CODEWITHSUNDEEP
pythonpython-3.xlistfor-loop
Barbod5412
Barbod5412

Reputation: 13

skipping part of a loop

I'm trying to make a function that returns only the unique number in a list by checking each number in the lists count. but even if it already knows a numbers count is more than 1 it will check that number over and over again which will of course cause a problem when checking a large list, I want it to stop checking counts of a number if it already has been checked, but I can't find any way to do it.

this is my code:

def find_uniq(arr):
    for i in arr:
        if arr.count(i) == 1:
            return i

Upvotes: 1

Views: 102

Answers (7)

James_481
James_481

Reputation: 176

I think one of the easiest ways to do what you're asking is to create a temporary list that stores the numbers that you have already checked, and then check to see if you have checked that number yet.

def find_uniq(arr):
    checked_nums = []  #list of all checked numbers
    for i in arr:
        if i not in checked_nums:  #if the number hasn't been checked yet:
            if arr.count(i) == 1:
                return i
            checked_nums.append(i)  #add the number to the checked numbers list

Upvotes: 0

Kavindu Nilshan
Kavindu Nilshan

Reputation: 811

    def find_uniq(arr):
        
        uniqueList = list(set(arr))  # This list contains unique elements in the list "arr".

        return uniqueList

In this way, you can easily find unique elements. In fact,

    def find_uniq(arr):

        return list(set(arr))

This function is also doing the same thing but the first code is easy to understand.

Upvotes: 1

Sam
Sam

Reputation: 331

This approach should work if you need to find first number that only occurs once. tempSet is here to shrink down the array to smallest way possible.

def find_uniq(arr):

    tempSet = list(set(arr))

    for i in tempSet:
        if arr.count(i) == 1:
            return i

Upvotes: 0

Metatrix
Metatrix

Reputation: 279

You can use a dictionary to store the count of each element in the array. More about dictionaries is given in the documentation here: https://docs.python.org/3/tutorial/datastructures.html#dictionaries

def find_uniq(arr):
    count = {}
    for i in arr:
        if i in count:
            count[i] += 1
        else:
            count[i] = 1
    return count

The returned dictionary gives a count of each element in the array. To find out which element occurs once, simply compare count[elem] to 1.

This is also more efficient as a dictionary is a 'lookup' table whose elements can be accessed in O(1) time. The arr.count function is O(n), because it needs to check the entire array to count the number of times an element is in it.

Upvotes: 0

I'mahdi
I'mahdi

Reputation: 24049

use set and check count of unique value. from doc:

Sets : Python also includes a data type for sets. A set is an unordered collection with no duplicate elements.

like below:

def find_uniq(arr):
    ret_arr = []
    for i in set(arr):
        if arr.count(i) == 1:
            ret_arr.append(i)
    return ret_arr

Upvotes: 1

user96564
user96564

Reputation: 1607

use Counter and it will create dictionary:

for example:

from collections import Counter

After that, get the list element where the value of the dictionary is 1

lst = [k for k,v in mydict.items() if v == 1]

Upvotes: 2

Harxish
Harxish

Reputation: 459

You could just convert it into set, it'll only hold unique elements.

arr = list(set(arr))

Upvotes: 0

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