Count number of pairs in list disregarding order

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find total number of (i,j) pairs in array such that i<j and a[i]<a[j]
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count number of distinct pairs whose sum exists in the given array
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find total number of (i,j) pairs in array such that i<j and a[i]%a[j]==0

In example, if I have the following script:

import collections

lst = [['a','b'],['b','a'],['c','d'],['c','d'],['d','c']]

print([(a, b, v) for (a, b),v in collections.Counter(map(tuple,lst)).items()])

I get as output:

[('a', 'b', 1), ('b', 'a', 1), ('c', 'd', 2), ('d', 'c', 1)]

Can I adapt my code to yield the following output:

[('a', 'b', 2), ('c', 'd', 3)]

So a function that doesn't include the order of the pairs?

You could just sort each element in the list before counting, like so:

import collections

lst = [['a','b'],['b','a'],['c','d'],['c','d'],['d','c']]

sorted_lst = [sorted(x) for x in lst]

print([(a, b, v) for (a, b),v in collections.Counter(map(tuple,sorted_lst)).items()])


[('a', 'b', 2), ('c', 'd', 3)]

Count number of pairs (i, j) such that arr[i] * arr[j , Count of pairs violating BST property � Count of even and odd power pairs in an Array � Count pairs in array whose sum is divisible by K � Count� If x = 0, then the count of pairs for this x is 0. If x = 1, then the count of pairs for this x is equal to count of 0s in Y[]. x smaller than y means x^y is greater than y^x. x = 2, y = 3 or 4; x = 3, y = 2; Note that the case where x = 4 and y = 2 is not there. Following diagram shows all exceptions in tabular form.

Use a data structure that doesn't care about order. In this case you'll need frozenset instead of a regular set because Counter requires it to be hashable. But basically it's a simple substitution of tuple in your original code for frozenset:

print([(a, b, v) for (a, b),v in collections.Counter(map(frozenset,lst)).items()])


[('a', 'b', 2), ('d', 'c', 3)]

Count all distinct pairs with difference equal to k, Given an integer array and a positive integer k, count all distinct pairs with 1) Initialize count as 0 2) Sort all numbers in increasing order. 1) Initialize count as 0 2) Sort all numbers in increasing order. 3) Remove duplicates from array. 4) Do following for each element arr[i] a) Binary Search for arr[i] + k in subarray from i+1 to n-1.

Sorting the list before you get collections of it solves the problem.

import collections

lst = [['a','b'],['b','a'],['c','d'],['c','d'],['d','c']]

sort_list = sorted(x) for x in lst

print([(a, b, v) for (a, b),v in collections.Counter(map(tuple,sort_list)).items()])

[PDF] Basic Counting, multiset: a collection of objects (repeats allowed) where order does not matter. How many ways can we construct a k-list of distinct elements from the set? For example, ignoring capitalization, punctuation and spaces, “Madam I'm a list of 95 seats, and then pair up people and seats so that the first person gets the first� Count of pairs is 3. Time Complexity : O(n 2) Auxiliary Space : O(1) A better solution is possible in O(n) time.. Below is the Algorithm. Create a map to store frequency of each number in the array.

You could sort the values of the key a,b and use groupby in itertools and then sum all the elements in the group.

import itertools as it
lst = [['a','b'],['b','a'],['c','d'],['c','d'],['d','c']]
output = [(*group,sum(1 for i in elements)) for group,elements in it.groupby(lst,key=lambda x:sorted(x))]


[('a', 'b', 2), ('c', 'd', 3)]

Compressing two integers disregarding order, If we order these sequences lexicographically, then N counts the number of smaller The total number of unordered pairs of numbers in a set of N is N(N+1)/ 2. The COUNT(*) function returns a number of rows in a specified table or view that includes the number of duplicates and NULL values. To return the number of rows that excludes the number of duplicates and NULL values, you use the following form of the COUNT() function:

[PDF] mergesort ‣ counting inversions ‣ randomized , closest pair of points List RSS news items in reverse chronological order. T(n ) = max number of compares to mergesort a list of length n. Recurrence. Ignoring floors, this implies that C(n) ≤ C(n/5 + n − 3n/10) + 11/5 n. Efficient approach: An efficient approach is to use Sum over Subsets Dynamic Programming method and count the number of ordered pairs. In the SOS DP we find out the pairs whose bitwise & returned 0. Here we need to count the number of pairs. Some key observations are the constraints, the maximum that an array element can be is 10 4. Calculating

Built-in Types — Python 3.8.5 documentation, Negative shift counts are illegal and cause a ValueError to be raised. Return a pair of integers whose ratio is exactly equal to the original integer and with a The constructor builds a list whose items are the same and in the same order as the raw memory disregarding the item format and the logical array structure. Count number of distinct pairs whose sum exists in the given array Find the maximum cost of an array of pairs choosing at most K pairs Find all unique pairs of maximum and second maximum elements over all sub-arrays in O(NlogN)

Python Word Count (Filter out Punctuation, Dictionary Manipulation , Output a List of Word Count Pairs (Sorted from Highest to Lowest). Approach 1: counting number of times each word comes up in list of words (in dictionary) COUNTIFS is configured to count "pairs" of items. Only when the item in column G and the corresponding item from row 4 appear together in a cell is the pair counted. A wildcard (*) is concatenated to both sides of the item to ensure a match will be counted no matter where it appears in the cell.

  • Seems like the most sensible way to go; if you're trying to ignore sorting, then an unsorted datatype seems to be the most appropriate. The accepted answer ignores sorting by forcing it to always be the same, which seems like a slightly less literal translation of the actual problem