Apr 21, 2017 · I need to verify if a list is a subset of another - a boolean return is all I seek. Is testing equality on the smaller list after an intersection the fastest way to do this? Performance is of utm...

11 You can use either set.issubset() or set.issuperset() (or their operator based counterparts: <= and >=). Note that the methods will accept any iterable as an argument, not just a set:

6 I'm trying to use set().issubset() for comparison of sequences. As you can imagine, it's not working as expected ;) In advance: sorry for the long code-blob.

Mar 15, 2022 · i need to use issubset() and difference() on lists but i can't translate my list in sets, because my list has duplicates and they are very important. For exemple, i have list1 = [33, 33, 100, …

May 14, 2013 · Thank you for the answer -- your list comprehension is more pythonic than my for loop -- but your answer does not account for the subset part of the problem. row in searchKey does not …

Mar 29, 2016 · I have 2 columns from 2 different dataframes. I want to check if column 1 is a subset of column 2. I was using the following code: set(col1).issubset(set(col2)) The issue with this is that if col...

bool isSubset = t2.All(elem => t1.Contains(elem)); If you can find a single element in t2 that isn't in t1, then you know that t2 is not a subset of t1. The advantage of this method is that it is done all in …

Dec 28, 2014 · The complexity of B.issubset(A) is O(len(B)), assuming that e in A is constant-time. This a reasonable assumption generally, but can be easily violated with a bad hash function. If, for …

Oct 21, 2018 · Python issubset on a set of sets Asked 7 years, 3 months ago Modified 7 years, 3 months ago Viewed 452 times

all(i in bar for i in foo) where bar is a dictionary and foo is a list by using set.issubset (with a conversion to set of foo to be able to use foo.issubset(bar)), and didn't succeed to beat the times of the all …