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Categorize numbers of an array into different sets and count the occurences

I have a array of numbers. lets say A = [9,10,16,19]

A = [9,10,16,19]

I have four sets {8,9,10} , {11,12,13} ,{14,15,16}, {17,18,19,20}

{8,9,10} , {11,12,13} ,{14,15,16}, {17,18,19,20}

for each element in array A, categorize the element into the given sets and count the occurences correspond to each set.

considering the example i gave the output should be like [2,0,1,1]

what is the efficient way to do this??

{8,9,10} , {11,12,13} ,{14,15,16}, {17,18,19,20}

2 Answers
2

I've used the set.intersection() method of set to do this:

set.intersection()

set

# Sets are generated with curly brackets
# or the set() constructor
mysets = [{8,9,10} , {11,12,13} , {14,15,16}, {17,18,19,20}]

# A is a list, which is a very different datatype
# For this problem here, it wouldn't make a difference if A
# were a set as well
A = [9,10,16,19]

[len(s.intersection(A)) for s in mysets]
>> [2, 0, 1, 1]


s0,s1,s2,s3 = [8,9,10] , [11,12,13] ,[14,15,16], [17,18,19,20]
sets = [s0,s1,s2,s3] #(these aren't sets, but ok)
A= [9,12,16,19]

numbers =
for s in sets:
n = sum(1 for x in s if x in A)
numbers.append(n)

# numbers[i] is the number of shared elements between A and sets[i]
# you should consider extension to duplicate elements.
# as written this may not give you the result you want on duplicate elements


The same thing as a one liner would read

numbers = [sum(1 for x in s if x in A) for s in sets]


Consider what you want to do with duplicate elements carefully. You may need to modify this

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