New in Python Vectors Need Advice

Question

I have an exam in two days and we an example from teacher, but I'm new to Python, and I kind of blocked at this

This is the problem statement:

Write an algorithm which will get the parameters two array of integers x[1..n] if y[1..n] which returns number of common elements in the two arrays.

def apartine(a,e):
    e=1
    gasit=False
    n=len(a)
    i=0
    while i<=n and gasit==False:
        if a[i]==e:
            gasit=True
        else:
            i=i+1
    return gasit

def intersectie(a,b,c):
    e=0
    k=len(c)
    m=len(b)
    for i in range(1,m):
        if apartine(a,e)==True:
            k=k+1
            c.append(b[i])
            print(c)
    return c

a=[1,3,5,7]
b=[2,3,4,6]
c=[]
print intersectie(a,b,c)

I'm stuck with this.

I need some help to find out what I'm doing wrong.

That's a poorly-written problem statement. I suspect that your teacher means "list" rather than "array". (They're not the same thing!) — Mark Dickinson, Jan 26 '16 at 12:47
Also, Python lists are 0-based so it's strange that your teacher wrote `x[1..n]`. — interjay, Jan 26 '16 at 12:49
Hmm; maybe the problem statement isn't supposed to be specific to Python? — Mark Dickinson, Jan 26 '16 at 12:50
Please explain the question properly. Also, the code is unreadable too. Please write comments to explain it (What is `c`, `apartine`, `gasit` etc. ?) — Anmol Singh Jaggi, Jan 26 '16 at 12:50

score 3 · Answer 1 · answered Jan 26 '16 at 12:47

3

A simple list comprehension can do that :

c=[i for i in a if i in b]

answered Jan 26 '16 at 12:47

CoMartel

3,521
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That is O(N^2), so it won't scale very well. – juanchopanza Jan 26 '16 at 12:54
True, but this was not an issue here :) – CoMartel Jan 26 '16 at 13:19

Vlad · Answer 2 · 2016-01-26T12:58:04.970

One issue you have in apartine() is that a[len(a)] (a[i] in the last run through the while loop) will raise an exception. Valid indices are from 0 to len(a)-1.

e=1 means apartine() will only look to see if 1 is in a, rather than whatever was passed.

You want to call apartine(a, b[i]), e doesn't seem to have any meaningful value.

You're not using k.

A very simple way to implement the whole thing in Python would be:

c = list(set(a) & set(b))

but I assume that's not what you're looking for.

score 0 · Answer 3 · answered Jan 26 '16 at 12:52

If the position in the lists doesn't matter you can get the common elements with a simple comprehension list

commons = [x for x in a if x in b]

If the position matters, you could use the index

commons = [a[i] for i in range(len(a)) if a[i] == b[i]]

The number of elements in common would be the length of this list

print(len(commmons))

score 0 · Answer 4 · answered Jan 26 '16 at 13:05

If you're having difficulty following the list comprehensions, here's basically the same thing written as a for loop:

def intersect(a, b):
    """Return a list of all elements common to a & b
    """
    intersection = []
    for element in a:
        if element in b and element not in intersection:
            intersection.append(element)
    return intersection


if __name__ == '__main__':
    test_pairs = (
        ([1, 3, 5, 7], [2, 3, 4, 6]),
        ([3], [1, 2, 3]),
        ([1, 2, 3], [3]),
        ([3, 3, 3], [3]),
        ([3], [3, 3, 3]),
        ([1, 2, 3], [4, 5, 6])
    )
    for a, b in test_pairs:
        print "Intersection of %s and %s is %s" % (a, b, intersect(a,b))

The output is:

Intersection of [1, 3, 5, 7] and [2, 3, 4, 6] is [3]
Intersection of [3] and [1, 2, 3] is [3]
Intersection of [1, 2, 3] and [3] is [3]
Intersection of [3, 3, 3] and [3] is [3]
Intersection of [3] and [3, 3, 3] is [3]
Intersection of [1, 2, 3] and [4, 5, 6] is []

The methodology is to take each element from one list (it doesn't matter which one) and see if it exists in the other list. Collect these elements in a third list which is your result. As you can see, this duplicates are removed by the element not in intersection clause.

New in Python Vectors Need Advice

4 Answers4