, which is a quite a good trick.
rand.to_s[2..11]
=> "5950281724"
This is a fast way to generate a 10-sized string of digits:
10.times.map{rand(10)}.join # => "3401487670"
The most straightforward answer would probably be
rand(1e9...1e10).to_i
The to_i part is needed because 1e9 and 1e10 are actually floats:
irb(main)> 1e9.class
=> Float
DON'T USE rand.to_s[2..11].to_i
Why? Because here's what you can get:
rand.to_s[2..9] #=> "04890612"
and then:
"04890612".to_i #=> 4890612
Note that:
4890612.to_s.length #=> 7
Which is not what you've expected!
To check that error in your own code, instead of .to_i you may wrap it like this:
Integer(rand.to_s[2..9])
and very soon it will turn out that:
ArgumentError: invalid value for Integer(): "02939053"
So it's always better to stick to .center, but keep in mind that:
rand(9)
sometimes may give you 0.
To prevent that:
rand(1..9)
which will always return something withing 1..9 range.
I'm glad that I had good tests and I hope you will avoid breaking your system.
Random number generation
Use Kernel#rand method:
rand(1_000_000_000..9_999_999_999) # => random 10-digits number
Random string generation
Use times + map + join combination:
10.times.map { rand(0..9) }.join # => random 10-digit string (may start with 0!)
Number to string conversion with padding
Use String#% method:
"%010d" % 123348 # => "0000123348"
Password generation
Use KeePass password generator library, it supports different patterns for generating random password:
KeePass::Password.generate("d{10}") # => random 10-digit string (may start with 0!)
A documentation for KeePass patterns can be found here.
Just because it wasn't mentioned, the Kernel#sprintf method (or it's alias Kernel#format in the Powerpack Library) is generally preferred over the String#% method, as mentioned in the Ruby Community Style Guide.
Of course this is highly debatable, but to provide insight:
The syntax of #quackingduck's answer would be
# considered bad
'%010d' % rand(10**10)
# considered good
sprintf('%010d', rand(10**10))
The nature of this preference is primarily due to the cryptic nature of %. It's not very semantic by itself and without any additional context it can be confused with the % modulo operator.
Examples from the Style Guide:
# bad
'%d %d' % [20, 10]
# => '20 10'
# good
sprintf('%d %d', 20, 10)
# => '20 10'
# good
sprintf('%{first} %{second}', first: 20, second: 10)
# => '20 10'
format('%d %d', 20, 10)
# => '20 10'
# good
format('%{first} %{second}', first: 20, second: 10)
# => '20 10'
To make justice for String#%, I personally really like using operator-like syntaxes instead of commands, the same way you would do your_array << 'foo' over your_array.push('123').
This just illustrates a tendency in the community, what's "best" is up to you.
More info in this blogpost.
I ended up with using Ruby kernel srand
srand.to_s.last(10)
Docs here: Kernel#srand
Here is an expression that will use one fewer method call than quackingduck's example.
'%011d' % rand(1e10)
One caveat, 1e10 is a Float, and Kernel#rand ends up calling to_i on it, so for some higher values you might have some inconsistencies. To be more precise with a literal, you could also do:
'%011d' % rand(10_000_000_000) # Note that underscores are ignored in integer literals
('%010d' % rand(0..9999999999)).to_s
or
"#{'%010d' % rand(0..9999999999)}"
I just want to modify first answer. rand (10**10) may generate 9 digit random no if 0 is in first place. For ensuring 10 exact digit just modify
code = rand(10**10)
while code.to_s.length != 10
code = rand(11**11)
end
Try using the SecureRandom ruby library.
It generates random numbers but the length is not specific.
Go through this link for more information: http://ruby-doc.org/stdlib-2.1.2/libdoc/securerandom/rdoc/SecureRandom.html
Simplest way to generate n digit random number -
Random.new.rand((10**(n - 1))..(10**n))
generate 10 digit number number -
Random.new.rand((10**(10 - 1))..(10**10))
This technique works for any "alphabet"
(1..10).map{"0123456789".chars.to_a.sample}.join
=> "6383411680"
Just use straightforward below.
rand(10 ** 9...10 ** 10)
Just test it on IRB with below.
(1..1000).each { puts rand(10 ** 9...10 ** 10) }
To generate a random, 10-digit string:
# This generates a 10-digit string, where the
# minimum possible value is "0000000000", and the
# maximum possible value is "9999999999"
SecureRandom.random_number(10**10).to_s.rjust(10, '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 = 10,000,000,000
upper_bound = 10**10
# n will be an integer with a minimum possible value of 0,
# and a maximum possible value of 9,999,999,999
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 "9999999999" if n == 9_999_999_999
unpadded_str = n.to_s
# Pad the string with leading zeroes if it is less than
# 10 digits long.
# "0" would be padded to "0000000000"
# "123" would be padded to "0000000123"
# "9999999999" would not be padded, and remains unchanged as "9999999999"
padded_str = unpadded_str.rjust(10, '0')
rand(9999999999).to_s.center(10, rand(9).to_s).to_i
is faster than
rand.to_s[2..11].to_i
You can use:
puts Benchmark.measure{(1..1000000).map{rand(9999999999).to_s.center(10, rand(9).to_s).to_i}}
and
puts Benchmark.measure{(1..1000000).map{rand.to_s[2..11].to_i}}
in Rails console to confirm that.
An alternative answer, using the regexp-examples ruby gem:
require 'regexp-examples'
/\d{10}/.random_example # => "0826423747"
There's no need to "pad with zeros" with this approach, since you are immediately generating a String.
This will work even on ruby 1.8.7:
rand(9999999999).to_s.center(10, rand(9).to_s).to_i
A better approach is use Array.new() instead of .times.map. Rubocop recommends it.
Example:
string_size = 9
Array.new(string_size) do
rand(10).to_s
end
Rubucop, TimesMap:
https://www.rubydoc.info/gems/rubocop/RuboCop/Cop/Performance/TimesMap
In my case number must be unique in my models, so I added checking block.
module StringUtil
refine String.singleton_class do
def generate_random_digits(size:)
proc = lambda{ rand.to_s[2...(2 + size)] }
if block_given?
loop do
generated = proc.call
break generated if yield(generated) # check generated num meets condition
end
else
proc.call
end
end
end
end
using StringUtil
String.generate_random_digits(3) => "763"
String.generate_random_digits(3) do |num|
User.find_by(code: num).nil?
end => "689"(This is unique in Users code)
I did something like this
x = 10 #Number of digit
(rand(10 ** x) + 10**x).to_s[0..x-1]
Random 10 numbers:
require 'string_pattern'
puts "10:N".gen
Related
Is it possible to use destructuring in ruby to extract the end and beginning from a range?
module PriceHelper
def price_range_human( range )
"$%s to $%s" % [range.begin, range.end].map(:number_to_currency)
end
end
I know that I can use array coercion as a really bad hack:
first, *center, last = *rng
"$%s to $%s" % [first, last].map(:number_to_currency)
But is there a syntactical way to get begin and end without actually manually creating an array?
min, max = (1..10)
Would have been awesome.
You can use minmax to destructure ranges:
min, max = (1..10).minmax
min # => 1
max # => 10
If you are using Ruby before 2.7, avoid using this on large ranges.
The beginning and end? I'd use:
foo = 1..2
foo.min # => 1
foo.max # => 2
Trying to use destructuring for a range is a bad idea. Imagine the sizes of the array that could be generated then thrown away, wasting CPU time and memory. It's actually a great way to DOS your own code if your range ends with Float::INFINITY.
end is not the same as max: in 1...10, end is 10, but max is 9
That's because start_val ... end_val is equivalent to start_val .. (end_val - 1):
start_value = 1
end_value = 2
foo = start_value...end_value
foo.end # => 2
foo.max # => 1
foo = start_value..(end_value - 1)
foo.end # => 1
foo.max # => 1
max reflects the reality of the values actually used by Ruby when iterating over the range or testing for inclusion in the range.
In my opinion, end should reflect the actual maximum value that will be considered inside the range, not the value used at the end of the definition of the range, but I doubt that'll change otherwise it'd affect existing code.
... is more confusing and leads to increased maintenance problems so its use is not recommended.
No, Until I am proven incorrect by Cary Swoveland, Weekly World News or another tabloid, I'll continue believing without any evidence that the answer is "no"; but it's easy enough to make.
module RangeWithBounds
refine Range do
def bounds
[self.begin, self.end]
end
end
end
module Test
using RangeWithBounds
r = (1..10)
b, e = *r.bounds
puts "#{b}..#{e}"
end
Then again, I'd just write "#{r.begin.number_to_currency}..#{r.end.number_to_currency}" in the first place.
Amadan's answer is fine. you just need to remove the splat (*) when using it since it is not needed
eg,
> "%s to %s" % (1..3).bounds.map{|x| number_to_currency(x)}
=> "$1.00 to $3.00"
I need to convert an input string to an integer using a get_integer_from_string() function.
I am using Ruby
and I have this schedule.
based on the schedule we need to covert from base 7 to base 10
Please note :
1- If you encounter a string with no numerical sequences, it should return zero (0).
2- The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.
Thank you for helping me with my homework.
As Tim pointed out above, your table shows base 7, not 6. Both String#to_i and Fixnum#to_s take an optional radix (base) argument:
p "020".to_i(7)
# => 14
p 14.to_s(7)
# => "20"
In order to deal with trailing characters that aren't 0-6, String#slice (a.k.a. String#[]) works well:
expr = /^[0-6]*/
p "72"[expr].to_i(7)
# => 0
p "0202xyz"[expr].to_i(7)
# => 100
p "27"[expr].to_i(7)
# => 2
What you want to do is convert the string (which represents a base 7 number) to a base 10 integer. This can be done via
"021".to_i(7)
# => 15
Please see the Documentation of Ruby String class for the method to_i()
i made this not sure if it's right
def get_integer_from_string(str)
return 0 if (/^(?<num>\d+)$/ =~ str).nil?
result = 0
str.reverse.each_char.with_index do |char, index|
tmp = char.to_i * 7**index
result += tmp
end
result rescue 0
end
It would be nice to have an equivalent of R's signif function in Ruby.
For example:
>> (11.11).signif(1)
10
>> (22.22).signif(2)
22
>> (3.333).signif(2)
3.3
>> (4.4).signif(3)
4.4 # It's usually 4.40 but that's OK. R does not print the trailing 0's
# because it returns the float data type. For Ruby we want the same.
>> (5.55).signif(2)
5.6
There is probably better way, but this seems to work fine:
class Float
def signif(signs)
Float("%.#{signs}g" % self)
end
end
(1.123).signif(2) # => 1.1
(11.23).signif(2) # => 11.0
(11.23).signif(1) # => 10.0
Here's an implementation that doesn't use strings or other libraries.
class Float
def signif(digits)
return 0 if self.zero?
self.round(-(Math.log10(self).ceil - digits))
end
end
I don't see anything like that in Float. Float is mostly a wrapper for the native double type and given the usual binary/decimal issues, I'm not that surprised that Float doesn't allow you to manipulate the significant digits.
However, BigDecimal in the standard library does understand significant digits but again, I don't see anything that allows you to directly alter the significant digits in a BigDecimal: you can ask for it but you can't change it. But, you can kludge around that by using a no-op version of the mult or add methods:
require 'bigdecimal'
a = BigDecimal.new('11.2384')
a.mult(1, 2) # the result is 0.11E2 (i.e. 11)
a.add(0, 4) # the result is 0.1124E2 (i.e. 11.24)
The second argument to these methods:
If specified and less than the number of significant digits of the result, the result is rounded to that number of digits, according to BigDecimal.mode.
Using BigDecimal will be slower but it might be your only choice if you need fine grained control or if you need to avoid the usual floating point problems.
Some of the previous answers and comments have alluded to this solution but this is what worked for me:
# takes in a float value and returns another float value rounded to
# given significant figures.
def round_to_sig_figs(val, sig_figs)
BigDecimal.new(val, sig_figs).to_f
end
You are probably looking for Ruby's Decimal.
You could then write:
require 'decimal/shortcut'
num = 1.23541764
D.context.precision = 2
num_with_2_significant_digits = +D(num.to_s) # => Decimal('1.2')
num_with_2_significant_digits.to_f # => 1.2000000000000002
Or if you prefer to use the same syntax add this as a function to class Float like this:
class Float
def signif num_digits
require 'decimal/shortcut'
D.context.precision = num_digits
(+D(self.to_s)).to_f
end
end
Usage would then be the same, i.e.
(1.23333).signif 3
# => 1.23
To use it, install the gem
gem install ruby-decimal
#Blou91's answer is nearly there, but it returns a string, instead of a float. This below works for me:
(sprintf "%.2f", 1.23456).to_f
So as a function,
def round(val, sig_figs)
(sprintf "%.#{sig_figs}f", val).to_f
end
Use sprintf if you want to print trailing zeros
2.0.0-p353 :001 > sprintf "%.3f", 500
=> "500.000"
2.0.0-p353 :002 > sprintf "%.4f", 500
=> "500.0000"
2.0.0-p353 :003 >
I looking for ruby code to convert BCD to Ascii.I have tried with many codes but i am not getting proper result.
Any suggestions or code samples?
I actually wrote a gem for this sort of thing.
https://rubygems.org/gems/bcd
The source code is at https://github.com/dafyddcrosby/ruby_bcd
Although this question is long dead, I just dealt with this. Assuming you've already converted to bcd:
bcd_val.to_s(16)
This converts the bcd value to hex, but because each nibble is a digit, and all digits are between 0-9, it effectively displays as an integer string.
Note that if your bcd number is stored LSB to MSB (depending on how you converted to bcd) the number will display backwards. At that point you can always user str.reverse as necessary.
It is not too clear what do you want to do, but hope that the following can help:
def to_bcd(n)
str = n.to_s
bin = ""
str.each_char do |c|
bin << c.to_i.to_s(2).rjust(4,'0')
end
bin
end
def to_dec(bcd)
n = ""
(bcd.length / 4).times do |i|
n << Integer('0b'+bcd[i*4..(i*4+3)]).to_s
end
n
end
result = to_bcd(120)
p result #=> "000100100000"
p to_dec(result) #=> "120"
p to_dec(result).to_i.chr #=> "x"
If you like, you could extend the Integer and String class with the two methods above, respectively, without arguments and substituting the parameter name with self. But someone does not like the idea to extend standard classes, because it is not a clean/safe programming habit.
Is there any worthy Ruby method to count the number of digits in a float? Also, how do I specify the precise when to_s float numbers?
# Number of digits
12345.23.to_s.split("").size -1 #=> 7
# The precious part
("." + 12345.23.to_s.split(".")[1]).to_f #=> .023
# I would rather used
# 12345.23 - 12345.23.to_i
# but this gives 0.22999999999563
to specify precision of a float in Ruby. you can use the round method.
number.round(2)
2 is the precision.
53.819.round(2) -> 53.82
I think you should check out the number_with_precision helper.
number_with_precision(13, :precision => 5) # => 13.00000