## Count total number of occurrences of given list of integers in another

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How do I count the number of times the same integer occurs?

My code so far:

def searchAlgorithm (target, array): i = 0 #iterating through elements of target list q = 0 #iterating through lists sublists via indexes while q < 4: x = 0 #counting number of matches for i in target: if i in array[q]: x += 1 else: x == 0 print(x) q += 1 a = [8, 12, 14, 26, 27, 28] b = [[4, 12, 17, 26, 30, 45], [8, 12, 19, 24, 33, 47], [3, 10, 14, 31, 39, 41], [4, 12, 14, 26, 30, 45]] searchAlgorithm(a, b)

The output of this is:

2 2 1 3

What I want to achieve is counting the number of times '1', '2' '3' matches occurs.

I have tried:

v = 0 if searchAlgorithm(a, b) == 2: v += 1 print(v)

But that results in `0`

You can use intersection of sets to find elements that are common in both lists. Then you can get the length of the sets. Here is how it looks:

num_common_elements = (len(set(a).intersection(i)) for i in b)

You can then iterate over the generator `num_common_elements`

to use the values. Or you can cast it to a list to see the results:

print(list(num_common_elements)) [Out]: [2, 2, 1, 3]

If you want to implement the intersection functionality yourself, you can use the sum method to implement your own version. This is equivalent to doing `len(set(x).intersection(set(y))`

sum(i in y for i in x)

This works because it generates values such as `[True, False, False, True, True]`

representing where the values in the first list are present in the second list. The `sum`

method then treats the `True`

s as 1s and `False`

s as 0s, thus giving you the size of the intersection set

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This is based on what I understand from your question. Probably you are looking for this:

from collections import Counter def searchAlgorithm (target, array): i = 0 #iterating through elements of target list q = 0 #iterating through lists sublists via indexes lst = [] while q < 4: x = 0 #counting number of matches for i in target: if i in array[q]: x += 1 else: x == 0 lst.append(x) q += 1 print(Counter(lst)) a = [8, 12, 14, 26, 27, 28] b = [[4, 12, 17, 26, 30, 45], [8, 12, 19, 24, 33, 47], [3, 10, 14, 31, 39, 41], [4, 12, 14, 26, 30, 45]] searchAlgorithm(a, b) # Counter({2: 2, 1: 1, 3: 1})

**Count number of occurrences (or frequency) in a sorted array ,** C++ Exercises, Practice and Solution: Write a C++ program to count the number of occurrences of given number in a sorted array of integers. Given a list of integers, count and output the number of times each value appears as a list of space-separated integers. Function Description. Complete the countingSort function in the editor below. It should return an array of integers where each value is the number of occurrences of the element's index value in the original array.

Thanks to some for their helpful feedback, I have since come up a more simplified solution that does exactly what I want.

By storing the `results`

of the matches in a list, I can then return the list out of the `searchAlgorithm`

function and simple use `.count()`

to count all the matches of a specific number within the list.

def searchAlgorithm (target, array): i = 0 q = 0 results = [] while q < 4: x = 0 #counting number of matches for i in target: if i in array[q]: x += 1 else: x == 0 results.append(x) q += 1 return results a = [8, 12, 14, 26, 27, 28] b = [[4, 12, 17, 26, 30, 45], [8, 12, 19, 24, 33, 47], [3, 10, 14, 31, 39, 41], [4, 12, 14, 26, 30, 45]] searchAlgorithm(a, b) d2 = (searchAlgorithm(winNum, lotto).count(2))

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##### Comments

- I don't understand your question. What is it that you are trying to do exactly, and what is the difficulty?
- Oh. So, yeah, your functions doesn't
*do anything except print*. you probably want to accumulate your results into a list and return that list form your function. As an aside, you probably don't want to use a`while`

loop. You already seem to know how to use a for-loop, so just stick with that when you know how many times you are goping to iterate - I am wanting to count the number of times a match has occurred.
- So do you want to count the number of twos, threes and ones in the output? Is that what you are looking for?
- You forgot to return
`q`

out of your`searchAlgorithm`

function. That is why v is never incremented.`None`

(no return) is never equal to 2. - I am trying to avoid the use of sets, as I'm wanting to write it as an algorithm. So while sets would work it's not my preferred solution.
- Note, your
`sum`

and`set`

based solutions*are not equivalent*. As an aside, you don't need`set(i)`

, the`.intersection`

method takes any iterable. - Yes you are correct. They would only be equivalent if there are no elements that are duplicates in both lists. In which case, the
`sum`

solution is more correct