How do I do this type of for loop in Ruby?
for(int i=0; i<array.length; i++) {
}
array.each do |element|
element.do_stuff
end
or
for element in array do
element.do_stuff
end
If you need index, you can use this:
array.each_with_index do |element,index|
element.do_stuff(index)
end
limit = array.length;
for counter in 0..limit
--- make some actions ---
end
the other way to do that is the following
3.times do |n|
puts n;
end
thats will print 0, 1, 2, so could be used like array iterator also
Think that variant better fit to the author's needs
I keep hitting this as a top link for google "ruby for loop", so I wanted to add a solution for loops where the step wasn't simply '1'. For these cases, you can use the 'step' method that exists on Numerics and Date objects. I think this is a close approximation for a 'for' loop.
start = Date.new(2013,06,30)
stop = Date.new(2011,06,30)
# step back in time over two years, one week at a time
start.step(stop, -7).each do |d|
puts d
end
The equivalence would be
for i in (0...array.size)
end
or
(0...array.size).each do |i|
end
or
i = 0
while i < array.size do
array[i]
i = i + 1 # where you may freely set i to any value
end
array.each_index do |i|
...
end
It's not very Rubyish, but it's the best way to do the for loop from question in Ruby
To iterate a loop a fixed number of times, try:
n.times do
#Something to be done n times
end
If you don't need to access your array, (just a simple for loop) you can use upto or each :
Upto:
2.upto(4) {|i| puts i}
2
3
4
Each:
(2..4).each {|i| puts i}
2
3
4
What? From 2010 and nobody mentioned Ruby has a fine for /in loop (it's just nobody uses it):
ar = [1,2,3,4,5,6]
for item in ar
puts item
end
['foo', 'bar', 'baz'].each_with_index {|j, i| puts "#{i} #{j}"}
Ruby's enumeration loop syntax is different:
collection.each do |item|
...
end
This reads as "a call to the 'each' method of the array object instance 'collection' that takes block with 'blockargument' as argument". The block syntax in Ruby is 'do ... end' or '{ ... }' for single line statements.
The block argument '|item|' is optional but if provided, the first argument automatically represents the looped enumerated item.
Related
I have a for loop that I would like to have increment forever.
My code:
for a in (0...Float::INFINITY).step(2)
puts a
end
Output:
0.0
2.0
4.0
Etc. Always with "#{a}.0"
Is there any way to express infinity as an integer, so that the output does not have a .0 at the end without preforming any operations on the contents of the loop?
Addendum
Could you also explain how your loop works? I am trying to find the most efficient solution, because since this loop will be iterating infinity, a few milliseconds shaved off will improve the performance greatly.
Also...
I will accept the solution that takes to shortest time to run to 1000000
According to benchmark both #Sefan and the while loop answers take the same ammount of timeFruity the while loop answers take a bit shorter, with the for loop answers in second, but the multiple loop do answers take far longer.
Since the reason why is out of the scope of this question, I have created another question that addresses why some loops are faster than others (https://stackoverflow.com/questions/33088764/peddle-to-the-metal-faster-loop-faster).
You can use Numeric#step without passing a limit:
0.step(by: 2) { |i| puts i }
Output:
0
2
4
6
...
You can also build your own Enumerator:
step2 = Enumerator.new do |y|
a = 0
loop do
y << a
a += 2
end
end
step2.each { |i| puts i }
You can use while true for that:
puts a = 0
puts a+=2 while true
BTW,
Is there any way to express infinity as an integer
NO
require 'bigdecimal'
(0..BigDecimal('Infinity')).step(2).each{ |n| puts n }
OR
require 'bigdecimal'
for a in (0...BigDecimal::INFINITY).step(2)
puts a
end
This is what the loop method is designed for. loop has no condition for which to run. It will run indefinitely and the only way to exit is to use the keyword break. (or raise a StopIteration)
a = 0
loop { puts a += 2}
This loop will be infinite as there is no break specified.
break can be specified very similarly to how the other answers use the while condition if needed:
a = 0
loop do
puts a += 2
break if a > 1_000_000
end
This loop will now exit once the value of a exceeds 1M.
That being said #Stefan's answer is more efficient as it does not store this integral value or have to perform any additional assignment but rather the number is simply yielded from an Enumerator and discarded it afterwards. The usefulness of this becomes more a matter of your implementation and purpose for this loop.
Try this:
arr = [0]
arr.cycle(1000000) { |i| puts arr[0] +=2 }
If you want infinite loop, then, don't pass any parameter to cycle
arr = [0]
arr.cycle { |i| puts arr[0] +=2 }
a = [-2]
puts a.unshift(a.shift+2) while 'loop forever'
I have an array containing a large number of contents
Names = ["one", "two", ......."forty"]
Now, I would like to loop through sections of the array. For example, from records 10 to 20, How can I go about this?
I tried this approach -
Names.each_with_index do |val,index|
break if index == 10
puts "#{val}"
end
In this way, I can print the first ten records. What should I do for getting next set of 10 names?
Any help is appreciated.
Cheers!
You can use each_slice to get successive sections of the array. For example:
require 'enumerator'
Names.each_slice(10) do |slice|
slice.each do |x|
puts x
end
puts "----"
end
You can e.g. use the each_slice method:
Names.each_slice(10) {|part|
puts part
}
this is the question
Shuffle. Now that you’ve finished your
new sorting algorithm, how about the
opposite? Write a shuffle method that
takes an array and returns a totally
shuffled version. As always, you’ll
want to test it, but testing this one
is trickier: How can you test to make
sure you are getting a perfect
shuffle? What would you even say a
perfect shuffle would be? Now test for
it.
This is my code answer:
def shuffle arr
x = arr.length
while x != 0
new_arr = []
rand_arr = (rand(x))
x--
new_arr.push rand_arr
arr.pop rand_arr
end
new_arr
end
puts (shuffle ([1,2,3]))
What are my mistakes? Why doesn't this code work?
Here's a far more Rubyish version:
class Array
def shuffle!
size.downto(1) { |n| push delete_at(rand(n)) }
self
end
end
puts [1,2,3].shuffle!
Here's a more concise way of writing it:
def shuffle(arr)
new_arr = []
while (arr.any?) do
new_arr << arr.delete_at(rand(arr.length))
end
new_arr
end
And some tests:
5.times do
puts shuffle((1..5).to_a).join(',')
end
>> 4,2,1,3,5
>> 3,2,1,4,5
>> 4,2,5,1,3
>> 5,2,1,4,3
>> 4,3,1,5,2
Beside minor other errors you seems not to understand what pop and push are doing (taking or adding some items from the end of the array).
You are probably trying to write something like below.
def shuffle arr
x = arr.length
new_arr = []
while x != 0
randpos = rand(x)
x = x-1
item = arr[randpos]
new_arr.push item
arr[randpos] = arr[x]
arr.pop
end
new_arr
end
puts (shuffle ([1,2,3]))
You're getting your indexes mixed up with your values. When you do new_arr.push rand_arr, you're putting whatever random index you came up with as a value on the end of new_arr. What you meant to do is new_arr.push arr[rand_arr], where arr[rand_arr] is the value at the index rand_arr in arr.
Ruby 1.8.7 and 1.9.2 have a built-in Array#shuffle method.
A variant of Mark Thomas's answer. His algorithm can be quite slow with a large array, due to delete operation performance.
class Array
def shuffle!
size.downto(1) do |n|
index=rand(n)
# swap elements at index and the end
self[index], self[size-1] = self[size-1],self[index]
end
self
end
end
puts [1,2,3].shuffle!
This algorithm is O(size), while Mark's algorithm is O(size^2). On my computer, Mark's answer takes 400 seconds to shuffle an array of 1,000,000 elements on my machine, versus 0.5 seconds with my method.
code:
c = 0
items.each { |i|
puts i.to_s
# if c > 9 escape the each iteration early - and do not repeat
c++
}
I want to grab the first 10 items then leave the "each" loop.
What do I replace the commented line with? is there a better approach? something more Ruby idiomatic?
While the break solution works, I think a more functional approach really suits this problem. You want to take the first 10 elements and print them so try
items.take(10).each { |i| puts i.to_s }
There is no ++ operator in Ruby. It's also convention to use do and end for multi-line blocks. Modifying your solution yields:
c = 0
items.each do |i|
puts i.to_s
break if c > 9
c += 1
end
Or also:
items.each_with_index do |i, c|
puts i.to_s
break if c > 9
end
See each_with_index and also Programming Ruby Break, Redo, and Next.
Update: Chuck's answer with ranges is more Ruby-like, and nimrodm's answer using take is even better.
break works for escaping early from a loop, but it's more idiomatic just to do items[0..9].each {|i| puts i}. (And if all you're doing is literally printing the items with no changes at all, you can just do puts items[0..9].)
Another option would be
items.first(10).each do |i|
puts i.to_s
end
That reads a little more easily to me than breaking on an iterator, and first will return only as many items as available if there aren't enough.
Another variant:
puts items.first(10)
Note that this works fine with arrays of less than 10 items:
>> nums = (1..5).to_a
=> [1, 2, 3, 4, 5]
>> puts nums.first(10)
1
2
3
4
5
(One other note, a lot of people are offering some form of puts i.to_s, but in such a case, isn't .to_s redundant? puts will automatically call .to_s on a non-string to print it out, I thought. You would only need .to_s if you wanted to say puts 'A' + i.to_s or the like.)
Does this look like what you want?
10.times { |i|
puts items[i].to_s
}
items.each_with_index { |i, c| puts i and break if c <= 9 }
It was asked:
I want to grab the first 10 items then leave the "each" loop.
Use throw and catch to accomplish this, with few changes to the example:
catch(:done) do
c = 0
collected = []
items.each do |item|
collected << item
throw(:done, collected) if c == 9 # started at 0
c += 1
end
collected # if the list is less than 10 long, return what was collected
end
Simply throw the label :done with collected and the catch which is waiting for :done will return collected.
And to "ruby" this up a bit:
catch(:done) do
items.inject([]) do |collected, item|
throw(:done, collected) if collected.size == 10
collected << item # collected gets returned here and populates the first argument of this block
end
end
I do not know why some people refuse to use inject and use reduce instead (they are equivalent) when clearly the empty array given to inject([]) is being injected with items! Anyhow, the inject will return collected if there are less than 10 items.
Most answers are trying to answer what might be the intent of the question instead of what was asked and items.take(10) does make perfect sense in that case. But I can imagine wanting to grab the first items that fit within my $100 budget. Then you can simply:
catch(:done) do
items.inject({items: [], budget: 100}) do |ledger, item|
remainder = ledger[:budget] - item.price
if remainder < 0
throw(:done, ledger)
else
ledger.tap do |this|
this[:items] << item
this[:budget] = remainder
end # tap just returns what is being tapped into, in this case, ledger
end
end
end
I want to use a for-each and a counter:
i=0
for blah in blahs
puts i.to_s + " " + blah
i+=1
end
Is there a better way to do it?
Note: I don't know if blahs is an array or a hash, but having to do blahs[i] wouldn't make it much sexier. Also I'd like to know how to write i++ in Ruby.
Technically, Matt's and Squeegy's answer came in first, but I'm giving best answer to paradoja so spread around the points a bit on SO. Also his answer had the note about versions, which is still relevant (as long as my Ubuntu 8.04 is using Ruby 1.8.6).
Should've used puts "#{i} #{blah}" which is a lot more succinct.
As people have said, you can use
each_with_index
but if you want indices with an iterator different to "each" (for example, if you want to map with an index or something like that) you can concatenate enumerators with the each_with_index method, or simply use with_index:
blahs.each_with_index.map { |blah, index| something(blah, index)}
blahs.map.with_index { |blah, index| something(blah, index) }
This is something you can do from ruby 1.8.7 and 1.9.
[:a, :b, :c].each_with_index do |item, i|
puts "index: #{i}, item: #{item}"
end
You can't do this with for. I usually like the more declarative call to each personally anyway. Partly because its easy to transition to other forms when you hits the limit of the for syntax.
Yes, it's collection.each to do loops, and then each_with_index to get the index.
You probably ought to read a Ruby book because this is fundamental Ruby and if you don't know it, you're going to be in big trouble (try: http://poignantguide.net/ruby/).
Taken from the Ruby source code:
hash = Hash.new
%w(cat dog wombat).each_with_index {|item, index|
hash[item] = index
}
hash #=> {"cat"=>0, "wombat"=>2, "dog"=>1}
If you don't have the new version of each_with_index, you can use the zip method to pair indexes with elements:
blahs = %w{one two three four five}
puts (1..blahs.length).zip(blahs).map{|pair|'%s %s' % pair}
which produces:
1 one
2 two
3 three
4 four
5 five
As to your question about doing i++, well, you cannot do that in Ruby. The i += 1 statement you had is exactly how you're supposed to do it.
If you want to get index of ruby for each, then you can use
.each_with_index
Here is an example to show how .each_with_index works:
range = ('a'..'z').to_a
length = range.length - 1
range.each_with_index do |letter, index|
print letter + " "
if index == length
puts "You are at last item"
end
end
This will print:
a b c d e f g h i j k l m n o p q r s t u v w x y z You are at last item
The enumerating enumerable series is pretty nice.
If blahs is a class that mixes in Enumerable, you should be able to do this:
blahs.each_with_index do |blah, i|
puts("#{i} #{blah}")
end