I am trying to access a value from an array of hash. An example array looks like this:
family = [
[
{ "Homer" => 1, "Marge" => 2, "Lisa" => 3, "Maggie" => 4,
"Abe" => 5, "Santa's Little Helper" => 6
}
],
[
{ "Homer" => 2, "Marge" => 4, "Lisa" => 6,
"Maggie" => 8, "Abe" => 10, "Santa's Little Helper" => 12
}
]
]
If I try to access the hash value for key "Homer" in array indexed 0 (family[0]) using the statement below and hoping to get the value 1:
family[0]["Homer"]
I get an error which says
"test.rb:4:in `[]': can't convert String into Integer (TypeError)"
Any suggestions on how one might be able to access a hash value in such an array, in a simple statement?
You should try family[0][0]["Homer"].
In your case family[0] gives you :
[{ "Homer" => 1, "Marge" => 2, "Lisa" => 3, "Maggie" => 4, "Abe" => 5,"Santa Little Helper" => 6}]
which is an array. The hash you want is inside it and can be got with family[0][0] :
{ "Homer" => 1, "Marge" => 2, "Lisa" => 3, "Maggie" => 4, "Abe" => 5,"Santa Little Helper" => 6}
So you can now use family[0][0]["Homer"] which will give you the value 1.
The array indices are always numeric values. If you get a can't convert String into Integer (TypeError) error message an exception is thrown because you are trying to access array element using a string that can't be converted to an integer.
You don't actually have an array of hashes. You have an array of arrays of hashes.
Your error occurs because you dereference your structure with [0] which gives you the first array of hashes, now you try to access the key 'homer' which doesnt exist because arrays are keyed by integers.
Here is an example of how you could see all of the values, see if you can get 'homer' on your own:
family.each do |a| # this is an array of arrays of hashes
a.each do |h| # this is an array of hashes
h.each do |k,v| # this is a hash
puts "#{k} => #{v}"
end
end
end
Output:
Homer => 1
Marge => 2
Lisa => 3
Maggie => 4
Abe => 5
Santa's Little Helper => 6
Homer => 2
Marge => 4
Lisa => 6
Maggie => 8
Abe => 10
Santa's Little Helper => 12
#Arup Rakshit is absolutely correct about how to get your value. But you should also know that you don't have an array of hashes, you have an array of arrays, and those sub-arrays contain hashes. Based on your title I'm concluding that you probably want a structure more like
family = [
{ "Homer" => 1, "Marge" => 2, "Lisa" => 3, "Maggie" => 4,
"Abe" => 5, "Santa's Little Helper" => 6
},
{ "Homer" => 2, "Marge" => 4, "Lisa" => 6,
"Maggie" => 8, "Abe" => 10, "Santa's Little Helper" => 12
}
]
Related
From a hash like the below one, need to extract the unique values per key
array_of_hashes = [ {'a' => 1, 'b' => 2 , 'c' => 3} ,
{'a' => 4, 'b' => 5 , 'c' => 3},
{'a' => 6, 'b' => 5 , 'c' => 3} ]
Need to extract the unique values per key in an array
unique values for 'a' should give
[1,4,6]
unique values for 'b' should give
[2,5]
unique values for 'c' should give
[3]
Thoughts ?
Use Array#uniq:
array_of_hashes = [ {'a' => 1, 'b' => 2 , 'c' => 3} ,
{'a' => 4, 'b' => 5 , 'c' => 3},
{'a' => 6, 'b' => 5 , 'c' => 3} ]
array_of_hashes.map { |h| h['a'] }.uniq # => [1, 4, 6]
array_of_hashes.map { |h| h['b'] }.uniq # => [2, 5]
array_of_hashes.map { |h| h['c'] }.uniq # => [3]
This is more generic:
options = {}
distinct_keys = array_of_hashes.map(&:keys).flatten.uniq
distinct_keys.each do |k|
options[k] = array_of_hashes.map {|o| o[k]}.uniq
end
I have these hashes:
{"a" => 1, "b" => 2, "c" => 3, "k" => 14}
{"b" => 51, "c" => 2, "d" => 8}
I need to write code, so that after manipulation, the result would be:
{"a" => 1, "b" => 51, "c" => 2, "k" => 14}
I tried:
h1.each do |h, j|
h2.each do |hh, jj|
if h == hh
j = jj
end
end
end
but it doesn't work. Also I think this is ugly code, so how would could it be written better/right?
I though I should compare the two hashes, and, if the second key is the same as the first, change the first hash value to the second hash's value.
Just iterate over the entries in h2 and update the corresponding entry in h1 only if it already exists:
h2.each { |k,v| h1[k]=v if h1.include?(k) }
h1 # => {"a"=>1, "b"=>51, "c"=>2, "k"=>14 }
Also, if you want to update the entries as above and also add new entries from h2 you can simply use the Hash#merge! method:
h1.merge!(h2)
h1 # => {"a"=>1, "b"=>51, "c"=>2, "k"=>14, "d"=>8}
I have something like this:
a = [{"group_id" => 1, "student_id" => 3, "candies" => 4},
{"group_id" => 2, "student_id" => 1, "candies" => 3},
{"group_id" => 1, "student_id" => 2, "candies" => 2},
{"group_id" => 3, "student_id" => 4, "candies" => 6},
{"group_id" => 1, "student_id" => 5, "candies" => 1},
{"group_id" => 3, "student_id" => 6, "candies" => 1},
{"group_id" => 4, "student_id" => 8, "candies" => 3}]
I have three groups of students and each student gets a certain number of candies. I wish to count the total number of candies in a group. For that, I need to know the students which belong to a certain group and accumulate their candy count. I can do it using loops and initializing counts to zero:
aa = a.group_by { |a| a["group_id"] }
# =>
{
1 => [
{"group_id"=>1, "student_id"=>3, "candies"=>4},
{"group_id"=>1, "student_id"=>2, "candies"=>2},
{"group_id"=>1, "student_id"=>5, "candies"=>1}
],
2 => [{"group_id"=>2, "student_id"=>1, "candies"=>3}],
3 => [
{"group_id"=>3, "student_id"=>4, "candies"=>6},
{"group_id"=>3, "student_id"=>6, "candies"=>1}
],
4 => [{"group_id"=>4, "student_id"=>8, "candies"=>3}]
}
But I'm not able to accumulate the values within the group_id. I wonder if there are any succinct ways of representing it. How do I sum the total number of candies that are present in a group?
The first your step (grouping) is correct. After that you can use the following:
a.group_by {|g| g['group_id']}.map do |g, students|
{group_id:g, candies:students.map {|st| st['candies']}.inject(&:+)}
end
map function is often used with collections instead of loops to make some operation on each element and return modified version of the collection.
Output:
[{:group_id=>1, :candies=>7},
{:group_id=>2, :candies=>3},
{:group_id=>3, :candies=>7},
{:group_id=>4, :candies=>3}]
Adding to #StasS answer, a more direct hash way to do (with a more cryptic code) is like this:
> Hash[a.group_by{|g| g['group_id']}.map{|g,s| [g, s.inject(0){|a,b| a + b["candies"]}]}]
=> {1=>7, 2=>3, 3=>7, 4=>3}
you can unfold the line like this:
groups = a.group_by{|g| g['group_id']}
id_candies_pairs = groups.map{|g,s| [g, s.inject(0){|a,b| a + b["candies"]}]}
id_candies_hash = Hash[id_candies_pairs]
return id_candies_hash
Riffing on the answer by #StasS, you can also just build a simpler looking hash like:
totals_by_group_id = {}
a.group_by {|g| g['group_id']}.map do |g, students|
totals_by_group_id[g] = students.map {|st| st['candies']}.inject(&:+)
end
The resulting totals_by_group_id hash is:
{1=>7, 2=>3, 3=>7, 4=>3}
I am trying to sort a document based on the number of times the word appears then alphabetically by the words so when it is outputted it will look something like this.
Unsorted:
'the', '6'
'we', '7'
'those', '5'
'have', '3'
Sorted:
'we', '7'
'the', '6'
'those', '5'
'have', '3'
Try this:
Assuming:
a = {
'the' => '6',
'we' => '7',
'those' => '5',
'have' => '3',
'hav' => '3',
'haven' => '3'
}
then after doing this:
b = a.sort_by { |x, y| [ -Integer(y), x ] }
b will look like this:
[
["we", "7"],
["the", "6"],
["those", "5"],
["hav", "3"],
["have", "3"],
["haven", "3"]
]
Edited to sort by reverse frequencies.
words = {'the' => 6,'we' => 7,'those' => 5,'have' => 3}
sorted_words = words.sort { |a,b| b.last <=> a.last }
sorted_words.each { |k,v| puts "#{k} #{v}"}
produces:
we 7
the 6
those 5
have 3
You probably want the values to be integers rather than strings for comparison purposes.
EDIT
Oops, overlooked the requirement that it needs to be sorted by the key too. So:
words = {'the' => 6,'we' => 7,'those' => 5,'have' => 3,'zoo' => 3,'foo' => 3}
sorted_words = words.sort do |a,b|
a.last == b.last ? a.first <=> b.first : b.last <=> a.last
end
sorted_words.each { |k,v| puts "#{k} #{v}"}
produces:
we 7
the 6
those 5
foo 3
have 3
zoo 3
When you use the sort method on a hash, you receive two element arrays in your comparison block, with which you can do comparisons in one pass.
hsh = { 'the' => '6', 'we' => '6', 'those' => '5', 'have' => '3'}
ary = hsh.sort do |a,b|
# a and b are two element arrays in the format [key,value]
value_comparison = a.last <=> b.last
if value_comparison.zero?
# compare keys if values are equal
a.first <=> b.first
else
value_comparison
end
end
# => [['have',3],['those',5],['the',6],['we',6]]
Note that the result is an array of arrays because hashes do not have intrinsic order in ruby
Try this:
words = {'the' => 6,'we' => 7,'those' => 5,'have' => 3}
words.sort { |(x_k, x_v), (y_k, y_v)| [y_v, y_k] <=> [x_v, x_k]}
#=> [["we", 7], ["the", 6], ["those", 5], ["have", 3]]
histogram = { 'the' => 6, 'we' => 7, 'those' => 5, 'have' => 3, 'and' => 6 }
Hash[histogram.sort_by {|word, freq| [-freq, word] }]
# {
# 'we' => 7,
# 'and' => 6,
# 'the' => 6,
# 'those' => 5,
# 'have' => 3
# }
Note: this assumes that you use numbers to store the numbers. In your data model, you appear to use strings to store the numbers. I have no idea why you would want to do this, but if you do want to do this, you would obviously have to convert them to numbers before sorting and then back to strings.
Also, this assumes Ruby 1.9. In Ruby 1.8, hashes aren't ordered, so you cannot convert the sorted result back to a hash since that would lose the ordering information, you would have to keep it as an array.
1.9.1
>> words = {'the' => 6,'we' => 7, 'those' => 5, 'have' => 3}
=> {"the"=>6, "we"=>7, "those"=>5, "have"=>3}
>> words.sort_by{ |x| x.last }.reverse
=> [["we", 7], ["the", 6], ["those", 5], ["have", 3]]
word_counts = {
'the' => 6,
'we' => 7,
'those' => 5,
'have' => 3,
'and' => 6
};
word_counts_sorted = word_counts.sort do
|a,b|
# sort on last field descending, then first field ascending if necessary
b.last <=> a.last || a.first <=> b.first
end
puts "Unsorted\n"
word_counts.each do
|word,count|
puts word + " " + count.to_s
end
puts "\n"
puts "Sorted\n"
word_counts_sorted.each do
|word,count|
puts word + " " + count.to_s
end
It does not seem to be documented very much:
hsh.merge(other_hash){|key, oldval, newval| block} → a_hash
http://ruby-doc.org/core/classes/Hash.html#M002880
As it might be expected, the resulting hash will contain the value returned by a block for every key which exists in both hashes being merged:
>> h1 = {:a => 3, :b => 5, :c => 6}
=> {:a=>3, :b=>5, :c=>6}
>> h2 = {:a => 4, :b => 7, :d => 8}
=> {:a=>4, :b=>7, :d=>8}
>> h1.merge h2
=> {:a=>4, :b=>7, :c=>6, :d=>8}
>> h1.merge(h2){|k,v1,v2| v1}
=> {:a=>3, :b=>5, :c=>6, :d=>8}
>> h1.merge(h2){|k,v1,v2| v1+v2}
=> {:a=>7, :b=>12, :c=>6, :d=>8}