I've found this similar two questions to the one I'm about to ask:
Split array up into n-groups of m size?
and
Need to split arrays to sub arrays of specified size in Ruby
This splits array into three arrays with each array having three elements :
a.each_slice(3) do |x,y,z|
p [x,y,z]
end
So if I do this (my array size is 1000) :
a.each_slice(200) do |a,b,c,d,e|
p "#{a} #{b} #{c} #{d} #{e}"
end
This should split my array into 5 arrays each having 200 members? But it doesn't?
What I actually need to do is to put 200 random elements into 5 arrays, am I on the right track here, how can I do this?
Enumerable#each_slice
If you provide a single argument to the block of each_slice then it will fill that argument with an array of values less than or equal to the given argument. On the last iteration if there are less than n values left then the array size will be whatever is left.
If you provide multiple arguments to the block of each_slice then it will fill those values with the values from the source array. If the slice size is greater than the number of arguments given then some values will be ignored. If it is less than the number of arguments than the excess arguments will be nil.
a = (1..9).to_a
a.each_slice(3) {|b| puts b.inspect }
[1,2,3]
[4,5,6]
[7,8,9]
a.each_slice(4) {|b| puts b.inspect }
[1,2,3,4]
[5,6,7,8]
[9]
a.each_slice(3) {|b,c,d| puts (b + c + d)}
6 # 1 + 2 + 3
15 # 4 + 5 + 6
24 # 7 + 8 + 9
a.each_slice(3) {|b,c| puts (b + c)}
3 # 1 + 2, skips 3
9 # 4 + 5, skips 6
15 # 7 + 8, skips 9
a.each_slice(2) {|b,c| puts c.inspect}
2
4
6
8
nil
a.each_slice(3) {|b,c,d,e| puts e.inspect}
nil
nil
nil
irb(main):001:0> a= (1..10).to_a
=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
irb(main):002:0> a.sample(3)
=> [5, 10, 1]
irb(main):003:0> (1..3).map{ a.sample(3) }
=> [[6, 2, 5], [8, 7, 3], [4, 5, 7]]
irb(main):004:0>
Actually you will return a string with the five elements inserted in it.
You can try something:
a1 = [], a2 = [], a3 = [], a4 = [], a5 = []
a.each_slice(5) do |a,b,c,d,e|
a1 << a
a2 << b
a3 << c
a4 << d
a5 << e
end
You will end up with five arrays containing 200 elements each.
I used the simplest possible syntax to make it clear, you can
make it much more condensed.
If you want to assign that result to 5 different arrays, you could use the splat operator,like this:
a,b,c,d,e = *(1..1000).each_slice(200).to_a
Related
x = {1, 2, 3}
y = {4, 5, 6}
z = x + y
I have two tables x and y and just want to create a third one which is just the sum of elements in them. I use the above LUA code in an effort but this gives error input:3: attempt to perform arithmetic on a table value (global 'x')...
Like, I want the result z = {5, 7, 9}
Please suggest functions that will be helpful, or please help me form such a function in LUA.
Thanks
Yes, iterate and check with table.concat()
do...
x = {1, 2, 3}
y = {4, 5, 6}
z = {}
-- First check same table length and if so then add sums to z table
if #x==#y then
for i=1,#x do
z[i]=x[i]+y[i]
end
end
print(table.concat(z,' '))
-- puts out: 5 7 9
...end
You cannot add tables in Lua unless you implement the __add metamethod.
For an element wise sum of two sequences simply do this:
function sumElements(t1,t2)
local result = {}
for i = 1, math.min(#t1, #t2) do
result[i] = t1[i] + t2[i]
end
return result
end
of course you should verify your inputs and think about how you want to deal with mismatching table sizes. Let's say t1 has 3 elements and t2 has 5, will you just have 3 result values or will you add 0 to the remaining 2?
I'm using Ruby 2.4 and Rails 5. I have an array of indexes within a line
[5, 8, 10]
How do I take the above array, and a string, and form anotehr array of strings that are split by the above indexes? FOr instance, if the string is
abcdefghijklmn
and split it based ont eh above indexes, I would have an array with the following strings
abcde
fgh
ij
klmn
Try this
str = "abcdefghijklmn"
positions = [5, 8, 10]
parts = [0, *positions, str.size].each_cons(2).map { |a, b| str[a...b] }
# => ["abcde", "fgh", "ij", "klmn"]
Or,
If the positions are constant and known ahead of runtime (for example if they were the format for a phone number or credit card) just use a regexp
str.match(/(.....)(...)(..)(.*)/).captures
# => ["abcde", "fgh", "ij", "klmn"]
This will get the Job done
str = "abcdefghijklmn"
arr_1 = [5, 8, 10]
arr_2, prev = [], 0
(arr_1.length + 1).times do |x|
if arr_1[x] == nil then arr_1[x] = str.size end
arr_2 << str[prev..arr_1[x] -1]
prev = arr_1[x]
end
p arr_2
---------------------------------------
Program Run Output
["abcde", "fgh", "ij", "klmn"]
---------------------------------------
I hope this Helps
This question already has answers here:
How to find an item in array which has the most occurrences [duplicate]
(11 answers)
Closed 6 years ago.
I'm trying to figure out how to find a count of the most frequent element in an array of integers. I can think of a few methods that might be helpful but when I get to writing an expression inside the block I get complete lost on how to compare an element with the next and previous element. Any ideas? All help is really really appreciated!!!
An easy was is to determine all the unique values, convert each to its count in the array, then determine the largest count.
def max_count(arr)
arr.uniq.map { |n| arr.count(n) }.max
end
For example:
arr = [1,2,4,3,2,6,3,4,2]
max_count(arr)
#=> 3
There are three steps:
a = arr.uniq
#=> [1, 2, 4, 3, 6]
b = a.map { |n| arr.count(n) }
#=> [1, 3, 2, 2, 1]
b.max
#=> 3
A somewhat more efficient way (because the elements of arr are enumerated only once) is to use a counting hash:
def max_count(arr)
arr.each_with_object(Hash.new(0)) { |n,h| h[n] += 1 }.values.max
end
max_count(arr)
#=> 3
We have:
a = arr.each_with_object(Hash.new(0)) { |n,h| h[n] += 1 }
#=> {1=>1, 2=>3, 4=>2, 3=>2, 6=>1}
b = a.values
#=> [1, 3, 2, 2, 1]
b.max
#=> 3
See Hash::new for an explanation of Hash.new(0). Briefly, if h = Hash.new(0) and h does not have a key k, h[k] will return the default value, which here is zero. h[k] += 1 expands to h[k] = h[k] + 1, so if h does not have a key k, this becomes h[k] = 0 + 1. On the other hand, if, say, h[k] => 2, then h[k] = h[k] + 1 #=> h[k] = 3 + 1.
please somebody if you can explain the 3 rd line of this code. This method is to subtract an array of numbers starting from 2nd no. subtracted from 1st and the 3rd no. subtracted from the resultant and so on...
def subtract(*numbers)
sum = numbers.shift
numbers.inject(sum) { |sum, number| sum - number }
end
http://ruby-doc.org/core-1.9.3/Enumerable.html#method-i-inject
What inject does is takes some initial value (sum) and applies some operation to it for each element of your enumerable. So here, we take 1, and subtract the first element from 1, and the subtract the second element from that result, etc...
So below we have 1 - 2 = -1, then -1 - 3 = -4.
>> numbers=[1,2,3]
=> [1, 2, 3]
>> sum = numbers.shift
=> 1
>> numbers
=> [2, 3]
>> numbers.inject(sum) { |sum, number| sum - number }
=> -4
how do we count idential values on after appending value in to array
such that
a=[]
a<<1 count of 1 is 1
a<<1 count of 1 is 2
thanks
You could do:
a.select{|v| v == 1}.size
It's only one solution
Someone will probably come up with a more specialized solution, but I would just reduce it
counts = [1,3,3].reduce({}) do |acc,n|
acc.tap do |a|
a[n] ||= 0
a[n] += 1
end
end
counts.each {|k,v| puts "#{k} was found #{v} times"}
(note that tap is ruby 1.9, and is backported in activesupport)
output of that will be
1 was found 1 times
3 was found 2 times
a = [1,2,3,4,5,1,2,2,3,4]
=> [1, 2, 3, 4, 5, 1, 2, 2, 3, 4]
a.uniq.each do |i|
?> puts i.to_s + ' has appeared ' + a.count(i).to_s + ' times'
end
1 has appeared 2 times
2 has appeared 3 times
3 has appeared 2 times
4 has appeared 2 times
5 has appeared 1 times
=> [1, 2, 3, 4, 5]