The JSON generator converts symbols to strings because JSON does not support symbols. Since JSON keys are all strings, parsing a JSON document will produce a Ruby hash with string keys by default. You can tell the parser to use symbols instead of strings by using the symbolize_names option. Example: original_hash = {:info => [{:from … Read more
You can just do key, value = hash.first or if you prefer: key = hash.keys[0] value = hash.values[0] Then maybe: new_hash = {key => value}
I did not have luck with memoise because it gave a ‘too deep recursive’ problem to some functions of a package I tried it with. With R.cache I had better luck. Following is more annotated code I adapted from the R.cache documentation. The code shows different options for doing caching: # Workaround to avoid question … Read more
As already stated, you cannot decrypt MD5 without attempting something like brute force hacking which is extremely resource intensive, not practical, and unethical. However you could use something like this to encrypt / decrypt passwords/etc safely: $input = “SmackFactory”; $encrypted = encryptIt( $input ); $decrypted = decryptIt( $encrypted ); echo $encrypted . ‘<br />’ . … Read more
It has been noted by many well respected developers such as Brian Goetz and Josh Bloch that : If an object’s hashCode() value can change based on its state, then we must be careful when using such objects as keys in hash-based collections to ensure that we don’t allow their state to change when they … Read more
Are the 160 bit hash values generated by SHA-1 large enough to ensure the fingerprint of every block is unique? Assuming random hash values with a uniform distribution, a collection of n different data blocks and a hash function that generates b bits, the probability p that there will be one or more collisions is … Read more
If you want to convert a Nokogiri XML document to a hash, just do the following: require ‘active_support/core_ext/hash/conversions’ hash = Hash.from_xml(nokogiri_document.to_s)
First, if you are using ruby < 1.9: array = [ {:price => 1, :count => 3}, {:price => 2, :count => 3}, {:price => 3, :count => 3} ] Then to get what you need: array.map{|x| x[:price]}
You don’t need andorid for this. You can just do it in simple java. Have you tried a simple java example and see if this returns the right sha1. import java.io.UnsupportedEncodingException; import java.security.MessageDigest; import java.security.NoSuchAlgorithmException; public class AeSimpleSHA1 { private static String convertToHex(byte[] data) { StringBuilder buf = new StringBuilder(); for (byte b : data) … Read more
For very small collections the difference is going to be negligible. At the low end of your range (500k items) you will start to see a difference if you’re doing lots of lookups. A binary search is going to be O(log n), whereas a hash lookup will be O(1), amortized. That’s not the same as … Read more