I feel like this should be really easy and I'm just running into a wall.
I need to loop through an array until a condition is met. For example:
count = 0
array = ["","","test","demo"]
I want to loop through this array incrementing count by 1 until the first non-empty value is found. So I want the index value of "test", but when "test" is reached I want to stop the loop.
Also, as a side note, how can I just find the index of the first non-empty value in an array? I want to know both methods as they both have potential application.
You can have both :)
This finds you the index of the first non-empty string:
array = ["","","test","demo"]
array.index {|str| !str.empty?}
#=> 2
You can update count in the {|str| !str.empty?}-block if you like, because Array#index loops from start to end through the array.
FYI: The index method is an alias of find_index
better use :
first_non_empty_index = array.index{ |string| !string.empty? }
do/while:
begin
puts arr[count]
count+=1
end while(arr[count].empty?)
puts count #2
puts arr[count] #test
Related
So the goal here is to print the index of the element if the element is in the array or print -1 if the element is not in the array. I have to do this using loops. PLEASE HELP!
def element_index(element, my_array)
while my_array.map.include? element do
puts my_array.index(element)
break
end
until my_array.include? element do
puts -1
break
end
end
p element_index("c", ["a","b","c"])
If it's OK to use Array#index, then
def element_index(elem, collection)
collection.index(elem) || -1
end
Or if it's a homework that you should not use Array#index, or you want to do this on arbitrary collections, then
def element_index(elem, collection)
collection.each_with_index.reduce(-1) do |default, (curr, index)|
curr == elem ? (return index) : default
end
end
By the way, I always turn to Enumerable#reduce when I want to iterate over a collection (array, map, set, ...) to compute one value.
This is an easy way but maybe it doesn't meet the criteria for "using loops":
def element_index(x, arr)
arr.index(x) || -1
end
element_index("c", ["a","b","c"]) #=> 2
element_index("d", ["a","b","c"]) #=> -1
To explicitly use a loop:
def element_index(x, arr)
arr.each_index.find { |i| arr[i] == x } || -1
end
As pointed out in the comments, we could instead write
arr.each_index.find(->{-1}) { |i| arr[i] == x }
element_index("c", ["a","b","c"]) #=> 2
element_index("d", ["a","b","c"]) #=> -1
I know this is an assignment, but I'll first cover this as if it were real code because it's teaching you some not-so-great Ruby.
Ruby has a method for doing this, Array#index. It returns the index of the first matching element (there can be more than one), or nil.
p ["a","b","c"].index("c") # 2
p ["a","b","c"].index("d") # nil
Returning -1 is inadvisable. nil is a safer "this thing does not exist" value because its never a valid value, always false (-1 and 0 are true in Ruby), and does not compare equal to anything but itself. Returning -1 indicates whomever came up with this exercise is converting it from another language like C.
If you must, a simple wrapper will do.
def element_index(element, array)
idx = array.index(element)
if idx == nil
return -1
else
return idx
end
end
I have to do this using loops.
Ok, it's homework. Let's rewrite Array#index.
The basic idea is to loop through each element until you find one which matches. Iterating through each element of an array is done with Array#each, but you need each index, that's done with Array#each_index. The element can be then gotten with array[idx].
def index(array, want)
# Run the block for each index of the array.
# idx will be assigned the index: 0, 1, 2, ...
array.each_index { |idx|
# If it's the element we want, return the index immediately.
# No need to spend more time searching.
if array[idx] == want
return idx
end
}
# Otherwise return -1.
# nil is better, but the assignment wants -1.
return -1
end
# It's better to put the thing you're working on first,
# and the thing you're looking for second.
# Its like verb( subject, object ) or subject.verb(object) if this were a method.
p index(["a","b","c"], "c")
p index(["a","b","c"], "d")
Get used to using list.each { |thing| ... }, that's how you loop in Ruby, along with many other similar methods. There's little call for while and for loops in Ruby. Instead, you ask the object to loop and tell it what to do with each thing. It's very powerful.
I have to do this using loops.
You approach is very creative. You have re-created an if statement using a while loop:
while expression do
# ...
break
end
Is equivalent to:
if expression
# ...
end
With expression being something like array.include? element.
How can I do the opposite?
To invert a (boolean) expression, you just prepend !:
if !expression
# ...
end
Applied to your while-hack:
while !expression do
# ...
break
end
The whole method would look like this:
def element_index(element, my_array)
while my_array.include? element do
puts my_array.index(element)
break
end
while !my_array.include? element do
puts -1
break
end
end
element_index("c", ["a","b","c"])
# prints 2
element_index("d", ["a","b","c"])
# prints -1
As I said at the beginning, this approach is very "creative". You are probably supposed to find the index using a loop (see Schwern's answer) instead of calling the built-in index.
I'm currently working through the Coderbyte series to get better at Ruby programming. Maybe this is just a bug in their site (I don't know), but my code works for me everywhere else besides on Coderbyte.
The purpose of the method is to return the 2nd smallest and the 2nd largest elements in any inputted array.
Code:
def SecondGreatLow(arr)
arr=arr.sort!
output=[]
j=1
i=(arr.length-1)
secSmall=''
secLarge=''
while output.length < 1
unless arr.length <= 2
#Get second largest here
while (j<arr.length)
unless arr[j]==arr[j-1]
unless secSmall != ''
secSmall=arr[j]
output.push(secSmall)
end
end
j+=1
end
#get second smallest here
while i>0
unless arr[i-1] == arr[i]
unless secLarge != ''
secLarge=arr[i-1]
output.push(secLarge)
end
end
i-=1
end
end
end
# code goes here
return output
end
# keep this function call here
# to see how to enter arguments in Ruby scroll down
SecondGreatLow(STDIN.gets)
Output
Input: [1,2,3,100] => Output: [2,3] (correct)
Input: [1,42,42,180] => Output: [42,42] (correct)
Input: [4,90] => Output: [90,4] (correct)
The problem is that I'm awarded 0 points and it tells me that my output was incorrect for every test. Yet, when I actually put any inputs in, it gives me the output that I expect. Can someone please assist with what the problem might be? Thanks!
Update
Thanks to #pjs answer below, I realized this could be done in just a few lines:
def SecondGreatLow(arr)
arr=arr.sort!.uniq
return "#{arr[1]} #{arr[-2]}"
end
# keep this function call here
# to see how to enter arguments in Ruby scroll down
SecondGreatLow(STDIN.gets)
It's important to pay close attention to the problem's specification. Coderbyte says the output should be the values separated by a space, i.e., a string, not an array. Note that they even put quotes around their "Correct Sample Outputs".
Spec aside, you're doing way too much work to achieve this. Once the array is sorted, all you need is the second element, a space, and the second-to-last element. Hint: Ruby allows both positive and negative indices for arrays. Combine that with .to_s and string concatenation, and this should only take a couple of lines.
If you are worried about non-unique numbers for the max and min, you can trim the array down using .uniq after sorting.
You need to check condition for when array contains only two elements. Here is the complete code:
def SecondGreatLow(arr)
arr.uniq!
arr.sort!
if arr.length == 2
sec_lowest = arr[1]
sec_greatest = arr[0]
else
sec_lowest = arr[1]
sec_greatest = arr[-2]
end
return "#{sec_lowest} #{sec_greatest}"
end
I'm trying to create my own .sort method as an exercise in a ruby book, using recursion, and for some reason they haven't taught me the spaceship operator yet. My code works to get the smallest value - apple - and puts it in the sorted array, and it even repeats using the recursion, and resets the array to repeat the process to add the second smallest word. The problem is for some reason it removes the smallest word -apple- and I can't figure out why. I know where I think - in the else myArray.length == 1 statement when I pop the element off the array, but why is it removing from the sortedArray too?
sortedArray ends up with value apple, then when it does recursion it SHOULD be sortedArray = ['apple', 'banana' …] but it removes apple, then it removes banana etc… until I end up with sortedArray = ['quincy']
I have tried moving my arrays to multiple places, and I've tried adding to the sortedWords array in multiple places but it is always deleting or resetting the sortedWords array.
It looks like I'm really close since I've got the alphabetizing working. How do I get it to add all the items to the sortedWords array?
ArrayofWords = ['cat', 'dog', 'bat', 'elephant', 'apple', 'banana', 'quincy', 'boo']
# Why is it deleting, or replacing my sortedWords array? If you run this code you will notice that the sortedWords array
# is giving me the smallest word in the array, but then I add the recursive part, and somehow the previous smallestword
#gets deleted... but I have never in any part of my code say delete or replace the sorted array...
def sortTheArray myArray
unsortedWords = []
sortedWords = []
smallestValue = ''
while myArray.length != 0
if myArray.first < myArray.last
unsortedWords.push(myArray.last)
myArray.pop
elsif myArray.first > myArray.last
unsortedWords.push(myArray.first)
myArray.delete_at(0)
else myArray.length == 1
sortedWords.push(myArray.first)
myArray.pop # This is my problem area I think???
end # if else
#puts 'sorted words'
#puts sortedWords
#puts 'unsortedWords'
#puts unsortedWords
end # while
puts 'sorted words'
puts sortedWords
puts 'unsortedWords'
puts unsortedWords
myArray = unsortedWords
while myArray.length > 0
sortTheArray myArray
end #while
end # sortTheArray
sortTheArray ArrayofWords
most of those puts's are not necessary, I was just trying to figure out where the problem was.
You've got numerous problems with your code. For example, you seem to want to accumulate sorted words across invocations of this method, but you reinitialize sorted_words to [] at the start of the method block.
I would suggest first trying to express your recursive solution in English as simply as possible and then seek to implement it.
For example, the following is an approach which seems to be in line with what you are trying to do:
def sorted_array(array)
lowest_value prepended to the sorted_value of the array with the lowest_value removed
end
I'm sharing the above because it appears that you're new to Ruby and just implementing the above in an idiomatic fashion will be a good challenge.
taglist = [{:name=>"Daniel_Xu_Forever", :tag=>["helo", "world"]},
{:name=>"kcuf", :tag=>["hhe"]},
{:name=>"fine", :tag=>[]},
{:name=>"how hare you", :tag=>[]},
{:name=>"heki", :tag=>["1", "2", "3"]},
{:name=>"railsgirls", :tag=>[]},
{:name=>"_byoy", :tag=>[]},
{:name=>"ajha", :tag=>[]},
{:name=>"nimei", :tag=>[]}]
How to get specified name's tag from taglist
For example , I want to extract user "fine"'s tag?
Could this be achieved without do iterator?
This will return the contents of the :tag key for any users name which == 'fine'
taglist.select { |x| x[:name] == 'fine' }.map { |u| u[:tag] }
First you select out only the users you are interested with .select.
And then use .map to collect an array of only what you want.
In this case the end result will be: []
Is do really an iterator?
taglist.find{|tl| tl[:name] == 'fine'}[:tag]
Just to be silly how about:
eval taglist.to_s[/:name=>"fine", :tag=>(.*?)}/, 1]
#=> []
No, it cannot be done without a loop.
And even if you find a solution where your code avoids a loop, for sure the library function that you're calling will include a loop. Finding an element in an array requires a loop. Period.
For example, take this (contrived) example
pattern = "fine"
def pattern.===(h); self == h[:name]; end
taglist.grep(pattern)
which does not seem to use a loop, but calls grep which is implemented using a loop.
Or another, equally contrived, example
class Hash; def method_missing(sym); self[sym]; end; end
taglist.group_by(&:name)["fine"]
which again does seem to call group_by without a loop, but actually it does.
So the answer is, no.
So my first answer missed the no do rule.
Here is an answer that doesn't use a do block.
i=0
begin
if taglist[i][:name] == 'fine'
tag = taglist[i][:tag]
break
end
i+=1
end while i < taglist.length - 0
Technically I think this is still using a block. But probably satisfies the restriction.
So, here's the preface:
I'm quite a beginner in Ruby. I'm working on a anagram finding script (find all anagrams in a text file). The essence is: I create a dictionary where key is word code and value is an array with words that refer to this code. It is like that: {"abdeis"=>["abides", "biased"] ,"achr"=>["char"], ... }. In the end I just print out the values with length of >1. So good so far.
Here's the deal: I want to modify the output, so that such cases are omitted: ["Wood", "wood", "WooD"] - all are different in case only. But such cases should stay: ["Doom", "DooM", "mood"].
My current piece of code:
def print_anagram(anagram_dict)
anagram_list = anagram_dict.values
anagram_list.each { |i|
if i.length > 1
print i.join("\t")
print "\n"
else
next
end
}
end
anagram_dict is a dictionary mentioned above.
What checks should I make to throw these cases away? The things I think of seem way to bulky to me. Thanks in advance!
def print_anagram(anagram_dict)
anagram_list = anagram_dict.values
anagram_list.each do |i|
next if i.map(&:downcase).uniq.length == 1
if i.length > 1
print i.join("\t")
print "\n"
else
next
end
end
end
What this does:
make all Strings lowercase
get only unique elements from the array
if you only have one unique element, all elements are the same
map(&:downcase) is a shorter way of doing: map { |element| element.downcase }
Does this do what you want?
def print_anagram(anagram_dict)
anagram_list = anagram_dict.values.uniq{|word| word.downcase}
# rest of your code