## 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

Python, Given a list in Python and a number x, count number of occurrences of x in the Counter method returns a dictionary with occurrences of all elements as a See your article appearing on the GeeksforGeeks main page and help other Geeks. Given a list in Python and a number x, count number of occurrences of x in the given list. Examples: Input : lst = [15, 6, 7, 10, 12, 20, 10, 28, 10] x = 10 Output : 3 10 appears three times in given list.

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))
```

C++ Exercises: Count the number of occurrences of given number in , Counting the number of times a value exists within a list is a common task, take a look at three different options: list comprehension, filter() , and the The second line is the for loop that'll assign each item in numbers to the  Count occurrences of a number in a sorted array with duplicates using Binary Search - Duration: 12:25. mycodeschool 177,406 views

Count occurrences of a number in sorted array with duplicates , I would like to be able to return the count of occurences of each element in a vector. I want to count the different number of elements in an array, whether unique or repeating but different, for example I numbers=unique(v); %list of elements. Java Basic: Exercise-92 with Solution. Write a Java program to count the number of even and odd elements in a given array of integers. Pictorial Presentation:

How To Count Occurrences in a Python List, This is a Java Program to Count the Number of Occurrence of an Element in an Here's the list of Best Reference Books in Java Programming, Data Structures if a given Integer X appears more than N/2 times in a Sorted Array of N Integers. I actually need to count the occurance of pairs in the second column given that they have a similar number in the first column. That means column one defines clusters and I would like to count the frequency of the pairs in the clusters.

How can I count the occurrences of each element in a vector in , The count() method returns the number of occurrences of an element in a list. In simple terms, count() method counts how many times an element has occurred in a  Previous: Write a Python program to find whether a given number (accept from the user) is even or odd, print out an appropriate message to the user. Next : Write a Python program to get the n (non-negative integer) copies of the first 2 characters of a given string.

• 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
• 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.
• 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