How to check if all elements of 1 list are in the *same quantity* and in any order, in the list2? ...
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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?
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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
edited 4 hours ago
petezurich
3,76581936
asked 4 hours ago
Vitor OliveiraVitor Oliveira
475
add a comment |
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
edited 4 hours ago
petezurich
3,76581936
asked 4 hours ago
Vitor OliveiraVitor Oliveira
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
add a comment |
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
edited 4 hours ago
petezurich
3,76581936
asked 4 hours ago
Vitor OliveiraVitor Oliveira
475
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
python python-3.x
edited 4 hours ago
petezurich
3,76581936
asked 4 hours ago
Vitor OliveiraVitor Oliveira
475
edited 4 hours ago
petezurich
3,76581936
asked 4 hours ago
Vitor OliveiraVitor Oliveira
475
edited 4 hours ago
petezurich
3,76581936
edited 4 hours ago
petezurich
3,76581936
edited 4 hours ago
petezurich
3,76581936
3,76581936
asked 4 hours ago
Vitor OliveiraVitor Oliveira
475
asked 4 hours ago
Vitor OliveiraVitor Oliveira
475
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
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.
answered 4 hours ago
gilchgilch
4,3101716
1
that's slick @gilch
– modesitt
4 hours ago
Does this work for something likea_all_in_b([1], [1, 1])?
– Tomothy32
4 hours ago
@Tomothy32 It should returnFalsein 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
3 hours ago
|
show 3 more comments
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.
answered 3 hours ago
Patrick ArtnerPatrick Artner
26k62544
It works too. Thanks !
– Vitor Oliveira
2 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered 4 hours ago
gilchgilch
4,3101716
1
that's slick @gilch
– modesitt
4 hours ago
Does this work for something likea_all_in_b([1], [1, 1])?
– Tomothy32
4 hours ago
@Tomothy32 It should returnFalsein 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
3 hours ago
|
show 3 more comments
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.
answered 4 hours ago
gilchgilch
4,3101716
1
that's slick @gilch
– modesitt
4 hours ago
Does this work for something likea_all_in_b([1], [1, 1])?
– Tomothy32
4 hours ago
@Tomothy32 It should returnFalsein 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
3 hours ago
|
show 3 more comments
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.
answered 4 hours ago
gilchgilch
4,3101716
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.
answered 4 hours ago
gilchgilch
4,3101716
edited 3 hours ago
answered 4 hours ago
gilchgilch
4,3101716
answered 4 hours ago
gilchgilch
4,3101716
answered 4 hours ago
gilchgilch
4,3101716
4,3101716
1
that's slick @gilch
– modesitt
4 hours ago
Does this work for something likea_all_in_b([1], [1, 1])?
– Tomothy32
4 hours ago
@Tomothy32 It should returnFalsein 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
3 hours ago
|
show 3 more comments
1
that's slick @gilch
– modesitt
4 hours ago
Does this work for something likea_all_in_b([1], [1, 1])?
– Tomothy32
4 hours ago
@Tomothy32 It should returnFalsein 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
3 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
3 hours ago
Also it should be
<=, not <.– user2357112
3 hours ago
|
show 3 more comments
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.
answered 3 hours ago
Patrick ArtnerPatrick Artner
26k62544
It works too. Thanks !
– Vitor Oliveira
2 hours ago
add a comment |
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.
answered 3 hours ago
Patrick ArtnerPatrick Artner
26k62544
It works too. Thanks !
– Vitor Oliveira
2 hours ago
add a comment |
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.
answered 3 hours ago
Patrick ArtnerPatrick Artner
26k62544
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.
answered 3 hours ago
Patrick ArtnerPatrick Artner
26k62544
answered 3 hours ago
Patrick ArtnerPatrick Artner
26k62544
answered 3 hours ago
Patrick ArtnerPatrick Artner
26k62544
answered 3 hours ago
Patrick ArtnerPatrick Artner
26k62544
26k62544
It works too. Thanks !
– Vitor Oliveira
2 hours ago
add a comment |
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
add a comment |
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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