I have an array s_ary that contains all the lines of a text document. The last line may be something like return some_string.
When the last line starts with return, I want the line to become sv1 = some_string.
My original code does exactly what I need:
if s_ary.last =~ /^[\s]*return/
t = s_ary.last.gsub(/^[\s]*return/, 'sv1 = ')
s_ary.pop
s_ary << t
else
s_ary << 'sv1 = last'
end
I tried to improve it with:
if s_ary.last =~ /^[\s]*return/
s_ary.map! {|x| (x =~/return/) ? x.gsub(/return/, 'sv1 = ') : x }
else
s_ary << 'sv1 = last'
end
But this version will change all the lines that start with return when the last line starts with return. I could still use this last version as this is not supposed to happen in my application, but I have the feeling that there must a better, more compact way of accomplishing this. Somehow, I can't find it.
Thanks for any suggestion.
EDIT: The original code that does exactly what I need is actually (in agreement with the second paragraph of this post):
if s_ary.last =~ /^[\s]*return/
t = s_ary.last.gsub(/^[\s]*return/, 'sv1 = ')
s_ary.pop
s_ary << t
end
I can't believe I wrote something so confusing. My apologies.
Why not just:
s_ary[-1].gsub!(/^return /, 'sv1 = ')
EDIT
Looking closer at your code, maybe what you want is kind of:
unless s_ary[-1].gsub!(/^return /, 'sv1 = ')
s_ary << 'sv1 = last'
end
The first snippet will only change the last line if it starts with return, doing nothing otherwise. The second snippet will change the last line if it starts with return, and will append sv1 = last otherwise.
Not sure which one you need.
arr = ["Foo is good","Bar is bad","return v1 + v2"]
arr[-1] = arr.last.sub("return","sv = ") if arr.last.start_with? "return"
p arr #=> "Foo is good", "Bar is bad", "sv = v1 + v2"]
Related
I'm trying to reverse a string using the code:
puts("Hi now it's going to be done!")
string = gets.chomp.to_s
i = string.length
while i >= 0
puts(string[i])
i = i - 1
end
It prints the string in backward order, but each word is on a single line. How can I keep all of them on a single line?
puts adds a newline to the end of the output if one isn't already present.
print does not. So do this:
while i >=0
print string[i]
i=i-1
end
puts
The final puts is because you want any further printing to be on a new line.
Try this:
"Hi now it's going to be done!".chars.inject([]) { |s, c| s.unshift(c) }.join
Or This is a little easier to follow:
string = 'Hi now it's going to be done!'
string.reverse!
I have a coding problem I solved and want to refactor. I know there has to be a cleaner way of doing what I did.
The goal is to write a method that takes a string of "!" and "?" and reduces the string by eliminating all odd groupings of each symbol.
Example - a string "????!!!" would have an odd grouping of "!!!" because there are three in a row. These would be deleted from the string.
If there is only one "!" or "?" its left because it is not in a group.
Ex -
remove("!????!!!?") answer == "!"
# => ("!????!!!?" --> "!?????" --> "!")
In the first string, the only odd grouping is "!!!", once removed, it leaves a new string with an odd grouping "?????". You remove the next odd grouping so you're left with "!". This fits the desired output.
Another example
remove("!???!!") == ""
# => ("!???!!" --> "!!!" --> "")
Current code:
def remove(s)
arr = [s]
i = 0
until i == arr[0].length
s = s.chars.chunk{|c|c}.map{ |n,a| a.join }.select{|x| x if x.length.even? || x.length <= 1}.join
arr << s
i += 1
end
return arr[-1]
end
My code solves this problem and all test cases. I have a suspicion that my until loop can be removed/refactored so that I could solve this problem in one line and have spent hours trying to figure it out with no luck.
Suppose
str = "???!!!???!"
If we first remove the two groups "???" we are left with "!!!!", which cannot be reduced further.
If we first remove the group "!!!" we are left with "??????!", which cannot be reduced further.
If we are permitted to remove all odd groups of either character without reference to the effect that either has on the other, we obtain !, which cannot be reduced further.
It's not clear what rule is to be used. Here are three possibilities and code to implement each.
I will use the following two regular expressions, and in the first two cases a helper method.
Rq = /
(?<!\?) # do not match a question mark, negative lookbehind
\? # match a question mark
(\?{2})+ # match two question marks one or more times
(?!\?) # do not match a question mark, negative lookahead
/x # free-spacing regex definition mode
which is commonly written /(?<!\?)\?(\?{2})+(?!\?)/.
Similarly,
Rx = /(?<!!)!(!{2})+(?!!)/
def sequential(str, first_regex, second_regex)
s = str.dup
loop do
size = s.size
s = s.gsub(first_regex,'').gsub(second_regex,'')
return s if s.size == size
end
end
I apply each of the three methods below to two example strings:
str1 = "???!!!???!"
str2 = 50.times.map { ['?', '!'].sample }.join
#=> "?!!!?!!!?!??????!!!?!!??!!???!?!????!?!!!?!?!???!?"
Replace all odd groups of "?" then odd groups of "!" then repeat until no further removals are possible
def question_before_exclamation(str)
sequential(str, Rq, Rx)
end
question_before_exclamation str1 #=> "!!!!"
question_before_exclamation str2 #=> "??!??!?!!?!?!!?"
Replace all odd groups of "!" then odd groups of "?" then repeat until no further removals are possible
def exclamation_before_question(str)
sequential(str, Rx, Rq)
end
exclamation_before_question str1 #=> "??????!"
exclamation_before_question str2 #=> "??!????!!?!?!!?!?!!?"
Replace all odd groups of both "?" and "!" then repeat until no further removals are possible
Rqx = /#{Rq}|#{Rx}/
#=> /(?-mix:(?<!\?)\?(\?{2})+(?!\?))|(?-mix:(?<!!)!(!{2})+(?!!))/
def question_and_explanation(str)
s = str.dup
loop do
size = s.size
s = s.gsub(Rqx,'')
return s if s.size == size
end
end
question_and_explanation str1 #=> "!"
question_and_explanation str2 #=> "??!?!!?!?!!?!?!!?"
I don't know the exact Ruby syntax for this, but you could simplify your solution by using regular expressions:
Gather all matches of consecutive characters
if all matches are of even length or 1 exit
Test if matches are an odd length
if an odd length, replace with the empty string
else do nothing
Goto step 1
A solution in Perl would be:
#!perl
use strict;
use warnings;
use feature qw(say);
my $string = '!????!!!?';
sub reduce {
my ($s) = #_;
while ( my #matches = $s =~ m/((.)\2+)/g ) {
last if ! grep { length($_) > 1 && length($_) % 2 == 1 } #matches;
foreach my $match ( #matches ) {
$s =~ s/\Q$match// if length($match) > 1 && length($match) % 2 == 1;
}
}
return $s;
}
say reduce($string);
I could be wrong (this is ruby, after all) but I don't think you'll find a one-liner for this because ruby's utility functions generally aren't recursive. But you can use regex to simplify your logic, at the very least:
def remove(s)
while s =~ /(?<!\!)\!([\!]{2})+(?!\!)/ || s =~ /(?<!\?)\?([\?]{2})+(?!\?)/
s.gsub! /(?<!\!)\!([\!]{2})+(?!\!)/, "" # remove odd !
s.gsub! /(?<!\?)\?([\?]{2})+(?!\?)/, "" # remove odd ?
end
return s
end
To make the regex less mind-boggling, it helps to look at them with 'a' instead of '?' and '!':
/(?<!a)a([a]{2})+(?!a)/ #regex for 'a'
(?<!a) #negative lookbehind: the match cannot start with an 'a'
a([a]{2})+ #the match should be an 'a' followed by 1 or more pairs
(?!a) #negative lookahead: the match cannot end with an 'a'
It should be simple enough with a regular expression replacement
def remove(string)
begin
original = string
string.gsub!(/(\!{3,})|(\?{3,})/) { |s| s.length.even? ? s : '' }
end until original == string
string
end
puts remove("!????!!!?").inspect # answer == "!"
puts remove("!???!!").inspect # answer == ""
puts remove("!????!!").inspect # answer == "!????!!"
I'm attempting to write a function that takes a string and returns it with all vowels removed. Below is my code.
def vowel(str)
result = ""
new = str.split(" ")
i = 0
while i < new.length
if new[i] == "a"
i = i + 1
elsif new[i] != "a"
result = new[i] + result
end
i = i + 1
end
return result
end
When I run the code, it returns the exact string that I entered for (str). For example, if I enter "apple", it returns "apple".
This was my original code. It had the same result.
def vowel(str)
result = ""
new = str.split(" ")
i = 0
while i < new.length
if new[i] != "a"
result = new[i] + result
end
i = i + 1
end
return result
end
I need to know what I am doing wrong using this methodology. What am I doing wrong?
Finding the bug
Let's see what's wrong with your original code by executing your method's code in IRB:
$ irb
irb(main):001:0> str = "apple"
#=> "apple"
irb(main):002:0> new = str.split(" ")
#=> ["apple"]
Bingo! ["apple"] is not the expected result. What does the documentation for String#split say?
split(pattern=$;, [limit]) → anArray
Divides str into substrings based on a delimiter, returning an array of these substrings.
If pattern is a String, then its contents are used as the delimiter when splitting str. If pattern is a single space, str is split on whitespace, with leading whitespace and runs of contiguous whitespace characters ignored.
Our pattern is a single space, so split returns an array of words. This is definitely not what we want. To get the desired result, i.e. an array of characters, we could pass an empty string as the pattern:
irb(main):003:0> new = str.split("")
#=> ["a", "p", "p", "l", "e"]
"split on empty string" feels a bit hacky and indeed there's another method that does exactly what we want: String#chars
chars → an_array
Returns an array of characters in str. This is a shorthand for str.each_char.to_a.
Let's give it a try:
irb(main):004:0> new = str.chars
#=> ["a", "p", "p", "l", "e"]
Perfect, just as advertised.
Another bug
With the new method in place, your code still doesn't return the expected result (I'm going to omit the IRB prompt from now on):
vowel("apple") #=> "elpp"
This is because
result = new[i] + result
prepends the character to the result string. To append it, we have to write
result = result + new[i]
Or even better, use the append method String#<<:
result << new[i]
Let's try it:
def vowel(str)
result = ""
new = str.chars
i = 0
while i < new.length
if new[i] != "a"
result << new[i]
end
i = i + 1
end
return result
end
vowel("apple") #=> "pple"
That looks good, "a" has been removed ("e" is still there, because you only check for "a").
Now for some refactoring.
Removing the explicit loop counter
Instead of a while loop with an explicit loop counter, it's more idiomatic to use something like Integer#times:
new.length.times do |i|
# ...
end
or Range#each:
(0...new.length).each do |i|
# ...
end
or Array#each_index:
new.each_index do |i|
# ...
end
Let's apply the latter:
def vowel(str)
result = ""
new = str.chars
new.each_index do |i|
if new[i] != "a"
result << new[i]
end
end
return result
end
Much better. We don't have to worry about initializing the loop counter (i = 0) or incrementing it (i = i + 1) any more.
Avoiding character indices
Instead of iterating over the character indices via each_index:
new.each_index do |i|
if new[i] != "a"
result << new[i]
end
end
we can iterate over the characters themselves using Array#each:
new.each do |char|
if char != "a"
result << char
end
end
Removing the character array
We don't even have to create the new character array. Remember the documentation for chars?
This is a shorthand for str.each_char.to_a.
String#each_char passes each character to the given block:
def vowel(str)
result = ""
str.each_char do |char|
if char != "a"
result << char
end
end
return result
end
The return keyword is optional. We could just write result instead of return result, because a method's return value is the last expression that was evaluated.
Removing the explicit string
Ruby even allows you to pass an object into the loop using Enumerator#with_object, thus eliminating the explicit result string:
def vowel(str)
str.each_char.with_object("") do |char, result|
if char != "a"
result << char
end
end
end
with_object passes "" into the block as result and returns it (after the characters have been appended within the block). It is also the last expression in the method, i.e. its return value.
You could also use if as a modifier, i.e.:
result << char if char != "a"
Alternatives
There are many different ways to remove characters from a string.
Another approach is to filter out the vowel characters using Enumerable#reject (it returns a new array containing the remaining characters) and then join the characters (see Nathan's answer for a version to remove all vowels):
def vowel(str)
str.each_char.reject { |char| char == "a" }.join
end
For basic operations like string manipulation however, Ruby usually already provides a method. Check out the other answers for built-in alternatives:
str.delete('aeiouAEIOU') as shown in Gagan Gami's answer
str.tr('aeiouAEIOU', '') as shown in Cary Swoveland's answer
str.gsub(/[aeiou]/i, '') as shown in Avinash Raj's answer
Naming things
Cary Swoveland pointed out that vowel is not the best name for your method. Choose the names for your methods, variables and classes carefully. It's desirable to have a short and succinct method name, but it should also communicate its intent.
vowel(str) obviously has something to do with vowels, but it's not clear what it is. Does it return a vowel or all vowels from str? Does it check whether str is a vowel or contains a vowel?
remove_vowels or delete_vowels would probably be a better choice.
Same for variables: new is an array of characters. Why not call it characters (or chars if space is an issue)?
Bottom line: read the fine manual and get to know your tools. Most of the time, an IRB session is all you need to debug your code.
I should use regex.
str.gsub(/[aeiou]/i, "")
> string= "This Is my sAmple tExt to removE vowels"
#=> "This Is my sAmple tExt to removE vowels"
> string.delete 'aeiouAEIOU'
#=> "Ths s my smpl txt t rmv vwls"
You can create a method like this:
def remove_vowel(str)
result = str.delete 'aeiouAEIOU'
return result
end
remove_vowel("Hello World, This is my sample text")
# output : "Hll Wrld, Ths s my smpl txt"
Live Demo
Assuming you're trying to learn about the basics of programming, rather than finding the quickest one-liner to do this (which would be to use a regular expression as Avinash has said), you have a number of problems with your code you need to change.
new = str.split(" ")
This line is likely the culprit, because it splits the string based on spaces. So your input string would have to be "a p p l e" to have the effect you're looking for.
new = str.split("")
You should also remove the duplicate i = i+1 once you've changed that.
As others have already identified the problems with the OP's code, I will merely suggest an alternative; namely, you could use String#tr:
"Now is the time for all good people...".tr('aeiouAEIOU', '')
#=> "Nw s th tm fr ll gd ppl..."
If regex is not allowed, you can do it this way:
def remove_vowels(string)
string.split("").delete_if { |letter| %w[a e i o u].include? letter }.join
end
This code capitalizes the first letter of each word in a string.
Eg "this is a sentence" becomes "This Is A Sentence".
def capitalize_words(string)
words = string.split(" ")
idx = 0
while idx < words.length
word = words[idx]
word[0] = word[0].upcase
words[idx] = word #this line of code can be made redundant, but why?
idx += 1
end
return words.join(" ")
end
In the while statement, I don't understand why the third line is unnecessary. The second line sets the first letter of a word to capital:
word[0] = word[0].upcase
how does the while statement know to refer back to the previous line
word = words[idx]
to put the new capitalised-letter word back into the words array? I thought that when codes are executed, it always works in a forward fashion, please let me know if this understanding is incorrect.
It's because word variable holds reference for object - the same object that is in words array. So if you modify this object, the object in array is modified also, because it's the same.
BTW what you're trying to do here can be done much easier:
string.split(' ').map(&:capitalize).join(' ')
As Stefan suggested: Keep in mind that capitalize not only converts first character to uppercase, but also converts all remaining chars to lowercase. If this is not what you want, you can also do:
string.split(' ').map { |word| word.slice(0, 1).upcase + word.slice(1..-1) }
or use Stefan's solution with regexp:
string.gsub(/\b\w/) { |ch| ch.upcase }
Keep in mind that \b in regexp will 'split' your word not only by spaces, but by any word boudary.
If you are only using ruby then use answer as per #Marek's answer :
string.split(' ').map(&:capitalize).join(' ')
and If you are using Ruby with Rails then use this:
"this is a sentence".titleize
I'd like to search through a txt file for a particular word. If I find that word, I'd like to retrieve the word that immediately follows it in the file. If my text file contained:
"My name is Jay and I want to go to the store"
I'd be searching for the word "want", and would want to add the word "to" to my array. I'll be looking through a very big text file, so any notes on performance would be great too.
The most literal way to read that might look like this:
a = []
str = "My name is Jack and I want to go to the store"
str.scan(/\w+/).each_cons(2) {|x, y| a << y if x == 'to'}
a
#=> ["go", "the"]
To read the file into a string use File.read.
This is one way:
Code
def find_next(fname, word)
enum = IO.foreach(fname)
loop do
e = (enum.next).scan(/\w+/)
ndx = e.index(word)
if ndx
return e[ndx+1] if ndx < e.size-1
loop do
e = enum.next
break if e =~ /\w+/
end
return e[/\w+/]
end
end
nil
end
Example
text =<<_
It was the best of times, it was the worst of times,
it was the age of wisdom, it was the age of foolishness,
. . . . .
it was the epoch of belief, it was the epoch of incredulity,
it was the season of light, it was the season of darkness,
it was the spring of hope, it was the winter of despair…
_
FName = "two_cities"
File.write(FName, text)
find_next(FName, "worst")
# of
find_next(FName, "wisdom")
# it
find_next(FName, "foolishness")
# it
find_next(FName, "dispair")
#=> nil
find_next(FName, "magpie")
#=> nil
Shorter, but less efficient, and problematic with large files:
File.read(FName)[/(?<=\b#{word}\b)\W+(\w+)/,1]
This is probably not the fastest way to do it, but something along these lines should work:
filename = "/path/to/filename"
target_word = "weasel"
next_word = ""
File.open(filename).each_line do |line|
line.split.each_with_index do |word, index|
if word == target_word
next_word = line.split[index + 1]
end
end
end
Given a File, String, or StringIO stored in file:
pattern, match = 'want', nil
catch :found do
file.each_line do |line|
line.split.each_cons(2) do |words|
if words[0] == pattern
match = words.pop
throw :found
end
end
end
end
match
#=> "to"
Note that this answer will find at most one match per file for speed, and linewise operation will save memory. If you want to find multiple matches per file, or find matches across line breaks, then this other answer is probably the way to go. YMMV.
This is the fastest I could come up with, assuming your file is loaded in a string:
word = 'want'
array = []
string.scan(/\b#{word}\b\s(\w+)/) do
array << $1
end
This will find ALL words that follow your particular word. So for example:
word = 'want'
string = 'My name is Jay and I want to go and I want a candy'
array = []
string.scan(/\b#{word}\b\s(\w+)/) do
array << $1
end
p array #=> ["to", "a"]
Testing this on my machine where I duplicated this string 500,000 times, I was able to reach 0.6 seconds execution time. I've also tried other approaches like splitting the string etc. but this was the fastest solution:
require 'benchmark'
Benchmark.bm do |bm|
bm.report do
word = 'want'
string = 'My name is Jay and I want to go and I want a candy' * 500_000
array = []
string.scan(/\b#{word}\b\s(\w+)/) do
array << $1
end
end
end