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
(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])
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.
collections.Counter is available to python v2.7 up.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.
a = [1,1,2], b = [1,1]
1 is in a - what else would you have expected?set will modify the value of a. When my variable a=[3,2,2,4] both 2 in b should match.set() does not modify the value of a, by the way.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
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
a = [1,1,2], b = [1,1]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?
a = [1,1,2], b = [1,1]
1 is an a And that is the required output1s in b.