to compare data of above 2 lists & add unique items only to 3rd list
I tried to get it this way, but it didn’t work:
unique
Find unique rows in numpy.array
I need to find unique rows in a numpy.array.
Generate ‘n’ unique random numbers within a range
I know how to generate a random number within a range in Python. random.randint(numLow, numHigh) And I know I can put this in a loop to generate n amount of these numbers for x in range (0, n): listOfNumbers.append(random.randint(numLow, numHigh)) However, I need to make sure each number in that list is unique. Other than … Read more
Count unique values per groups with Pandas
I need to count unique ID values in every domain. I have data: ID, domain 123, 'vk.com' 123, 'vk.com' 123, 'twitter.com' 456, 'vk.com' 456, 'facebook.com' 456, 'vk.com' 456, 'google.com' 789, 'twitter.com' 789, 'vk.com' I try df.groupby([‘domain’, ‘ID’]).count() But I want to get domain, count vk.com 3 twitter.com 2 facebook.com 1 google.com 1 Answers: Thank you … Read more
Checking if all elements in a list are unique
What is the best way (best as in the conventional way) of checking whether all elements in a list are unique?
How to sort and remove duplicates from Python list?
Given a list of strings, I want to sort it alphabetically and remove duplicates. I know I can do this: from sets import Set […] myHash = Set(myList) but I don’t know how to retrieve the list members from the hash in alphabetical order. I’m not married to the hash, so any way to accomplish … Read more
pandas unique values multiple columns
df = pd.DataFrame({'Col1': ['Bob', 'Joe', 'Bill', 'Mary', 'Joe'], 'Col2': ['Joe', 'Steve', 'Bob', 'Bob', 'Steve'], 'Col3': np.random.random(5)}) What is the best way to return the unique values of ‘Col1’ and ‘Col2’? The desired output is 'Bob', 'Joe', 'Bill', 'Mary', 'Steve' Answers: Thank you for visiting the Q&A section on Magenaut. Please note that all the answers … Read more
Python: Uniqueness for list of lists
I am curious what would be an efficient way of uniquefying such data objects: