CODEWITHSUNDEEP
python
ajpieri
ajpieri

Reputation: 347

Python - Loop over all instances of multiple lists

Is there a better way for looping over every combination of multiple lists in Python? For example...

list1 = [1,2,3,4]
list2 = [6,7,8]

for i in list1:
   for j in list2:
      print(str(i) + ", " + str(j))

1, 6
1, 7
1, 8
2, 6
2, 7
2, 8
3, 6
3, 7
3, 8
4, 6
4, 7
4, 8

I ask because I would like to break out of both loops once a value is found. I do not want to use a bool flag to break out of the top level loop. All answers I have seen so far say to use zip, but that is not the same thing. zip would produce the following.

1, 6
2, 7
3, 8

If you use map, you get the following, which is also not what I am looking for.

1, 6
2, 7
3, 8
4, None

Upvotes: 1

Views: 3228

Answers (6)

nigel222
nigel222

Reputation: 8222

If you need to know where you were in the scan when the break condition arose, you can use list comprehensions and enumerate together

for i,a,j,b in [ x+y for x in enumerate([11,12,13,14]) for y in enumerate([16,17,18]) ]:
    print( i,a,j,b)
    if a==13 and b==17: break
print(i,j) # get 2 1

Upvotes: 1

Farhan.K
Farhan.K

Reputation: 3485

If you don't want to use itertools.product as suggested in another answer, you can wrap it in a function and return:

list1 = [1,2,3,4]
list2 = [6,7,8]

def findNumbers(x, y):
    for i in list1:
       for j in list2:
          print(str(i) + ", " + str(j))
          if (x, y) == (i, j):
              return (x, y)

Output:

>>> findNumbers(2, 7)
1, 6
1, 7
1, 8
2, 6
2, 7
(2, 7)

Upvotes: 1

Bill Bell
Bill Bell

Reputation: 21663

You might like to see product used in a simple code.

  • In this case product is an iterator that returns elements of the cross-product of list1 and list2 one at a time.
  • Within the for-loop we watch for the appearance of a certain pair, and break out of the loop if and when we encounter it.

--

>>> list1 = [1,2,3,4]
>>> list2 = [6,7,8]
>>> from itertools import product
>>> for i, j in product(list1, list2):
...     if (i,j)==(2,7):
...         print (i,j)
...         break
...     
2 7

Upvotes: 0

kellymandem
kellymandem

Reputation: 1769

Have you tried using a list comprehension

[(x, y) for x in [1,2,3,4]
 for y in [6,7,8]]

Upvotes: 2

Remolten
Remolten

Reputation: 2682

You can use itertools.product like so:

list1 = [1,2,3,4]
list2 = [6,7,8]
find_this_value = (1, 8)

found_value = False
for permutation in itertools.product(list1, list2):
    if permutation == find_this_value:
        found_value = True
        break

if found_value:
    pass  # Take action

itertools.product returns a generator with all of the possible permutations of the 2 lists. Then, you simply iterate over those, and search until you find the value you want.

Upvotes: 2

omri_saadon
omri_saadon

Reputation: 10669

You can use the product function.

You can read more in here

Roughly equivalent to nested for-loops in a generator expression

import itertools
print (list(itertools.product(list1, list2)))
# [(1, 6), (1, 7), (1, 8), (2, 6), (2, 7), (2, 8), (3, 6), (3, 7), (3, 8), (4, 6), (4, 7), (4, 8)]

Upvotes: 0

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