Imagine I have this list:
list = ['a','a','b','a']
I would like to do something like this:
print(unique(list))
to retrieve the unique item(s), so python will output b. How could I do this?
5 Answers
Count the items, keep only those which have a count of 1:
>>> data = ['a','a','b','a']
>>> from collections import Counter
>>> [k for k,v in Counter(data).items() if v == 1]
['b']
answered May 23, 2021 at 9:13
juanpa.arrivillaga
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Comments
Using set() property of Python, we can easily check for the unique values. Insert the values of the list in a set. Set only stores a value once even if it is inserted more then once. After inserting all the values in the set by list_set=set(list1), convert this set to a list to print it.
for more ways check : https://www.geeksforgeeks.org/python-get-unique-values-list/#:~:text=Using%20set()%20property%20of,a%20list%20to%20print%20it.
Example to make a unique function :
# Python program to check if two
# to get unique values from list
# using set
# function to get unique values
def unique(list1):
# insert the list to the set
list_set = set(list1)
# convert the set to the list
unique_list = (list(list_set))
for x in unique_list:
print x,
# driver code
list1 = [10, 20, 10, 30, 40, 40]
print("the unique values from 1st list is")
unique(list1)
list2 =[1, 2, 1, 1, 3, 4, 3, 3, 5]
print("\nthe unique values from 2nd list is")
unique(list2)
output should be: the unique values from 1st list is 40 10 20 30 the unique values from 2nd list is 1 2 3 4 5
You can also use numpy.unique example:
`
#Ppython program to check if two
# to get unique values from list
# using numpy.unique
import numpy as np
# function to get unique values
def unique(list1):
x = np.array(list1)
print(np.unique(x))
# driver code
list1 = [10, 20, 10, 30, 40, 40]
print("the unique values from 1st list is")
unique(list1)
list2 =[1, 2, 1, 1, 3, 4, 3, 3, 5]
print("\nthe unique values from 2nd list is")
unique(list2)
` output should be : the unique values from 1st list is [10 20 30 40]
the unique values from 2nd list is [1 2 3 4 5]
Fore more please check "https://www.geeksforgeeks.org/python-get-unique-values-list/#:~:text=Using%20set()%20property%20of,a%20list%20to%20print%20it."
Comments
You could use collections.Counter() for this, e.g.:
from collections import Counter
my_list = ['a','a','b','a']
counts = Counter(my_list)
for element, count in counts.items():
if count == 1:
print(element)
would print b and nothing else in this case.
Or, if you'd like to store the result in a list:
from collections import Counter
my_list = ['a','a','b','a']
counts = Counter(my_list)
unique_elements = [element for element, count in counts.items() if count == 1]
unique_elements is ['b'] in this case.
Comments
- Count the appearance of an element in the list, and store it in a dictionary
- From the dictionary, check which elements has only one appearance
- Store the unique (one appearance) elements in a list
- Print the list
You can try this:
my_list = ['a','a','b','a']
my_dict = {}
only_one = []
#Step 1
for element in my_list:
if element not in my_dict:
my_dict[element] = 1
else:
my_dict[element] += 1
#Step 2 and Step 3
for key, value in my_dict.items():
if value == 1:
only_one.append(key)
#Step 4
for unique in only_one:
print(unique)
Output:
b
In case you are wondering what other variables contain:
my_dict = {'a': 3, 'b': 1}
only_one = ['b']
Comments
Most straight forward way, use a dict of counters.
a = ['a','a','b','a']
counts = {}
for element in a:
counts[element] = counts.get(element, 0) + 1
for element in a:
if counts[element] == 1:
print(element)
out:
b
count()method of lists.listfor a variable name in Python, as it overrides a built-in name (the list constructor). Same fordict,sum, and more.