This question already has answers here:
How do I count vowels?
(5 answers)
Test whether a variable equals either one of two values
(8 answers)
Closed 6 years ago.
I'm making a method in Ruby to count how many vowels are in a string. I figured it out, but at first I made the mistake of not repeating "variable==" after every '||'. The correct way seems inefficient. Is there a quicker way to do this and get the same result?
See example:
CORRECT:
def count_vowels(string)
vnum = 0
n = 0
while n < string.length
if string[n] == "a"||string[n] == "e"||string[n] =="i"||string[n] =="o"||string[n] =="u"||string[n] =="y"
vnum += 1
end
n += 1
end
return vnum
end
INCORRECT:
def count_vowels(string)
vnum = 0
n = 0
while n < string.length
if string[n] == "a"|| "e"||"i"||"o"||"u"||"y"
vnum += 1
end
n += 1
end
return vnum
end
Related
How can I get the number of iterations/steps that this method takes to find an answer?
def binary_search(array, n)
min = 0
max = (array.length) - 1
while min <= max
middle = (min + max) / 2
if array[middle] == n
return middle
elsif array[middle] > n
max = middle - 1
elsif array[middle] < n
min = middle + 1
end
end
"#{n} not found in this array"
end
One option to use instead of a counter is the .with_index keyword. To use this you'll need to use loop instead of while, but it should work the same. Here's a basic example with output.
arr = [1,2,3,4,5,6,7,8]
loop.with_index do |_, index| # The underscore is to ignore the first variable as it's not used
if (arr[index] % 2).zero?
puts "even: #{arr[index]}"
else
puts "odd: #{arr[index]}"
end
break if index.eql?(arr.length - 1)
end
=>
odd: 1
even: 2
odd: 3
even: 4
odd: 5
even: 6
odd: 7
even: 8
Just count the number of iterations.
Set a variable to 0 outside the loop
Add 1 to it inside the loop
When you return the index, return the count with it (return [middle, count]).
I assume the code to count numbers of interations required by binary_search is to be used for testing or optimization. If so, the method binary_search should be modified in such a way that to produce production code it is only necessary to remove (or comment out) lines of code, as opposed to modifying statements. Here is one way that might be done.
def binary_search(array, n)
# remove from production code lines marked -> #******
_bin_srch_iters = 0 #******
begin #******
min = 0
max = (array.length) - 1
loop do
_bin_srch_iters += 1 #******
middle = (min + max) / 2
break middle if array[middle] == n
break nil if min == max
if array[middle] > n
max = middle - 1
else # array[middle] < n
min = middle + 1
end
end
ensure #******
puts "binary_search reqd #{_bin_srch_iters} interations" #******
end #******
end
x = binary_search([1,3,6,7,9,11], 3)
# binary_search reqd 3 interations
#=> 1
binary_search([1,3,6,7,9,11], 5)
# binary_search reqd 3 interations
#=> nil
I am moving from Julia 0.7 to 1.0. It seems that Julia's rule for the scope of variables changed from 0.7 to 1.0. For example, I want to run a simple loop like this:
num = 0
for i = 1:5
if i == 3
num = num + 1
end
end
print(num)
In Julia 0.7 (and in most of other languages), we could expect num = 1 after the loop. However, it will incur UndefVarError: num not defined in Julia 1.0. I know that by using let I can do this
let
num = 0
for i = 1:5
if i == 3
num = num + 1
end
end
print(num)
end
It will print out 1. But I do want to get the num = 1 outside the loop and the let block. Some answers suggest putting all code in a let block, but it will incur other problems including UndefVarError while testing line-by-line. Is there any way instead of using let blocking? Thanks!
This is discussed here.
Add global as shown below inside the loop for the num variable.
num = 0
for i = 1:5
if i == 3
global num = num + 1
end
end
print(num)
Running in the Julia 1.0.0 REPL:
julia> num = 0
0
julia> for i = 1:5
if i == 3
global num = num + 1
end
end
julia> print(num)
1
Edit
For anyone coming here new to Julia, the excellent comment made in the answer below by vasja, should be noted:
Just remember that inside a function you won't use global, since the scope rules inside a function are as you would expect:
See that answer for a good example of using a function for the same code without the scoping problem.
Just remember that inside a function you won't use global, since the scope rules inside a function are as you would expect:
function testscope()
num = 0
for i = 1:5
if i == 3
num = num + 1
end
end
return num
end
julia> t = testscope()
1
The unexpected behaviour is only in REPL.
More on this here
I wrote code to find how many operations a number required under the Collatz Conjecture. However, my operations variable doesn't seem to be incrementing.
My code is:
puts "Please input a number"
number = gets.chomp
number = number.to_i
operations = 0
modulo = number % 2
while number =! 1
if modulo == 0
number = number / 2
operations = operations + 1
elsif modulo =! 0 && number =! 1
number = number * 3
number = number += 1
operations = operations + 2
else
puts "Uh oh, something went wrong."
end
end
puts "It took #{operations} operations!"
I am running this code on https://www.repl.it.
First of all, it's elsif; not elseif (I edited that in your question). And unequal sign is !=; not =!. But that has a somewhat different meaning. (i.e.: number =! 1 means number = !1)
In the 12th line, what is number = number += 1? I think you meant number += 1 or number = number + 1.
Now, the code works. :)
Here's the final version.
puts "Please input a number"
number = gets.chomp
number = number.to_i
operations = 0
modulo = number % 2
while number != 1
if modulo == 0
number = number / 2
operations = operations + 1
elsif modulo != 0 && number != 1
number = number * 3
number = number + 1
operations = operations + 2
else
puts "Uh oh, something went wrong."
end
end
puts "It took #{operations} operations!"
Usage:
Please input a number
256
It took 8 operations!
An optimal solution using functions:
def collatz(n)
if n % 2 == 0
return n / 2
else
return 3*n + 1
end
end
def chainLength(num)
count = 1
while num > 1
count += 1
num = collatz(num)
end
return count
end
puts "Please input a number"
number = gets.chomp
number = number.to_i
operations = chainLength(number)
puts "It took #{operations} operations!"
If you need more performance, read about dynamic programming and memoization techniques.
I am currently doing Project Euler problem 1. I have no idea why these two loops are not the same.
total = 0
for i in 0..1000
if (i % 3 == 0 || i % 5 == 0)
total += i
end
end
and
total = 0
(0...1000).each do |i|
total += i if (i % 3 == 0 || i % 5 == 0)
end
puts total
When you use three dots in range (0...1000), the end value is not part of the range - it is equivalent to (0..999)
So, in first case 1000 is part of the loop, but in second case it is not.
I am taking a lesson on codecademy in which I am currently stuck do not know how to proceed - it is regarding return values.
The instructions are:
Write a method that takes an integer as an argument, and returns that integer times ten. Call times_ten in your code after you define it and print out its return value.
What is given in the script is:
def times_ten(integer)
# your code here
end
# call times_ten here
This is the example it gives but I am having a hard time understanding:
def first_squares(number_of_squares)
squares = []
idx = 0
while idx < number_of_squares
squares.push(idx * idx)
idx = idx + 1
end
return squares
end
puts("How many square numbers do you want?")
number_of_squares = gets.to_i
squares = first_squares(number_of_squares)
idx = 0
while idx < squares.length
puts(squares[idx])
idx = idx + 1
end
Thanks for your help
It should be:
def ten_times(n)
n*10 # you don't have to use 'return' explicitly
end
ten_times(n) -- but put in an actual integer instead of n (or maybe you have to puts or print it, depending on what they want)
Your example is not really related to your outcome.
The example script should be like this:
def ten_times(integer)
# integer * 10 #for implicit return
return integer * 10 #for explicit return
end
print ten_times(any number you want goes in here)
You can run the following code at www.rubyplus.biz:
Implicit return:
def times_ten(integer)
integer * 10
end
p times_ten(1)
Explicit return:
def times_ten(integer)
return integer * 10
end
p times_ten(2)