On one of my log file I have the below hidden values and color ASCII codes,
...WAITING^H^H^H^H^H^H^H^H^H^H
I was able to remove the color ASCII codes using below method,
gsub(/\e\[(\d+)(;(\d+))?m/, '')
but am still unable to remove that mentioned above hidden characters. Is there any way to get them rid of?
Theory
Backspaces ?
If the ctrl-H characters are backspaces :
puts "foo\b\b\bbar"
#=> "bar"
puts "foo\b\b\bbar".delete("\b")
#=> "foobar"
NOTE: delete is fine here, because we use it with just one character.
Or "^H" substring ?
If the ctrl-H characters are "^H" :
puts "foo^H^H^Hbar".gsub(/\^H/,'')
#=> "foobar"
NOTE: delete wouldn't work here, because it would also remove every H character from the strings, not just the substring ^H. Also, using delete("^H") means delete every character that isn't a 'H'. So :
"foo^H^H^Hbar".delete("^H") => "HHH"
Test
With :
bytes = [46, 46, 46, 87, 65, 73, 84, 73, 78, 71, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 91, 32, 32, 32, 79, 75, 32, 32, 32, 93, 10, 27, 91, 63, 49, 50, 108, 27, 91, 63, 50, 53, 104, 68, 111, 110, 101, 33, 10, 10]
We get :
string = bytes.map(&:chr).join
string # => "...WAITING\b\b\b\b\b\b\b\b\b\b[ OK ]\n\e[?12l\e[?25hDone!\n\n"
puts string
# [ OK ]
# Done!
#
Bytes equal to 8 are backspaces, and they delete WAITING when displayed with puts.
The first alternative should work fine :
puts string.delete("\b")
# ...WAITING[ OK ]
# Done!
NOTE: This only works on the original data, in which backspaces are byte 8. Any copy-paste, use of cat, | or text editor might convert those to "^H" or other string.
Related
(Also open to other similar non-Rails methods)
Given (0..99), return entries that are randomly picked in-order.
Example results:
0, 5, 11, 13, 34..
3, 12, 45, 67, 87
0, 1, 2, 3, 4, 5.. (very unlikely, of course)
Current thought:
(0..99).step(rand(0..99)).each do |subindex|
array.push(subindex)
end
However, this sets a single random value for all the steps whereas I'm looking for each step to be random.
Get a random value for the number of elements to pick, randomly get this number of elements, sort.
(0..99).to_a.sample((0..99).to_a.sample).sort
#⇒ [7, 20, 22, 29, 45, 48, 57, 61, 62, 76, 80, 82]
Or, shorter (credits to #Stefan):
(0..99).to_a.sample(rand(0..99)).sort
#⇒ [7, 20, 22, 29, 45, 48, 57, 61, 62, 76, 80, 82]
Or, in more functional manner:
λ = (0..99).to_a.method(:sample)
λ.(λ.()).sort
To feed exactly N numbers:
N = 10
(0..99).to_a.sample(N).sort
#⇒ [1, 5, 8, 12, 45, 54, 60, 65, 71, 91]
There're many ways to achieve it.
For example here's slow yet simple one:
# given `array`
random_indexes = (0...array.size).to_a.sample(rand(array.size))
random_indexes.sort.each { |i| puts array[i] }
Or why don't you just:
array.each do |value|
next if rand(2).zero?
puts value
end
Or you could use Enumerator#next random number of times.
Below example returns a sorted array with random entries from given range based on randomly picked true or false from array [true, false]:
(0..99).select { [true, false].sample }
=> [0, 3, 12, 13, 14, 17, 20, 24, 26, 28, 30, 32, 34, 35, 36, 38, 39, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 53, 54, 55, 56, 58, 59, 60, 61, 62, 65, 67, 69, 70, 71, 79, 81, 84, 86, 91, 93, 94, 95, 98, 99]
To reduce the chances of a bigger array being returned, you can modify your true/false array to include more falsey values:
(0..99).select { ([true] + [false] * 9).sample }
=> [21, 22, 28, 33, 37, 58, 59, 63, 77, 85, 86]
in Ruby I have a string like this:
myString = "mystring"
I want to convert the string to a byte array taking only the first 16 bytes and pad with 0's if shorter.
I can do this the brute force way. But...
Care to share a 'cool' way?
Something like this? You should probably check for edge cases like multibyte chars.
"my string"[0..15].ljust(16,'0')
You can get the string as a byte array by calling bytes on it, then once you have it as a byte array, you can take the first 16 elements. Finally, you pad the array by filling it with a range as the second argument:
def padded_byte_array(string, length = 16)
bytes = string.bytes.take(length)
bytes.fill(0, bytes.length...length)
end
and then you can call it:
padded_byte_array('my string')
# => [109, 121, 32, 115, 116, 114, 105, 110, 103, 0, 0, 0, 0, 0, 0, 0]
padded_byte_array('some super long string longer than 16 bytes')
# => [115, 111, 109, 101, 32, 115, 117, 112, 101, 114, 32, 108, 111, 110, 103, 32]
padded_byte_array('本当に長いマルチバイト文字列')
# => [230, 156, 172, 229, 189, 147, 227, 129, 171, 233, 149, 183, 227, 129, 132, 227]
I assume that if arr.size < str.size, where str is the string and arr is the array to be returned, str.bytes is returned. If, in that case, str.bytes[0, [str.size, arr.size].min] is to be returned, that requires an obvious adjustment.
def padded_bytes(str, arr_size)
str_bytes = str.bytes
Array.new([arr_size, str.size].max) { |i| str_bytes.fetch(i, 0) }
end
padded_bytes("tiger", 8)
#=> [116, 105, 103, 101, 114, 0, 0, 0]
padded_bytes("tiger", 3)
#=> [116, 105, 103, 101, 114]
Thanks folks for your answers.
In the end, I implemented
ba = name[0..15].ljust(16, "\0").bytes.to_a
Aza gave the closest to what I asked.
My original looked like this:
ba = name[0..15].bytes.to_a
while ba.length < 16 do ba.push(0) end
Until I got your answers. Thanks again!
This question already has an answer here:
replacing an element in nested array ruby
(1 answer)
Closed 7 years ago.
Hello I have my following bingo code that marks X for numbers that get returned:
class BingoBoard
def initialize(board)
#bingo_board = board
end
def number_letter
#letter = ['B','I','N','G','O'].sample
#number = rand(1..100)
end
def checker
number_letter
#bingo_board.map! do |n|
if n.include?(#number) #cleaned up code from the initial solution.
n.map! { |x| x == #number ? 'X' : x}
else
n
end
end
end
end
My question is how do I change my code so that when I'm using the test code:
board = [[47, 44, 71, 8, 88],
[22, 69, 75, 65, 73],
[83, 85, 97, 89, 57],
[25, 31, 96, 68, 51],
[75, 70, 54, 80, 83]]
new_game = BingoBoard.new(board)
new_game.checker
It will appear neatly like a bingo board in irb.
Right now it looks like:
=>[[47, 44, 71, 8, 88], [22, 69, 75, 65, 73], ["X", 85, 97, 89, 57], [25, 31, 96, 68, 51], [75, 70, 54, 80, "X"]]
Append .map { |block| puts block.inspect } to the new_game.checker call.
Is there any cool way in Ruby to create an array with 1 to 100 with only odd entries (1, 3 etc). I now have a loop for this but that is obviously not a cool way to do it! Any suggestions?
My current code:
def create_1_to_100_odd_array
array = [1]
i = 3
while i < 100
array.push i
i += 2
end
array
end
Thanks in advance
The Range class comes with a very cool feature for that purpose:
1.9.3-p286 :005 > (1..10).step(2).to_a
=> [1, 3, 5, 7, 9]
May not be efficient, but a short piece of code:
(1..100).select(&:odd?)
# => [1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99]
Just toying...
(0...50).map(&:object_id)
#or
1.step(100,2).to_a
Since you need a function, then:
def odd_to(n)
(1..n).step(2).to_a
end
Not very effective solution, but quite elegant:
(1..100).select {|a| a%2 != 0}
You can do it as a one-liner when you instantiate the array:
def create_array_of_odds_to(n)
Array.new((n + 1) / 2) {|i| 2 * i + 1}
end
create_array_of_odds_to 10 # => [1, 3, 5, 7, 9]
I was under the impression that set() would order a collection much like .sort()
However it seems that it doesn't, what was peculiar to me was why it reorders the collection.
>>> h = '321'
>>> set(h)
set(['1', '3', '2'])
>>> h
'321'
>>> h = '22311'
>>> set(h)
set(['1', '3', '2'])
why doesn't it return set(['1', '2', '3']). I also seems that no matter how many instances of each number I user or in what order I use them it always return set(['1', '3', '2']). Why?
Edit:
So I have read your answers and my counter to that is this.
>>> l = [1,2,3,3]
>>> set(l)
set([1, 2, 3])
>>> l = [3,3,2,3,1,1,3,2,3]
>>> set(l)
set([1, 2, 3])
Why does it order numbers and not strings?
Also
import random
l = []
for itr in xrange(101):
l.append(random.randint(1,101))
print set(l)
Outputs
>>>
set([1, 2, 4, 5, 6, 8, 10, 11, 12, 14, 15, 16, 18, 19, 23, 24, 25, 26, 29, 30, 31, 32, 34, 40, 43, 45, 46, 47, 48, 49, 50, 51, 53, 54, 55, 57, 58, 59, 60, 61, 62, 63, 64, 66, 67, 69, 70, 74, 75, 77, 79, 80, 83, 84, 85, 87, 88, 89, 90, 93, 94, 96, 97, 99, 101])
python set is unordered, hence there is no guarantee that the elements would be ordered in the same way as you specify them
If you want a sorted output, then call sorted:
sorted(set(h))
Responding to your edit: it comes down to the implementation of set. In CPython, it boils down to two things:
1) the set will be sorted by hash (the __hash__ function) modulo a limit
2) the limit is generally the next largest power of 2
So let's look at the int case:
x=1
type(x) # int
x.__hash__() # 1
for ints, the hash equals the original value:
[x==x.__hash__() for x in xrange(1000)].count(False) # = 0
Hence, when all the values are ints, it will use the integer hash value and everything works smoothly.
for the string representations, the hashes dont work the same way:
x='1'
type(x)
# str
x.__hash__()
# 6272018864
To understand why the sort breaks for ['1','2','3'], look at those hash values:
[str(x).__hash__() for x in xrange(1,4)]
# [6272018864, 6400019251, 6528019634]
In our example, the mod value is 4 (3 elts, 2^1 = 2, 2^2 = 4) so
[str(x).__hash__()%4 for x in xrange(1,4)]
# [0, 3, 2]
[(str(x).__hash__()%4,str(x)) for x in xrange(1,4)]
# [(0, '1'), (3, '2'), (2, '3')]
Now if you sort this beast, you get the ordering that you see in set:
[y[1] for y in sorted([(str(x).__hash__()%4,str(x)) for x in xrange(1,4)])]
# ['1', '3', '2']
From the python documentation of the set type:
A set object is an unordered collection of distinct hashable objects.
This means that the set doesn't have a concept of the order of the elements in it. You should not be surprised when the elements are printed on your screen in an unusual order.
A set in Python tries to be a "set" in the mathematical sense of the term. No duplicates, and order shouldn't matter.