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How to find the largest number(s) in a list of elements, possibly non-unique?



2019 Community Moderator ElectionHow do I check if a list is empty?How do I check whether a file exists without exceptions?Finding the index of an item given a list containing it in PythonPHP: Delete an element from an arrayWhat's the simplest way to print a Java array?Convert bytes to a string?How to make a flat list out of list of lists?How do I get the number of elements in a list in Python?How do I list all files of a directory?How do I remove a particular element from an array in JavaScript?










13















Here is my program,



item_no = []
max = 0
for i in range(5):
input_no = int(input("Enter an item number: "))
item_no.append(input_no)
for i in item_no:
if i > max:
max = i
high = item_no.index(max)
print (item_no[high])


Example input: [5, 6, 7, 8, 8]



Example output: 8



How can I change my program to output the same highest numbers in an array?



Expected output: [8, 8]










share|improve this question









New contributor




user11206537 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.















  • 2





    Band indent at line 8

    – DirtyBit
    17 hours ago






  • 1





    You can use Dictionary to store the maximum number and its count as item_no = , if you max is different then original in item_no, reinitialize it and add that item and add count =1

    – dkb
    17 hours ago







  • 2





    ("Band" probably is a misspelling for "Bad")

    – tripleee
    17 hours ago






  • 1





    @tripleee Indeed. bulls-eye!

    – DirtyBit
    17 hours ago











  • As answers show, it's more Pythonic to do it with list comprehensions, rather than loops. But if you use loops to iterate over the list, a style tip: don't use the same variable name i both for indices i in range(5) then also for items/values: for i in item_no. Better to do for no in item_no

    – smci
    8 mins ago
















13















Here is my program,



item_no = []
max = 0
for i in range(5):
input_no = int(input("Enter an item number: "))
item_no.append(input_no)
for i in item_no:
if i > max:
max = i
high = item_no.index(max)
print (item_no[high])


Example input: [5, 6, 7, 8, 8]



Example output: 8



How can I change my program to output the same highest numbers in an array?



Expected output: [8, 8]










share|improve this question









New contributor




user11206537 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.















  • 2





    Band indent at line 8

    – DirtyBit
    17 hours ago






  • 1





    You can use Dictionary to store the maximum number and its count as item_no = , if you max is different then original in item_no, reinitialize it and add that item and add count =1

    – dkb
    17 hours ago







  • 2





    ("Band" probably is a misspelling for "Bad")

    – tripleee
    17 hours ago






  • 1





    @tripleee Indeed. bulls-eye!

    – DirtyBit
    17 hours ago











  • As answers show, it's more Pythonic to do it with list comprehensions, rather than loops. But if you use loops to iterate over the list, a style tip: don't use the same variable name i both for indices i in range(5) then also for items/values: for i in item_no. Better to do for no in item_no

    – smci
    8 mins ago














13












13








13


3






Here is my program,



item_no = []
max = 0
for i in range(5):
input_no = int(input("Enter an item number: "))
item_no.append(input_no)
for i in item_no:
if i > max:
max = i
high = item_no.index(max)
print (item_no[high])


Example input: [5, 6, 7, 8, 8]



Example output: 8



How can I change my program to output the same highest numbers in an array?



Expected output: [8, 8]










share|improve this question









New contributor




user11206537 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












Here is my program,



item_no = []
max = 0
for i in range(5):
input_no = int(input("Enter an item number: "))
item_no.append(input_no)
for i in item_no:
if i > max:
max = i
high = item_no.index(max)
print (item_no[high])


Example input: [5, 6, 7, 8, 8]



Example output: 8



How can I change my program to output the same highest numbers in an array?



Expected output: [8, 8]







python arrays python-3.x






share|improve this question









New contributor




user11206537 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




user11206537 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 4 mins ago









smci

15.4k677108




15.4k677108






New contributor




user11206537 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 17 hours ago









user11206537user11206537

7512




7512




New contributor




user11206537 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





user11206537 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






user11206537 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







  • 2





    Band indent at line 8

    – DirtyBit
    17 hours ago






  • 1





    You can use Dictionary to store the maximum number and its count as item_no = , if you max is different then original in item_no, reinitialize it and add that item and add count =1

    – dkb
    17 hours ago







  • 2





    ("Band" probably is a misspelling for "Bad")

    – tripleee
    17 hours ago






  • 1





    @tripleee Indeed. bulls-eye!

    – DirtyBit
    17 hours ago











  • As answers show, it's more Pythonic to do it with list comprehensions, rather than loops. But if you use loops to iterate over the list, a style tip: don't use the same variable name i both for indices i in range(5) then also for items/values: for i in item_no. Better to do for no in item_no

    – smci
    8 mins ago













  • 2





    Band indent at line 8

    – DirtyBit
    17 hours ago






  • 1





    You can use Dictionary to store the maximum number and its count as item_no = , if you max is different then original in item_no, reinitialize it and add that item and add count =1

    – dkb
    17 hours ago







  • 2





    ("Band" probably is a misspelling for "Bad")

    – tripleee
    17 hours ago






  • 1





    @tripleee Indeed. bulls-eye!

    – DirtyBit
    17 hours ago











  • As answers show, it's more Pythonic to do it with list comprehensions, rather than loops. But if you use loops to iterate over the list, a style tip: don't use the same variable name i both for indices i in range(5) then also for items/values: for i in item_no. Better to do for no in item_no

    – smci
    8 mins ago








2




2





Band indent at line 8

– DirtyBit
17 hours ago





Band indent at line 8

– DirtyBit
17 hours ago




1




1





You can use Dictionary to store the maximum number and its count as item_no = , if you max is different then original in item_no, reinitialize it and add that item and add count =1

– dkb
17 hours ago






You can use Dictionary to store the maximum number and its count as item_no = , if you max is different then original in item_no, reinitialize it and add that item and add count =1

– dkb
17 hours ago





2




2





("Band" probably is a misspelling for "Bad")

– tripleee
17 hours ago





("Band" probably is a misspelling for "Bad")

– tripleee
17 hours ago




1




1





@tripleee Indeed. bulls-eye!

– DirtyBit
17 hours ago





@tripleee Indeed. bulls-eye!

– DirtyBit
17 hours ago













As answers show, it's more Pythonic to do it with list comprehensions, rather than loops. But if you use loops to iterate over the list, a style tip: don't use the same variable name i both for indices i in range(5) then also for items/values: for i in item_no. Better to do for no in item_no

– smci
8 mins ago






As answers show, it's more Pythonic to do it with list comprehensions, rather than loops. But if you use loops to iterate over the list, a style tip: don't use the same variable name i both for indices i in range(5) then also for items/values: for i in item_no. Better to do for no in item_no

– smci
8 mins ago













5 Answers
5






active

oldest

votes


















20














Just get the maximum using max and then its count and combine the two in a list-comprehension.



item_no = [5, 6, 7, 8, 8]

max_no = max(item_no)
highest = [max_no for _ in range(item_no.count(max_no))]
print(highest) # -> [8, 8]


Note that this will return a list of a single item in case your maximum value appears only once.




A solution closer to your current programming style would be the following:



item_no = [5, 6, 7, 8, 8]
max_no = 0
for i in item_no:
if i > max_no:
max_no = i
high = [i]
elif i == max_no:
high.append(i)


with the same results as above of course.



Note that this last one is less efficient than the first by quite a margin. Python allows you to be more efficient than these explicit, fortran-like loops and it is more efficient itself when you use it properly.





share

























  • Actually, I would love to use the first code, but since this part of my school project, I am not allowed to use inbuilt functions, therefore I need to use the second code.

    – user11206537
    17 hours ago






  • 2





    @user11206537 I kinda saw this coming. Knowing what a better way would be is still of some value. Have fun!

    – Ev. Kounis
    16 hours ago











  • One last question, how do you find the index of the result (high) in item_no?

    – user11206537
    16 hours ago












  • You might want to start with a smaller number than 0.

    – Eric Duminil
    15 hours ago






  • 2





    @user11206537 To get the index as well, you can modify the code slightly and use enumerate on item_no. As you loop through, store the index in a 2nd list as you do for the max. Take a look at this

    – Ev. Kounis
    15 hours ago


















11














You can do it even shorter:



item_no = [5, 6, 7, 8, 8]
#compute once - use many
max_item = max(item_no)
print(item_no.count(max_item) * [max_item])


output:



[8, 8]





share|improve this answer

























  • One last question, how do you find the index of the result in item_no?

    – user11206537
    15 hours ago



















4














You could use list comprehension for that task following way:



numbers = [5, 6, 7, 8, 8]
maxnumbers = [i for i in numbers if i==max(numbers)]
print(*maxnumbers,sep=',')


output:



8,8


* operator in print is used to unpack values, sep is used to inform print what seperator to use: , in this case.



EDIT: If you want to get indices of biggest value and call max only once then do:



numbers = [5, 6, 7, 8, 8]
biggest = max(numbers)
positions = [inx for inx,i in enumerate(numbers) if i==biggest]
print(*positions,sep=',')


Output:



3,4


As you might check numbers[3] is equal to biggest and numbers[4] is equal to biggest.






share|improve this answer




















  • 2





    note that your solution calls max a total of len(numbers) times. Storing it outside the list and then using that would be better.

    – Ev. Kounis
    17 hours ago












  • How do you find the index of the result in item_no?

    – user11206537
    15 hours ago











  • @Ev.Kounis I assume the Python interpreter optimizes this, no?

    – user1717828
    8 hours ago


















2














This issue can be solved in one line, by finding an item which is equal to the maximum value:



[i for i in item_no if i==max(item_no)]






share|improve this answer

























  • How do you find the index of the result in item_no?

    – user11206537
    15 hours ago











  • By using enumerate function, we can get index position: [i for i, val in enumerate(item_no) if val in result]

    – Pradeep Pandey
    12 hours ago



















1














  1. Count the occurrence of max number


  2. iterate over the list to print the max number for the range of the count (1)


Hence:



item_no = [5, 6, 7, 8, 8]
counter = item_no.count(max(item_no)) # 2
print([max(item_no) for x in range(counter)])


OUTPUT:



[8, 8]





share|improve this answer























  • How do you find the index of the result in item_no?

    – user11206537
    15 hours ago










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5 Answers
5






active

oldest

votes








5 Answers
5






active

oldest

votes









active

oldest

votes






active

oldest

votes









20














Just get the maximum using max and then its count and combine the two in a list-comprehension.



item_no = [5, 6, 7, 8, 8]

max_no = max(item_no)
highest = [max_no for _ in range(item_no.count(max_no))]
print(highest) # -> [8, 8]


Note that this will return a list of a single item in case your maximum value appears only once.




A solution closer to your current programming style would be the following:



item_no = [5, 6, 7, 8, 8]
max_no = 0
for i in item_no:
if i > max_no:
max_no = i
high = [i]
elif i == max_no:
high.append(i)


with the same results as above of course.



Note that this last one is less efficient than the first by quite a margin. Python allows you to be more efficient than these explicit, fortran-like loops and it is more efficient itself when you use it properly.





share

























  • Actually, I would love to use the first code, but since this part of my school project, I am not allowed to use inbuilt functions, therefore I need to use the second code.

    – user11206537
    17 hours ago






  • 2





    @user11206537 I kinda saw this coming. Knowing what a better way would be is still of some value. Have fun!

    – Ev. Kounis
    16 hours ago











  • One last question, how do you find the index of the result (high) in item_no?

    – user11206537
    16 hours ago












  • You might want to start with a smaller number than 0.

    – Eric Duminil
    15 hours ago






  • 2





    @user11206537 To get the index as well, you can modify the code slightly and use enumerate on item_no. As you loop through, store the index in a 2nd list as you do for the max. Take a look at this

    – Ev. Kounis
    15 hours ago















20














Just get the maximum using max and then its count and combine the two in a list-comprehension.



item_no = [5, 6, 7, 8, 8]

max_no = max(item_no)
highest = [max_no for _ in range(item_no.count(max_no))]
print(highest) # -> [8, 8]


Note that this will return a list of a single item in case your maximum value appears only once.




A solution closer to your current programming style would be the following:



item_no = [5, 6, 7, 8, 8]
max_no = 0
for i in item_no:
if i > max_no:
max_no = i
high = [i]
elif i == max_no:
high.append(i)


with the same results as above of course.



Note that this last one is less efficient than the first by quite a margin. Python allows you to be more efficient than these explicit, fortran-like loops and it is more efficient itself when you use it properly.





share

























  • Actually, I would love to use the first code, but since this part of my school project, I am not allowed to use inbuilt functions, therefore I need to use the second code.

    – user11206537
    17 hours ago






  • 2





    @user11206537 I kinda saw this coming. Knowing what a better way would be is still of some value. Have fun!

    – Ev. Kounis
    16 hours ago











  • One last question, how do you find the index of the result (high) in item_no?

    – user11206537
    16 hours ago












  • You might want to start with a smaller number than 0.

    – Eric Duminil
    15 hours ago






  • 2





    @user11206537 To get the index as well, you can modify the code slightly and use enumerate on item_no. As you loop through, store the index in a 2nd list as you do for the max. Take a look at this

    – Ev. Kounis
    15 hours ago













20












20








20







Just get the maximum using max and then its count and combine the two in a list-comprehension.



item_no = [5, 6, 7, 8, 8]

max_no = max(item_no)
highest = [max_no for _ in range(item_no.count(max_no))]
print(highest) # -> [8, 8]


Note that this will return a list of a single item in case your maximum value appears only once.




A solution closer to your current programming style would be the following:



item_no = [5, 6, 7, 8, 8]
max_no = 0
for i in item_no:
if i > max_no:
max_no = i
high = [i]
elif i == max_no:
high.append(i)


with the same results as above of course.



Note that this last one is less efficient than the first by quite a margin. Python allows you to be more efficient than these explicit, fortran-like loops and it is more efficient itself when you use it properly.





share















Just get the maximum using max and then its count and combine the two in a list-comprehension.



item_no = [5, 6, 7, 8, 8]

max_no = max(item_no)
highest = [max_no for _ in range(item_no.count(max_no))]
print(highest) # -> [8, 8]


Note that this will return a list of a single item in case your maximum value appears only once.




A solution closer to your current programming style would be the following:



item_no = [5, 6, 7, 8, 8]
max_no = 0
for i in item_no:
if i > max_no:
max_no = i
high = [i]
elif i == max_no:
high.append(i)


with the same results as above of course.



Note that this last one is less efficient than the first by quite a margin. Python allows you to be more efficient than these explicit, fortran-like loops and it is more efficient itself when you use it properly.






share













share


share








edited 15 hours ago









idmean

10.7k73669




10.7k73669










answered 17 hours ago









Ev. KounisEv. Kounis

11.2k21651




11.2k21651












  • Actually, I would love to use the first code, but since this part of my school project, I am not allowed to use inbuilt functions, therefore I need to use the second code.

    – user11206537
    17 hours ago






  • 2





    @user11206537 I kinda saw this coming. Knowing what a better way would be is still of some value. Have fun!

    – Ev. Kounis
    16 hours ago











  • One last question, how do you find the index of the result (high) in item_no?

    – user11206537
    16 hours ago












  • You might want to start with a smaller number than 0.

    – Eric Duminil
    15 hours ago






  • 2





    @user11206537 To get the index as well, you can modify the code slightly and use enumerate on item_no. As you loop through, store the index in a 2nd list as you do for the max. Take a look at this

    – Ev. Kounis
    15 hours ago

















  • Actually, I would love to use the first code, but since this part of my school project, I am not allowed to use inbuilt functions, therefore I need to use the second code.

    – user11206537
    17 hours ago






  • 2





    @user11206537 I kinda saw this coming. Knowing what a better way would be is still of some value. Have fun!

    – Ev. Kounis
    16 hours ago











  • One last question, how do you find the index of the result (high) in item_no?

    – user11206537
    16 hours ago












  • You might want to start with a smaller number than 0.

    – Eric Duminil
    15 hours ago






  • 2





    @user11206537 To get the index as well, you can modify the code slightly and use enumerate on item_no. As you loop through, store the index in a 2nd list as you do for the max. Take a look at this

    – Ev. Kounis
    15 hours ago
















Actually, I would love to use the first code, but since this part of my school project, I am not allowed to use inbuilt functions, therefore I need to use the second code.

– user11206537
17 hours ago





Actually, I would love to use the first code, but since this part of my school project, I am not allowed to use inbuilt functions, therefore I need to use the second code.

– user11206537
17 hours ago




2




2





@user11206537 I kinda saw this coming. Knowing what a better way would be is still of some value. Have fun!

– Ev. Kounis
16 hours ago





@user11206537 I kinda saw this coming. Knowing what a better way would be is still of some value. Have fun!

– Ev. Kounis
16 hours ago













One last question, how do you find the index of the result (high) in item_no?

– user11206537
16 hours ago






One last question, how do you find the index of the result (high) in item_no?

– user11206537
16 hours ago














You might want to start with a smaller number than 0.

– Eric Duminil
15 hours ago





You might want to start with a smaller number than 0.

– Eric Duminil
15 hours ago




2




2





@user11206537 To get the index as well, you can modify the code slightly and use enumerate on item_no. As you loop through, store the index in a 2nd list as you do for the max. Take a look at this

– Ev. Kounis
15 hours ago





@user11206537 To get the index as well, you can modify the code slightly and use enumerate on item_no. As you loop through, store the index in a 2nd list as you do for the max. Take a look at this

– Ev. Kounis
15 hours ago













11














You can do it even shorter:



item_no = [5, 6, 7, 8, 8]
#compute once - use many
max_item = max(item_no)
print(item_no.count(max_item) * [max_item])


output:



[8, 8]





share|improve this answer

























  • One last question, how do you find the index of the result in item_no?

    – user11206537
    15 hours ago
















11














You can do it even shorter:



item_no = [5, 6, 7, 8, 8]
#compute once - use many
max_item = max(item_no)
print(item_no.count(max_item) * [max_item])


output:



[8, 8]





share|improve this answer

























  • One last question, how do you find the index of the result in item_no?

    – user11206537
    15 hours ago














11












11








11







You can do it even shorter:



item_no = [5, 6, 7, 8, 8]
#compute once - use many
max_item = max(item_no)
print(item_no.count(max_item) * [max_item])


output:



[8, 8]





share|improve this answer















You can do it even shorter:



item_no = [5, 6, 7, 8, 8]
#compute once - use many
max_item = max(item_no)
print(item_no.count(max_item) * [max_item])


output:



[8, 8]






share|improve this answer














share|improve this answer



share|improve this answer








edited 13 hours ago

























answered 17 hours ago









AllanAllan

7,90731534




7,90731534












  • One last question, how do you find the index of the result in item_no?

    – user11206537
    15 hours ago


















  • One last question, how do you find the index of the result in item_no?

    – user11206537
    15 hours ago

















One last question, how do you find the index of the result in item_no?

– user11206537
15 hours ago






One last question, how do you find the index of the result in item_no?

– user11206537
15 hours ago












4














You could use list comprehension for that task following way:



numbers = [5, 6, 7, 8, 8]
maxnumbers = [i for i in numbers if i==max(numbers)]
print(*maxnumbers,sep=',')


output:



8,8


* operator in print is used to unpack values, sep is used to inform print what seperator to use: , in this case.



EDIT: If you want to get indices of biggest value and call max only once then do:



numbers = [5, 6, 7, 8, 8]
biggest = max(numbers)
positions = [inx for inx,i in enumerate(numbers) if i==biggest]
print(*positions,sep=',')


Output:



3,4


As you might check numbers[3] is equal to biggest and numbers[4] is equal to biggest.






share|improve this answer




















  • 2





    note that your solution calls max a total of len(numbers) times. Storing it outside the list and then using that would be better.

    – Ev. Kounis
    17 hours ago












  • How do you find the index of the result in item_no?

    – user11206537
    15 hours ago











  • @Ev.Kounis I assume the Python interpreter optimizes this, no?

    – user1717828
    8 hours ago















4














You could use list comprehension for that task following way:



numbers = [5, 6, 7, 8, 8]
maxnumbers = [i for i in numbers if i==max(numbers)]
print(*maxnumbers,sep=',')


output:



8,8


* operator in print is used to unpack values, sep is used to inform print what seperator to use: , in this case.



EDIT: If you want to get indices of biggest value and call max only once then do:



numbers = [5, 6, 7, 8, 8]
biggest = max(numbers)
positions = [inx for inx,i in enumerate(numbers) if i==biggest]
print(*positions,sep=',')


Output:



3,4


As you might check numbers[3] is equal to biggest and numbers[4] is equal to biggest.






share|improve this answer




















  • 2





    note that your solution calls max a total of len(numbers) times. Storing it outside the list and then using that would be better.

    – Ev. Kounis
    17 hours ago












  • How do you find the index of the result in item_no?

    – user11206537
    15 hours ago











  • @Ev.Kounis I assume the Python interpreter optimizes this, no?

    – user1717828
    8 hours ago













4












4








4







You could use list comprehension for that task following way:



numbers = [5, 6, 7, 8, 8]
maxnumbers = [i for i in numbers if i==max(numbers)]
print(*maxnumbers,sep=',')


output:



8,8


* operator in print is used to unpack values, sep is used to inform print what seperator to use: , in this case.



EDIT: If you want to get indices of biggest value and call max only once then do:



numbers = [5, 6, 7, 8, 8]
biggest = max(numbers)
positions = [inx for inx,i in enumerate(numbers) if i==biggest]
print(*positions,sep=',')


Output:



3,4


As you might check numbers[3] is equal to biggest and numbers[4] is equal to biggest.






share|improve this answer















You could use list comprehension for that task following way:



numbers = [5, 6, 7, 8, 8]
maxnumbers = [i for i in numbers if i==max(numbers)]
print(*maxnumbers,sep=',')


output:



8,8


* operator in print is used to unpack values, sep is used to inform print what seperator to use: , in this case.



EDIT: If you want to get indices of biggest value and call max only once then do:



numbers = [5, 6, 7, 8, 8]
biggest = max(numbers)
positions = [inx for inx,i in enumerate(numbers) if i==biggest]
print(*positions,sep=',')


Output:



3,4


As you might check numbers[3] is equal to biggest and numbers[4] is equal to biggest.







share|improve this answer














share|improve this answer



share|improve this answer








edited 14 hours ago

























answered 17 hours ago









DaweoDaweo

1,03525




1,03525







  • 2





    note that your solution calls max a total of len(numbers) times. Storing it outside the list and then using that would be better.

    – Ev. Kounis
    17 hours ago












  • How do you find the index of the result in item_no?

    – user11206537
    15 hours ago











  • @Ev.Kounis I assume the Python interpreter optimizes this, no?

    – user1717828
    8 hours ago












  • 2





    note that your solution calls max a total of len(numbers) times. Storing it outside the list and then using that would be better.

    – Ev. Kounis
    17 hours ago












  • How do you find the index of the result in item_no?

    – user11206537
    15 hours ago











  • @Ev.Kounis I assume the Python interpreter optimizes this, no?

    – user1717828
    8 hours ago







2




2





note that your solution calls max a total of len(numbers) times. Storing it outside the list and then using that would be better.

– Ev. Kounis
17 hours ago






note that your solution calls max a total of len(numbers) times. Storing it outside the list and then using that would be better.

– Ev. Kounis
17 hours ago














How do you find the index of the result in item_no?

– user11206537
15 hours ago





How do you find the index of the result in item_no?

– user11206537
15 hours ago













@Ev.Kounis I assume the Python interpreter optimizes this, no?

– user1717828
8 hours ago





@Ev.Kounis I assume the Python interpreter optimizes this, no?

– user1717828
8 hours ago











2














This issue can be solved in one line, by finding an item which is equal to the maximum value:



[i for i in item_no if i==max(item_no)]






share|improve this answer

























  • How do you find the index of the result in item_no?

    – user11206537
    15 hours ago











  • By using enumerate function, we can get index position: [i for i, val in enumerate(item_no) if val in result]

    – Pradeep Pandey
    12 hours ago
















2














This issue can be solved in one line, by finding an item which is equal to the maximum value:



[i for i in item_no if i==max(item_no)]






share|improve this answer

























  • How do you find the index of the result in item_no?

    – user11206537
    15 hours ago











  • By using enumerate function, we can get index position: [i for i, val in enumerate(item_no) if val in result]

    – Pradeep Pandey
    12 hours ago














2












2








2







This issue can be solved in one line, by finding an item which is equal to the maximum value:



[i for i in item_no if i==max(item_no)]






share|improve this answer















This issue can be solved in one line, by finding an item which is equal to the maximum value:



[i for i in item_no if i==max(item_no)]







share|improve this answer














share|improve this answer



share|improve this answer








edited 15 hours ago









376writer

132




132










answered 16 hours ago









Pradeep PandeyPradeep Pandey

15417




15417












  • How do you find the index of the result in item_no?

    – user11206537
    15 hours ago











  • By using enumerate function, we can get index position: [i for i, val in enumerate(item_no) if val in result]

    – Pradeep Pandey
    12 hours ago


















  • How do you find the index of the result in item_no?

    – user11206537
    15 hours ago











  • By using enumerate function, we can get index position: [i for i, val in enumerate(item_no) if val in result]

    – Pradeep Pandey
    12 hours ago

















How do you find the index of the result in item_no?

– user11206537
15 hours ago





How do you find the index of the result in item_no?

– user11206537
15 hours ago













By using enumerate function, we can get index position: [i for i, val in enumerate(item_no) if val in result]

– Pradeep Pandey
12 hours ago






By using enumerate function, we can get index position: [i for i, val in enumerate(item_no) if val in result]

– Pradeep Pandey
12 hours ago












1














  1. Count the occurrence of max number


  2. iterate over the list to print the max number for the range of the count (1)


Hence:



item_no = [5, 6, 7, 8, 8]
counter = item_no.count(max(item_no)) # 2
print([max(item_no) for x in range(counter)])


OUTPUT:



[8, 8]





share|improve this answer























  • How do you find the index of the result in item_no?

    – user11206537
    15 hours ago















1














  1. Count the occurrence of max number


  2. iterate over the list to print the max number for the range of the count (1)


Hence:



item_no = [5, 6, 7, 8, 8]
counter = item_no.count(max(item_no)) # 2
print([max(item_no) for x in range(counter)])


OUTPUT:



[8, 8]





share|improve this answer























  • How do you find the index of the result in item_no?

    – user11206537
    15 hours ago













1












1








1







  1. Count the occurrence of max number


  2. iterate over the list to print the max number for the range of the count (1)


Hence:



item_no = [5, 6, 7, 8, 8]
counter = item_no.count(max(item_no)) # 2
print([max(item_no) for x in range(counter)])


OUTPUT:



[8, 8]





share|improve this answer













  1. Count the occurrence of max number


  2. iterate over the list to print the max number for the range of the count (1)


Hence:



item_no = [5, 6, 7, 8, 8]
counter = item_no.count(max(item_no)) # 2
print([max(item_no) for x in range(counter)])


OUTPUT:



[8, 8]






share|improve this answer












share|improve this answer



share|improve this answer










answered 17 hours ago









DirtyBitDirtyBit

9,04821540




9,04821540












  • How do you find the index of the result in item_no?

    – user11206537
    15 hours ago

















  • How do you find the index of the result in item_no?

    – user11206537
    15 hours ago
















How do you find the index of the result in item_no?

– user11206537
15 hours ago





How do you find the index of the result in item_no?

– user11206537
15 hours ago










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