I am trying to create a function that returns a unique value on a python list. So for example:
List = [1,1,1,1,2,1,1,1,1,]
How would I go about returning the unique value, 2.
I have tried the following:
import numpy as np
def find_uniq(arr):
abc = np.array(arr)
return list(np.unique(abc))
However when I try this ^^^ , it returns (2) two values , the 1 and the 2. Since they are both unique values to the list. Though I only want the 2.
Try this:
>>> from collections import Counter
>>> c = Counter( [1,1,1,1,2,1,1,1,1,])
>>> c
Counter({1: 8, 2: 1})
>>> print ([k for k,v in c.items() if v==1])
[2]
This will tell you quickly if there is no unique element, or more than one.
Return counts from numpy.unique and then select values that have count of 1:
lst = [1,1,1,1,2,1,1,1,1]
val, cnt = np.unique(lst, return_counts=True)
val[cnt == 1]
# [2]
You can loop through the unique elements using set, and return the element which appears only once:
for ele in set(arr):
if arr.count(ele) == 1:
print(ele)
break
For every unique element append it to list
def find_unique(nums_list):
result = []
for i in nums_list :
if nums_list.count(i) == 1 :
result.append(i)
return result
print(find_unique([1,1,1,1,2,1,1,1,1]))
If you want to find only first unique element
def find_unique(nums_list):
for i in nums_list :
if nums_list.count(i) == 1 :
return i
print(find_unique([1,9,1,1,2,1,1,1,1]))
def check_window(i,a):
return all(a[i+d]!=a[i] for d in (-1,1))
l = [1,1,1,1,2,1,1,1,1]
ls = sorted(l)
ls.append(None)
uniques = [ls[i] for i in range(1,len(ls)-1) if check_window(i, ls)]
for i in numList:
if numList.count(i) == 1 :
print ("unique", i)
The first solution I came up with below, has an issue where it takes too long to return what I need. It doesn't create additional arrays, but I guess iterating through a giant list would be a problem.
for n in arr:
if arr.count(n) == 1:
return n
else: None
The method below worked perfectly, and leads me to believe that python sorts faster than iterating.
x,y = sorted(arr)[0], sorted(arr)[-1]
return x if arr.count(x) == 1 else y
countmethod for each unique value and return the element with count equal to 1.Counterobject from thecollectionsmodule