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 Mar 30 at 20:31
petezurich
3,86081936
asked Mar 30 at 19:52
Vitor OliveiraVitor Oliveira
816
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 Mar 30 at 20:31
petezurich
3,86081936
asked Mar 30 at 19:52
Vitor OliveiraVitor Oliveira
816
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
Mar 30 at 20:38
Possible duplicate of Checking if list is a sublist
– Noctis Skytower
Mar 31 at 4:04
Would it be correct to title this "Unordered comparison between lists"?
– Ben Voigt
Mar 31 at 6:12
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 Mar 30 at 20:31
petezurich
3,86081936
asked Mar 30 at 19:52
Vitor OliveiraVitor Oliveira
816
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 Mar 30 at 20:31
petezurich
3,86081936
asked Mar 30 at 19:52
Vitor OliveiraVitor Oliveira
816
edited Mar 30 at 20:31
petezurich
3,86081936
asked Mar 30 at 19:52
Vitor OliveiraVitor Oliveira
816
edited Mar 30 at 20:31
petezurich
3,86081936
edited Mar 30 at 20:31
petezurich
3,86081936
edited Mar 30 at 20:31
petezurich
3,86081936
3,86081936
asked Mar 30 at 19:52
Vitor OliveiraVitor Oliveira
816
asked Mar 30 at 19:52
Vitor OliveiraVitor Oliveira
816
asked Mar 30 at 19:52
Vitor OliveiraVitor Oliveira
816
816
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
Mar 30 at 20:38
Possible duplicate of Checking if list is a sublist
– Noctis Skytower
Mar 31 at 4:04
Would it be correct to title this "Unordered comparison between lists"?
– Ben Voigt
Mar 31 at 6:12
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
Mar 30 at 20:38
Possible duplicate of Checking if list is a sublist
– Noctis Skytower
Mar 31 at 4:04
Would it be correct to title this "Unordered comparison between lists"?
– Ben Voigt
Mar 31 at 6:12
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
Mar 30 at 20:38
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
Mar 30 at 20:38
Possible duplicate of Checking if list is a sublist
– Noctis Skytower
Mar 31 at 4:04
Possible duplicate of Checking if list is a sublist
– Noctis Skytower
Mar 31 at 4:04
Would it be correct to title this "Unordered comparison between lists"?
– Ben Voigt
Mar 31 at 6:12
Would it be correct to title this "Unordered comparison between lists"?
– Ben Voigt
Mar 31 at 6:12
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 works on hashable elements only, because it's a subclass of dict.
answered Mar 30 at 20:14
gilchgilch
4,4101817
1
that's slick @gilch
– modesitt
Mar 30 at 20:17
Does this work for something likea_all_in_b([1], [1, 1])?
– Tomothy32
Mar 30 at 20:20
@Tomothy32 It should returnFalsein that case, because the 1's are not "in the same quantity".
– gilch
Mar 30 at 20:28
@gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.
– Tomothy32
Mar 30 at 20:32
1
Also it should be<=, not<.
– user2357112
Mar 30 at 20:48
|
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 Mar 30 at 20:48
Patrick ArtnerPatrick Artner
26.6k62544
It works too. Thanks !
– Vitor Oliveira
Mar 30 at 21:51
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 works on hashable elements only, because it's a subclass of dict.
answered Mar 30 at 20:14
gilchgilch
4,4101817
1
that's slick @gilch
– modesitt
Mar 30 at 20:17
Does this work for something likea_all_in_b([1], [1, 1])?
– Tomothy32
Mar 30 at 20:20
@Tomothy32 It should returnFalsein that case, because the 1's are not "in the same quantity".
– gilch
Mar 30 at 20:28
@gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.
– Tomothy32
Mar 30 at 20:32
1
Also it should be<=, not<.
– user2357112
Mar 30 at 20:48
|
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 works on hashable elements only, because it's a subclass of dict.
answered Mar 30 at 20:14
gilchgilch
4,4101817
1
that's slick @gilch
– modesitt
Mar 30 at 20:17
Does this work for something likea_all_in_b([1], [1, 1])?
– Tomothy32
Mar 30 at 20:20
@Tomothy32 It should returnFalsein that case, because the 1's are not "in the same quantity".
– gilch
Mar 30 at 20:28
@gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.
– Tomothy32
Mar 30 at 20:32
1
Also it should be<=, not<.
– user2357112
Mar 30 at 20:48
|
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 works on hashable elements only, because it's a subclass of dict.
answered Mar 30 at 20:14
gilchgilch
4,4101817
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 works on hashable elements only, because it's a subclass of dict.
answered Mar 30 at 20:14
gilchgilch
4,4101817
edited Mar 31 at 3:55
answered Mar 30 at 20:14
gilchgilch
4,4101817
answered Mar 30 at 20:14
gilchgilch
4,4101817
answered Mar 30 at 20:14
gilchgilch
4,4101817
4,4101817
1
that's slick @gilch
– modesitt
Mar 30 at 20:17
Does this work for something likea_all_in_b([1], [1, 1])?
– Tomothy32
Mar 30 at 20:20
@Tomothy32 It should returnFalsein that case, because the 1's are not "in the same quantity".
– gilch
Mar 30 at 20:28
@gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.
– Tomothy32
Mar 30 at 20:32
1
Also it should be<=, not<.
– user2357112
Mar 30 at 20:48
|
show 3 more comments
1
that's slick @gilch
– modesitt
Mar 30 at 20:17
Does this work for something likea_all_in_b([1], [1, 1])?
– Tomothy32
Mar 30 at 20:20
@Tomothy32 It should returnFalsein that case, because the 1's are not "in the same quantity".
– gilch
Mar 30 at 20:28
@gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.
– Tomothy32
Mar 30 at 20:32
1
Also it should be<=, not<.
– user2357112
Mar 30 at 20:48
1
1
that's slick @gilch
– modesitt
Mar 30 at 20:17
that's slick @gilch
– modesitt
Mar 30 at 20:17
Does this work for something like
a_all_in_b([1], [1, 1])?– Tomothy32
Mar 30 at 20:20
Does this work for something like
a_all_in_b([1], [1, 1])?– Tomothy32
Mar 30 at 20:20
@Tomothy32 It should return
False in that case, because the 1's are not "in the same quantity".– gilch
Mar 30 at 20:28
@Tomothy32 It should return
False in that case, because the 1's are not "in the same quantity".– gilch
Mar 30 at 20:28
@gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.
– Tomothy32
Mar 30 at 20:32
@gilch The question is a bit fuzzy regarding this, but I have to admit that you probably interpreted it correctly.
– Tomothy32
Mar 30 at 20:32
1
1
Also it should be
<=, not <.– user2357112
Mar 30 at 20:48
Also it should be
<=, not <.– user2357112
Mar 30 at 20:48
|
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 Mar 30 at 20:48
Patrick ArtnerPatrick Artner
26.6k62544
It works too. Thanks !
– Vitor Oliveira
Mar 30 at 21:51
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 Mar 30 at 20:48
Patrick ArtnerPatrick Artner
26.6k62544
It works too. Thanks !
– Vitor Oliveira
Mar 30 at 21:51
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 Mar 30 at 20:48
Patrick ArtnerPatrick Artner
26.6k62544
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 Mar 30 at 20:48
Patrick ArtnerPatrick Artner
26.6k62544
answered Mar 30 at 20:48
Patrick ArtnerPatrick Artner
26.6k62544
answered Mar 30 at 20:48
Patrick ArtnerPatrick Artner
26.6k62544
answered Mar 30 at 20:48
Patrick ArtnerPatrick Artner
26.6k62544
26.6k62544
It works too. Thanks !
– Vitor Oliveira
Mar 30 at 21:51
add a comment |
It works too. Thanks !
– Vitor Oliveira
Mar 30 at 21:51
It works too. Thanks !
– Vitor Oliveira
Mar 30 at 21:51
It works too. Thanks !
– Vitor Oliveira
Mar 30 at 21:51
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
Mar 30 at 20:38
Possible duplicate of Checking if list is a sublist
– Noctis Skytower
Mar 31 at 4:04
Would it be correct to title this "Unordered comparison between lists"?
– Ben Voigt
Mar 31 at 6:12