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How to check if all elements of 1 list are in the *same quantity* and in any order, in the list2?



The Next CEO of Stack OverflowChecking if list is a sublistHow do I check if a list is empty?How to generate all permutations of a list in PythonHow do you remove duplicates from a list whilst preserving order?How do I remove an element from a list by index in Python?How do I get the number of elements in a list in Python?How do I list all files of a directory?check if all elements in a list are identicalHow to check if all items in a list are there in another list?Checking for sublists in list of lists preserving sequencePrint 'x' amount of items 'y' times?










6















I know its a very common question at first, but I haven't found one that specific. (If you do, please tell me.) And all ways I found didnt work for me.
I need to check if all elements of list 1 appears in the same amount in the list2.



Ex :



#If list1 = [2,2,2,6] 
# and list2 =[2,6,2,5,2,4]
#then all list1 are in list2.
#If list2 = [2,6] then all list1 are not in list2.


i'm trying this way :



list1 = [6,2]

import itertools

for i in itertools.product((2,4,5,1), repeat=3) :
asd = i[0] + i[1]
asd2= i[1] + i[2]

list2 = [asd, asd2]
if all(elem in list2 for elem in list1):
print (i,list2)


It works when the elements are not repeated in the list1, like [1,2]. But when they are repeated, all repeated elements is beeing counted as only 1 : [2,2,2] its beeing understanded as [2]. Or so i think.










share|improve this question



















  • 1





    After reading both questions, it does not look like a duplicate to me. This question cares about quantity, but not order. The other one cares about order, but not quantity.

    – gilch
    4 hours ago















6















I know its a very common question at first, but I haven't found one that specific. (If you do, please tell me.) And all ways I found didnt work for me.
I need to check if all elements of list 1 appears in the same amount in the list2.



Ex :



#If list1 = [2,2,2,6] 
# and list2 =[2,6,2,5,2,4]
#then all list1 are in list2.
#If list2 = [2,6] then all list1 are not in list2.


i'm trying this way :



list1 = [6,2]

import itertools

for i in itertools.product((2,4,5,1), repeat=3) :
asd = i[0] + i[1]
asd2= i[1] + i[2]

list2 = [asd, asd2]
if all(elem in list2 for elem in list1):
print (i,list2)


It works when the elements are not repeated in the list1, like [1,2]. But when they are repeated, all repeated elements is beeing counted as only 1 : [2,2,2] its beeing understanded as [2]. Or so i think.










share|improve this question



















  • 1





    After reading both questions, it does not look like a duplicate to me. This question cares about quantity, but not order. The other one cares about order, but not quantity.

    – gilch
    4 hours ago













6












6








6








I know its a very common question at first, but I haven't found one that specific. (If you do, please tell me.) And all ways I found didnt work for me.
I need to check if all elements of list 1 appears in the same amount in the list2.



Ex :



#If list1 = [2,2,2,6] 
# and list2 =[2,6,2,5,2,4]
#then all list1 are in list2.
#If list2 = [2,6] then all list1 are not in list2.


i'm trying this way :



list1 = [6,2]

import itertools

for i in itertools.product((2,4,5,1), repeat=3) :
asd = i[0] + i[1]
asd2= i[1] + i[2]

list2 = [asd, asd2]
if all(elem in list2 for elem in list1):
print (i,list2)


It works when the elements are not repeated in the list1, like [1,2]. But when they are repeated, all repeated elements is beeing counted as only 1 : [2,2,2] its beeing understanded as [2]. Or so i think.










share|improve this question
















I know its a very common question at first, but I haven't found one that specific. (If you do, please tell me.) And all ways I found didnt work for me.
I need to check if all elements of list 1 appears in the same amount in the list2.



Ex :



#If list1 = [2,2,2,6] 
# and list2 =[2,6,2,5,2,4]
#then all list1 are in list2.
#If list2 = [2,6] then all list1 are not in list2.


i'm trying this way :



list1 = [6,2]

import itertools

for i in itertools.product((2,4,5,1), repeat=3) :
asd = i[0] + i[1]
asd2= i[1] + i[2]

list2 = [asd, asd2]
if all(elem in list2 for elem in list1):
print (i,list2)


It works when the elements are not repeated in the list1, like [1,2]. But when they are repeated, all repeated elements is beeing counted as only 1 : [2,2,2] its beeing understanded as [2]. Or so i think.







python python-3.x






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 4 hours ago









petezurich

3,76581936




3,76581936










asked 4 hours ago









Vitor OliveiraVitor Oliveira

475




475







  • 1





    After reading both questions, it does not look like a duplicate to me. This question cares about quantity, but not order. The other one cares about order, but not quantity.

    – gilch
    4 hours ago












  • 1





    After reading both questions, it does not look like a duplicate to me. This question cares about quantity, but not order. The other one cares about order, but not quantity.

    – gilch
    4 hours ago







1




1





After reading both questions, it does not look like a duplicate to me. This question cares about quantity, but not order. The other one cares about order, but not quantity.

– gilch
4 hours ago





After reading both questions, it does not look like a duplicate to me. This question cares about quantity, but not order. The other one cares about order, but not quantity.

– gilch
4 hours ago












2 Answers
2






active

oldest

votes


















7














Use collections.Counter to convert to a dict_items view Set of (value, count) pairs. Then you can use normal set operations.



from collections import Counter

def a_all_in_b(a, b):
"""True only if all elements of `a` are in `b` in the *same quantity* (in any order)."""
return Counter(a).items() <= Counter(b).items()


Note that Counter only works on hashable elements because it's a subclass of dict.






share|improve this answer




















  • 1





    that's slick @gilch

    – modesitt
    4 hours ago











  • Does this work for something like a_all_in_b([1], [1, 1])?

    – Tomothy32
    4 hours ago











  • @Tomothy32 It should return False in that case, because the 1's are not "in the same quantity".

    – gilch
    4 hours ago












  • @gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.

    – Tomothy32
    4 hours ago






  • 1





    Also it should be <=, not <.

    – user2357112
    4 hours ago


















1














Modify this answer to Checking if list is a sublist to check for equality of occurences:



from collections import Counter 

list1 = [2,2,2,6]
list2 =[2,6,2,5,2,4]

def same_amount(a,b):
c1 = Counter(a)
c2 = Counter(b)

for key,value in c1.items():
if c2[key] != value:
return False
return True


print(same_amount(list1,list2))
print(same_amount(list1 + [2],list2))


Output:



True
False


There is almost no transfere-knowledge needed to create this answer, thats why I suggested it as dupe. This question is simply a more specific case of what Checking if list is a sublist discussed.






share|improve this answer























  • It works too. Thanks !

    – Vitor Oliveira
    2 hours ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









7














Use collections.Counter to convert to a dict_items view Set of (value, count) pairs. Then you can use normal set operations.



from collections import Counter

def a_all_in_b(a, b):
"""True only if all elements of `a` are in `b` in the *same quantity* (in any order)."""
return Counter(a).items() <= Counter(b).items()


Note that Counter only works on hashable elements because it's a subclass of dict.






share|improve this answer




















  • 1





    that's slick @gilch

    – modesitt
    4 hours ago











  • Does this work for something like a_all_in_b([1], [1, 1])?

    – Tomothy32
    4 hours ago











  • @Tomothy32 It should return False in that case, because the 1's are not "in the same quantity".

    – gilch
    4 hours ago












  • @gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.

    – Tomothy32
    4 hours ago






  • 1





    Also it should be <=, not <.

    – user2357112
    4 hours ago















7














Use collections.Counter to convert to a dict_items view Set of (value, count) pairs. Then you can use normal set operations.



from collections import Counter

def a_all_in_b(a, b):
"""True only if all elements of `a` are in `b` in the *same quantity* (in any order)."""
return Counter(a).items() <= Counter(b).items()


Note that Counter only works on hashable elements because it's a subclass of dict.






share|improve this answer




















  • 1





    that's slick @gilch

    – modesitt
    4 hours ago











  • Does this work for something like a_all_in_b([1], [1, 1])?

    – Tomothy32
    4 hours ago











  • @Tomothy32 It should return False in that case, because the 1's are not "in the same quantity".

    – gilch
    4 hours ago












  • @gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.

    – Tomothy32
    4 hours ago






  • 1





    Also it should be <=, not <.

    – user2357112
    4 hours ago













7












7








7







Use collections.Counter to convert to a dict_items view Set of (value, count) pairs. Then you can use normal set operations.



from collections import Counter

def a_all_in_b(a, b):
"""True only if all elements of `a` are in `b` in the *same quantity* (in any order)."""
return Counter(a).items() <= Counter(b).items()


Note that Counter only works on hashable elements because it's a subclass of dict.






share|improve this answer















Use collections.Counter to convert to a dict_items view Set of (value, count) pairs. Then you can use normal set operations.



from collections import Counter

def a_all_in_b(a, b):
"""True only if all elements of `a` are in `b` in the *same quantity* (in any order)."""
return Counter(a).items() <= Counter(b).items()


Note that Counter only works on hashable elements because it's a subclass of dict.







share|improve this answer














share|improve this answer



share|improve this answer








edited 3 hours ago

























answered 4 hours ago









gilchgilch

4,3101716




4,3101716







  • 1





    that's slick @gilch

    – modesitt
    4 hours ago











  • Does this work for something like a_all_in_b([1], [1, 1])?

    – Tomothy32
    4 hours ago











  • @Tomothy32 It should return False in that case, because the 1's are not "in the same quantity".

    – gilch
    4 hours ago












  • @gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.

    – Tomothy32
    4 hours ago






  • 1





    Also it should be <=, not <.

    – user2357112
    4 hours ago












  • 1





    that's slick @gilch

    – modesitt
    4 hours ago











  • Does this work for something like a_all_in_b([1], [1, 1])?

    – Tomothy32
    4 hours ago











  • @Tomothy32 It should return False in that case, because the 1's are not "in the same quantity".

    – gilch
    4 hours ago












  • @gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.

    – Tomothy32
    4 hours ago






  • 1





    Also it should be <=, not <.

    – user2357112
    4 hours ago







1




1





that's slick @gilch

– modesitt
4 hours ago





that's slick @gilch

– modesitt
4 hours ago













Does this work for something like a_all_in_b([1], [1, 1])?

– Tomothy32
4 hours ago





Does this work for something like a_all_in_b([1], [1, 1])?

– Tomothy32
4 hours ago













@Tomothy32 It should return False in that case, because the 1's are not "in the same quantity".

– gilch
4 hours ago






@Tomothy32 It should return False in that case, because the 1's are not "in the same quantity".

– gilch
4 hours ago














@gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.

– Tomothy32
4 hours ago





@gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.

– Tomothy32
4 hours ago




1




1





Also it should be <=, not <.

– user2357112
4 hours ago





Also it should be <=, not <.

– user2357112
4 hours ago













1














Modify this answer to Checking if list is a sublist to check for equality of occurences:



from collections import Counter 

list1 = [2,2,2,6]
list2 =[2,6,2,5,2,4]

def same_amount(a,b):
c1 = Counter(a)
c2 = Counter(b)

for key,value in c1.items():
if c2[key] != value:
return False
return True


print(same_amount(list1,list2))
print(same_amount(list1 + [2],list2))


Output:



True
False


There is almost no transfere-knowledge needed to create this answer, thats why I suggested it as dupe. This question is simply a more specific case of what Checking if list is a sublist discussed.






share|improve this answer























  • It works too. Thanks !

    – Vitor Oliveira
    2 hours ago















1














Modify this answer to Checking if list is a sublist to check for equality of occurences:



from collections import Counter 

list1 = [2,2,2,6]
list2 =[2,6,2,5,2,4]

def same_amount(a,b):
c1 = Counter(a)
c2 = Counter(b)

for key,value in c1.items():
if c2[key] != value:
return False
return True


print(same_amount(list1,list2))
print(same_amount(list1 + [2],list2))


Output:



True
False


There is almost no transfere-knowledge needed to create this answer, thats why I suggested it as dupe. This question is simply a more specific case of what Checking if list is a sublist discussed.






share|improve this answer























  • It works too. Thanks !

    – Vitor Oliveira
    2 hours ago













1












1








1







Modify this answer to Checking if list is a sublist to check for equality of occurences:



from collections import Counter 

list1 = [2,2,2,6]
list2 =[2,6,2,5,2,4]

def same_amount(a,b):
c1 = Counter(a)
c2 = Counter(b)

for key,value in c1.items():
if c2[key] != value:
return False
return True


print(same_amount(list1,list2))
print(same_amount(list1 + [2],list2))


Output:



True
False


There is almost no transfere-knowledge needed to create this answer, thats why I suggested it as dupe. This question is simply a more specific case of what Checking if list is a sublist discussed.






share|improve this answer













Modify this answer to Checking if list is a sublist to check for equality of occurences:



from collections import Counter 

list1 = [2,2,2,6]
list2 =[2,6,2,5,2,4]

def same_amount(a,b):
c1 = Counter(a)
c2 = Counter(b)

for key,value in c1.items():
if c2[key] != value:
return False
return True


print(same_amount(list1,list2))
print(same_amount(list1 + [2],list2))


Output:



True
False


There is almost no transfere-knowledge needed to create this answer, thats why I suggested it as dupe. This question is simply a more specific case of what Checking if list is a sublist discussed.







share|improve this answer












share|improve this answer



share|improve this answer










answered 4 hours ago









Patrick ArtnerPatrick Artner

26k62544




26k62544












  • It works too. Thanks !

    – Vitor Oliveira
    2 hours ago

















  • It works too. Thanks !

    – Vitor Oliveira
    2 hours ago
















It works too. Thanks !

– Vitor Oliveira
2 hours ago





It works too. Thanks !

– Vitor Oliveira
2 hours ago

















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