How to find two items of a list with the same return value of a function on their attribute?

How to find two items of a list with the same return value of a function on their attribute?

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Given a basic class Item:

class Item(object):
    def __init__(self, val):
        self.val = val

a list of objects of this class (the number of items can be much larger):

items = [ Item(0), Item(11), Item(25), Item(16), Item(31) ]

and a function compute that process and return a value.

How to find two items of this list for which the function compute return the same value when using the attribute val? If nothing is found, an exception should be raised. If there are more than two items that match, simple return any two of them.

For example, let's define compute:

def compute( x ):
    return x % 10

The excepted pair would be: (Item(11), Item(31)).

Assuming the values returned by compute are hashable (e.g., float values), you can use a dict to store results.

And you don't need to do anything fancy, like a multidict storing all items that produce a result. As soon as you see a duplicate, you're done. Besides being simpler, this also means we short-circuit the search as soon as we find a match, without even calling compute on the rest of the elements.

def find_pair(items, compute):
    results = {}
    for item in items:
        result = compute(item.val)
        if result in results:
            return results[result], item
        results[result] = item
    raise ValueError('No pair of items')

How to call JavaScript code on multiple DIV elements without the ID , You can select these DIV elements, even though they don't have id attributes. In fact, they share the same class name “image”. Let's see a few  If there are three items that have the same output value for compute, only the two firsts will be returned. @AChampion Afterthought, it is not mandatory that the pair is the two first elements that match. It could be any two elements of the list.

You can check the length of the set of resulting values:

class Item(object):
  def __init__(self, val):
    self.val = val
  def __repr__(self):
    return f'Item({self.val})'

def compute(x):
  return x%10

items = [ Item(0), Item(11), Item(25), Item(16), Item(31)]
c = list(map(lambda x:compute(x.val), items))
if len(set(c)) == len(c): #no two or more equal values exist in the list
  raise Exception("All elements have unique computational results")

To find values with similar computational results, a dictionary can be used:

from collections import Counter
new_d = {i:compute(i.val) for i in items}
d = Counter(new_d.values())
multiple = [a for a, b in new_d.items() if d[b] > 1]


[Item(11), Item(31)]

A slightly more efficient way to find if multiple objects of the same computational value exist is to use any, requiring a single pass over the Counter object, whereas using a set with len requires several iterations:

if all(b == 1 for b in d.values()):
   raise Exception("All elements have unique computational results")

Return Multiple Match Results in Excel (2 methods), The following example demonstrates the Find method on a List<T> that get; set​; } public int PartId { get; set; } public override string ToString() { return "ID: " + public static void Main(string[] args) { FillList(); // Find a book by its ID. Elements​(); // Evaluate each element and set set values in the book object. Attribute("id"). Find example. Here we consider the Find() method on List. Find accepts a Predicate, which we can specify as a lambda expression. It returns the first match.Predicate Lambda. Here: This code loops through each int value in the List, starting at the beginning, and tests each one to see if it is greater than 20. Return: The value 23 is returned

A dictionary val_to_it that contains Items keyed by computed val can be used:

val_to_it = {}
for it in items:
    computed_val = compute(it.val)
    # Check if an Item in val_to_it has the same computed val
    dict_it = val_to_it.get(computed_val)
    if dict_it is None:
        # If not, add it to val_to_it so it can be referred to 
        val_to_it[computed_val] = it
        # We found the two elements!
        res = [dict_it, it]
    raise Exception( "Can't find two items" )

The for block can be rewrite to handle n number of elements:

for it in items:
    computed_val = compute(it.val)
    dict_lit = val_to_it.get(computed_val)
    if dict_lit is None:
        val_to_it[computed_val] = [it]
        # Check if we have the expected number of elements
        if len(dict_lit) == n:
            # Found n elements!
            res = dict_lit

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