My goal is to determine whether there is a blank in a hash like this:
{"departure_time_slots"=>{"date"=>[""], "time"=>{"start"=>[""], "end"=>[""]}}}
The strings are nested arbitrary times. I do not want to check each of them manually. It would be nice if I can extract all the strings regardless of their depth. Is there a simple way to do that?
Here's an example of how you can extract the values. However you will still have the problem of matching the values to the keys. You can't really have both at the same time.
# Extend the Hash class
class Hash
def recursive_values
self.values.collect { |v|
v.is_a?(Hash) ? v.recursive_values : v
}.flatten
end
end
Usage:
h = {"departure_time_slots"=>{"date"=>[""], "time"=>{"start"=>[""], "end"=>[""]}}}
h.recursive_values
=> ["", "", ""]
It will be better if you will use sth like that:
departure_time_slots = {:date => Time.new, :time_start => nil, :time_end => nil}
And when you use keys in Hash it is good practise to using Symbols for keys. (http://www.ruby-doc.org/core-1.9.3/Symbol.html)
No not possible. Because they are totally present in different scope with respect to each other.
For e.g.. keys start and end is totally unknown and masked from the departure_time_slots object in the example above.
One round abut way could be, getting all the values of the hashmap which are of type hashmap again and obtaining their keys recusively.
Fetch keys of departure_time_slots and then from the value list of that map, find all the keys from every value, if that were to be a hashmap. Other than that, I don't think there is another way.
P.S. On side note, see if u can modify your structure to an array where elements can also be arrays, and try and use flatten concept of arrays. :P
Related
I have an array of hashes:
arr = [{"id"=>"1", "name"=>"Alan"}, {"id"=>"2", "name"=>"Ben"}, {"id"=>"3", "name"=>"Carl"}, {"id"=>"4", "name"=>"Danny"}, {"id"=>"5", "name"=>"Eva"}]
If I were to find the name of id #4:
arr.find{ |a| a["id"] == "4" }["name"]
returns "Danny", which is what I want.
My question is, is there a shorter, more elegant way of accomplish the same search?
If you only need to look up one id, a linear search as you are doing is fine.
If you are doing many lookups on the same array, create a hash from the array, so the individual lookups are O(1)
h = Hash[arr.map{|a|[a["id"], a["name"]]}]
h["4"]
It is a perfectly reasonable way of doing it. The only problem I see is that your data structure is not lookup friendly. You need to search in array, instead of looking up by ID in hash. Searching in array is O(N) and is relatively more difficult code. Looking up in hash is O(1) and is no code at all.
So, I would convert your array to hash, and then look up in it. Especially, if you are planning to do many lookups.
people = arr.inject({}) {|memo,v| memo[v["id"].to_i]=v["name"]; memo}
people[4] # will return Danny
people[3] # will return Carl
Hope it helps!
Suppose I have a non-empty array ids of Thing object ids and I want to find the corresponding objects using things = Thing.find_all_by_id(ids). My impression is that things will not necessarily have an ordering analogous to that of ids.
Is my impression correct?
If so, what can I used instead of find_all_by_id that preserves order and doesn't hit the database unnecessarily many times?
Yes
Use Array#sort
Check it out:
Thing.where(:id => ids).sort! {|a, b| ids.index(a.id) <=> ids.index(b.id)}
where(:id => ids) will generate a query using an IN(). Then the sort! method will iterate through the query results and compare the positions of the id's in the ids array.
#tybro0103's answer will work, but gets inefficient for a large N of ids. In particular, Array#index is linear in N. Hashing works better for large N, as in
by_id = Hash[Thing.where(:id => ids).map{|thing| [thing.id, thing]}]
ids.map{|i| by_id[i]}
You can even use this technique to arbitrarily sort by any not-necessarily unique attribute, as in
by_att = Thing.where(:att => atts).group_by(&:att)
atts.flat_map{|a| by_att[a]}
find_all_by_id is deprecated in rails 4, which is why I use where here, but the behavior is the same.
I have a hash that will render my html differently based on a particular variable. The variable is within the hash. So I am wondering how I can pass a hash value to a group by. to sort the rest heres what I am trying, maybe this will explain it better than me wording it.
<% grouped = svcs.group_by { |svc| svc[val[:sorttype]] } %>
val is a multidimensional hash. the first 2 key value pairs sorttype and one other are simple key and value, the 3rd piece (svcs) contains the equivilent of a 2D hash. Which if I manually type the type of sort I want to apply to it for the group by it works ie:
<% grouped = svcs.group_by { |svc| svc[:service_name] } %>
in PHP i know in a similar instance I can pass a variable of some sort to something like this and have it work. I assume such is the case here. However Im not sure how to put the variable in. Cause all the ways Ive tried don't work
It depends a little.
Rails' has a HashWithIndifferentAccess that will not distinguish between string and symbol keys; if you're using one of those, it should work as-is.
If it's not, it depends what the val entries are--if they're strings, convert to a symbol using to_sym, e.g., svc[val[:sorttype].to_sym].
Is there a simple way, instead of looping my entire array to fetch the first value of every inner array.
so in essence I have the following;
array = [['test', 'test2'...], ['test' ....]]
so I want to grab array[#][0] and store the unqiue values.
EDIT
Is there a similar way to use the transpose method for arrays with Hash?
I essentially want to do the same thing
Hash = {1=> {1=> 'test', .....}, 2=> {1=> 'test',....}
so at the end I want to have something like new hash variable and leave my existing hash within hash alone.... = {1 => 'test', 2=> 'test2'}
Not sure if I fully understand the question, but if you have a 2 dimensional array (array in array), and you want to turn that into an array of the first element of the second dimension, you can use the map function
firsts = array.map {|array2| array2.first}
The way map works is that it turns one collection into a second collection by applying a function you provide (the block) to each element.
Maybe this?
array.transpose[0]
I'm building an xml document from a hash. The xml attributes need to be in order. How can this be accomplished?
hash.to_xml
Ruby 1.8's hash aren't in insertion order. With ruby 1.9, they will be.
However rails offers an alternative to that, the class OrderedHash.
my_hash = ActiveSupport::OrderedHash.new
my_hash[:key] = 'value'
my_hash[:second_key] = 'second value'
This hash is in fact an array of that format :
[[:key, 'value'], [:second_key, 'second value']]
The entries remains in the order you inserted them.
And you can access them like with any other hash.
h = Hash[:x,123,:a,553,:d,949,:e,5321]
=> {:e=>5321, :x=>123, :a=>553, :d=>949}
h.sort { |x,y| x[0].to_s <=> y[0].to_s }
=> [[:a, 553], [:d, 949], [:e, 5321], [:x, 123]]
The usual ways of sorting a hash is by key or value. Have a look here:
hash.sort
More complex sorts can be accomplised however by utilizing the spaceship operator
What order did you want them to be in? You shouldn't expect them to be in insertion order. From the docs for Hash:
The order in which you traverse a hash
by either key or value may seem
arbitrary, and will generally not be
in the insertion order.
If you need them to be in a specific order you can determine just from the keys/values (e.g. order the attribute names alphabetically), you'll need to apply that ordering explicitly.
This piece of code I've just made for i18n-js might help you out as it convert Hash to ActiveSupport::OrderedHash if needed then sort it's key by natural order.
http://seb.box.re/2010/1/15/deep-hash-ordering-with-ruby-1-8