Compare values of two arrays in python

Question

How can i check if item in b is in a and the found match item in a should not be use in the next matching?

Currently this code will match both 2 in b.

a = [3,2,5,4]
b = [2,4,2]

for i in b:
  if i in a:
    print "%d is in a" % i

This is the required output:

2 => 2 is in a
4 => 4 is in a
2 =>

EDIT: Example 2:

a = [3,2,2,4]
b = [2,4,2]

output should be

2 => 2 is in a
4 => 4 is in a
2 => 2 is in a

Mihai Maruseac · Accepted Answer · 2012-07-11T11:15:12.750

(long post but read it entirely, solution is at the end).

Remove the found value or register it in another dict.

Better though is to count the number of apparitions inside each array and test how many are common.

For the second case, you'd have

for a:

3 appears 1 times 2 appears 1 times 5 appears 1 times 4 appears 1 times
for b:

2 appears 2 times 4 appears 1 times

Keep these values in dictionaries:

a_app = {3:1, 2:1, 5:1, 4:1}
b_app = {2:2, 4:1}

And now, it is simple:

for i in b:
    if a_app.has_key(i) and a_app[i] > 0:
        a_app[i] -= 1

The b_app dictionary would be used in other case.

Here is a test script I wrote (testing all testcases issued here):

def f(a, b):
    a_app = {}
    for i in a:
        if not a_app.has_key(i):
            a_app[i] = 0
        a_app[i] += 1
    print a_app
    for i in b:
        print i, '=>',
        if a_app.has_key(i) and a_app[i] > 0:
            a_app[i] -= 1
            print i, ' is in a',
        print '.'

f([1,1,2],[1,1])
f([3,2,5,4],[2,4,2])
f([3,2,2,4],[2,4,2])
f([3,2,5,4],[2,3,2])

And here is the output:

$ python 1.py
{1: 2, 2: 1}
1 => 1  is in a .
1 => 1  is in a .
{2: 1, 3: 1, 4: 1, 5: 1}
2 => 2  is in a .
4 => 4  is in a .
2 => .
{2: 2, 3: 1, 4: 1}
2 => 2  is in a .
4 => 4  is in a .
2 => 2  is in a .
{2: 1, 3: 1, 4: 1, 5: 1}
2 => 2  is in a .
3 => 3  is in a .
2 => .

Everything is perfect and no order is lost :)

Edit: Updated with @Avaris's suggestions, this script looks like:

import collections

def f(a, b):
    a_app = collections.Counter(a)
    for i in b:
        print i, '=>',
        if i in a_app and a_app[i] > 0:
            a_app[i] -= 1
            print i, ' is in a',
        print '.'
    print ''

f([1,1,2],[1,1])
f([3,2,5,4],[2,4,2])
f([3,2,2,4],[2,4,2])
f([3,2,5,4],[2,3,2])

Nice but loses order of the values found - not sure if that is important. — Maria Zverina, Jul 11 '12 at 11:01
You should use `i in a_app` instead of `a_app.has_key(i)`. `has_key` is removed in 3.x. And for 2.7+ `collections.Counter` will do nicely for `a_app` or `b_app`. — Avaris, Jul 11 '12 at 11:10

Tim Pietzcker · Answer 2 · 2012-07-11T10:53:02.930

3

I would do this:

a = [3,2,5,4]
b = [2,4,2]
temp = set(a)
for item in b:
    if item in temp:
        print "{0} is in a".format(item)
        temp.remove(item)

The set makes the x in y check faster (O(1) instead of (worst-case) O(n)), and it also can be modified safely without destroying my original a.

edited Jul 11 '12 at 10:53

answered Jul 11 '12 at 10:42

Tim Pietzcker

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fails with `a = [1,1,2], b = [1,1]` – Maria Zverina Jul 11 '12 at 10:49
1

@MariaZverina: How do you mean? Output is `1 is in a` - what else would you have expected? – Tim Pietzcker Jul 11 '12 at 10:54
1

`set` will modify the value of `a`. When my variable `a=[3,2,2,4]` both 2 in `b` should match. – Bob Jul 11 '12 at 10:56
I guess `1` should be matched twice since `a` has 2 `1`s in it. Depends on what OP wants though. – Avaris Jul 11 '12 at 10:57
@Avaris: Well, he wrote that the found match should not be used again, so if `1` has been matched once, it should not be matched again. – Tim Pietzcker Jul 11 '12 at 10:58
2

@Bob: Then you should edit your question to specify your problem correctly. `set()` does not modify the value of `a`, by the way. – Tim Pietzcker Jul 11 '12 at 10:58
2

He stated it clearly: he wanted to match each value exactly once even though he had duplicates. – Mihai Maruseac Jul 11 '12 at 11:09
@Avaris in Maria's comment 1 should match twice. – Bob Jul 11 '12 at 11:10
@TimPietzcker Isn't it bad to modify a list during an iteration? https://unspecified.wordpress.com/2009/02/12/thou-shalt-not-modify-a-list-during-iteration/ – Mr. President Oct 16 '18 at 09:29

Maria Zverina · Answer 3 · 2012-07-11T10:58:03.313

3

Recursive solution:

a = [3,2,5,4]
b = [2,4,2]

def find_matches(x, y):
    if y == []:            # nothing more to find
        return
    n = y.pop()
    if n in x:
        print n, "matches"
        x.remove(n)
    find_matches(x, y)

find_matches(list(a), list(b))      # copy the list as they get consumed in process

edited Jul 11 '12 at 10:58

answered Jul 11 '12 at 10:45

Maria Zverina

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elssar · Answer 4 · 2012-07-11T11:27:13.593

0

This probably isn't the best way to do it, but you could make a list of all matched items and check using that.

a= [3, 2, 5, 4]
b= [2, 4, 2]
c= []

for i in b:
    if i in a and i not in c:
        print '%d is in a' % i
        c.append(i)

I think looping over a instead of b should solve the problem.

a= [3, 2, 2, 4]
b= [2, 4, 2]

for i in a:
    if i in b:
        print i, 'is in a'

Obviously i is in a, but it does solve the problem doesn't it?

edited Jul 11 '12 at 11:27

answered Jul 11 '12 at 10:44

elssar

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fails with `a = [1,1,2], b = [1,1]` – Maria Zverina Jul 11 '12 at 10:50
Fails? How? It gives the output `1 is an a` And that is the required output – elssar Jul 11 '12 at 11:09
2

The output should be given twice, you have two `1`s in `b`. – Mihai Maruseac Jul 11 '12 at 11:10
Ah, got it. The OP should edit the question to make that clear. – elssar Jul 11 '12 at 11:15
@elssar yes, already edited. My first example is not that appropriate – Bob Jul 11 '12 at 11:17
@Bob how about iterating over a instead of b. That should solve the problem, right? – elssar Jul 11 '12 at 11:24
it will not meet the requirements – Bob Jul 11 '12 at 11:42

score 0 · Answer 5 · answered Jul 11 '12 at 10:44

0

a = [3,2,5,4]
b = [2,4,2]

hadAlready = {}
for i in b:
    if i in a:
        if not (i in hadAlready):
            print "%d is in a" % i
            hadAlready[i] = 1

answered Jul 11 '12 at 10:44

1

fails with `a = [1,1,2], b = [1,1]` – Maria Zverina Jul 11 '12 at 10:59

Compare values of two arrays in python

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