How do you get a non repeated value in python?

Use Counter and defaultdict from collections:

from collections import defaultdict
from collections import Counter

A = [1 ,2 ,3 ,2 ,5, 1]
# create a mapping from each item to it's count
counter = Counter[A]

# now reverse the mapping by using a defaultdict for convenience
countmap = defaultdict[list]

for k, v in counter.iteritems[]:
    countmap[v].append[k]

# take all values that occurred once
print countmap[1] # [3, 5]

If you're interested in what the mappings are you can print them:

The counter:

Counter[{1: 2, 2: 2, 3: 1, 5: 1}]

The countmap:

defaultdict[, {1: [3, 5], 2: [1, 2]}]

To manually create the counter you can use this function:

def make_counter[lst]:
    counter = dict[]
    for item in lst:
        if item in counter:
            counter[item] += 1
        else:
            counter[item] = 1
    return counter

make_counter[A]

Output:

{1: 2, 2: 2, 3: 1, 5: 1}

Non-repeating elements in an array in python

Here, in this page we will discuss the program to print the non-repeating elements in python programming language. We are given with an integer array and need to print those elements which occurs only one time.

Example

Methods Discussed :

• Method 1 : Using Two for loops
• Method 2 : Using dictionary

Method 1 :

In this method we will count the frequency of each elements using two for loops and print those elements which have frequency equals to one.

• To check the status of visited elements create a array of size n.
• Run a loop from index 0 to n and check if [visited[i]==1] then skip that element.
• Otherwise create a variable count = 1 to keep the count of frequency.
• Run a loop from index i+1 to n
• Check if[arr[i]==arr[j]], then increment the count by 1 and set visited[j]=1.
• After complete iteration of inner for loop and check if count == 1 then print those element.

Time and Space Complexity :

• Time Complexity : O[n2]
• Space Complexity : O[n]

Method 1 : Code in Python

Run

# Python 3 program to count unique elements
def count[arr, n]:

   # Mark all array elements as not visited
   visited = [False for i in range[n]]

   # Traverse through array elements
   # and count frequencies
   for i in range[n]:

     # Skip this element if already
     # processed
     if [visited[i] == True]:
        continue

     # Count frequency
     count = 1
     for j in range[i + 1, n, 1]:
        if [arr[i] == arr[j]]:
          visited[j] = True
          count += 1
     if count == 1 :
        print[arr[i]];
   

# Driver Code
arr = [10, 30, 40, 20, 10, 20, 50, 10]
n = len[arr]
count[arr, n]

Method 2 :

In this method we will count use dictionary to count the frequency of each elements and then check if that frequency is equal to 1 or not.

• Declare a dictionary dict[] say, mp = dict[].
• Start iterating over the entire array
• If element is present in map, then increase the value of frequency by 1.
• Otherwise, insert that element in map.
• After complete iteration over array, start traversing map and check if value is equal to 1,then print the key value.

Time and Space Complexity:

• Time Complexity : O[n]
• Space Complexity : O[n]

Method 2 : Code in Python

Run

def count[arr, n]:

    mp = dict[]
  
    # Traverse through array elements
    # and count frequencies

    for i in range[n]:
        if arr[i] in mp.keys[]:
            mp[arr[i]] += 1
        else:
            mp[arr[i]] = 1

           
    # Traverse through map and print
    # frequencies

    for x in mp:
        if mp[x]==1 :
          print[x];
   
# Driver Code 
arr = [10, 30, 40, 20, 10, 20, 50, 10] 
n = len[arr] 
count[arr, n]

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