Python hashable In order to perform comparisons, a hashable needs an __eq__ method. Note: Hashable objects which compare equal must have the same hash value. Hashability makes an object usable as a dictionary key and a set member, because these data structures use the hash value internally.
We can merge two hashes using the merge() method. When using the merge() method: Each new entry is added to the end. Each duplicate-key entry's value overwrites the previous value.
You can compare hashes directly for equality:
hash1 = {'a' => 1, 'b' => 2}
hash2 = {'a' => 1, 'b' => 2}
hash3 = {'a' => 1, 'b' => 2, 'c' => 3}
hash1 == hash2 # => true
hash1 == hash3 # => false
hash1.to_a == hash2.to_a # => true
hash1.to_a == hash3.to_a # => false
You can convert the hashes to arrays, then get their difference:
hash3.to_a - hash1.to_a # => [["c", 3]]
if (hash3.size > hash1.size)
difference = hash3.to_a - hash1.to_a
else
difference = hash1.to_a - hash3.to_a
end
Hash[*difference.flatten] # => {"c"=>3}
Simplifying further:
Assigning difference via a ternary structure:
difference = (hash3.size > hash1.size) \
? hash3.to_a - hash1.to_a \
: hash1.to_a - hash3.to_a
=> [["c", 3]]
Hash[*difference.flatten]
=> {"c"=>3}
Doing it all in one operation and getting rid of the difference
variable:
Hash[*(
(hash3.size > hash1.size) \
? hash3.to_a - hash1.to_a \
: hash1.to_a - hash3.to_a
).flatten]
=> {"c"=>3}
You can try the hashdiff gem, which allows deep comparison of hashes and arrays in the hash.
The following is an example:
a = {a:{x:2, y:3, z:4}, b:{x:3, z:45}}
b = {a:{y:3}, b:{y:3, z:30}}
diff = HashDiff.diff(a, b)
diff.should == [['-', 'a.x', 2], ['-', 'a.z', 4], ['-', 'b.x', 3], ['~', 'b.z', 45, 30], ['+', 'b.y', 3]]
If you want to get what is the difference between two hashes, you can do this:
h1 = {:a => 20, :b => 10, :c => 44}
h2 = {:a => 2, :b => 10, :c => "44"}
result = {}
h1.each {|k, v| result[k] = h2[k] if h2[k] != v }
p result #=> {:a => 2, :c => "44"}
Rails is deprecating the diff
method.
For a quick one-liner:
hash1.to_s == hash2.to_s
You could use a simple array intersection, this way you can know what differs in each hash.
hash1 = { a: 1 , b: 2 }
hash2 = { a: 2 , b: 2 }
overlapping_elements = hash1.to_a & hash2.to_a
exclusive_elements_from_hash1 = hash1.to_a - overlapping_elements
exclusive_elements_from_hash2 = hash2.to_a - overlapping_elements
I developed this to compare if two hashes are equal
def hash_equal?(hash1, hash2)
array1 = hash1.to_a
array2 = hash2.to_a
(array1 - array2 | array2 - array1) == []
end
The usage:
> hash_equal?({a: 4}, {a: 4})
=> true
> hash_equal?({a: 4}, {b: 4})
=> false
> hash_equal?({a: {b: 3}}, {a: {b: 3}})
=> true
> hash_equal?({a: {b: 3}}, {a: {b: 4}})
=> false
> hash_equal?({a: {b: {c: {d: {e: {f: {g: {h: 1}}}}}}}}, {a: {b: {c: {d: {e: {f: {g: {h: 1}}}}}}}})
=> true
> hash_equal?({a: {b: {c: {d: {e: {f: {g: {marino: 1}}}}}}}}, {a: {b: {c: {d: {e: {f: {g: {h: 2}}}}}}}})
=> false
If you need a quick and dirty diff between hashes which correctly supports nil in values you can use something like
def diff(one, other)
(one.keys + other.keys).uniq.inject({}) do |memo, key|
unless one.key?(key) && other.key?(key) && one[key] == other[key]
memo[key] = [one.key?(key) ? one[key] : :_no_key, other.key?(key) ? other[key] : :_no_key]
end
memo
end
end
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