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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?Finding the index of an item given a list containing it in PythonWhat's the simplest way to print a Java array?Convert bytes to a string?Getting the last element of a list in PythonHow 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?Use of *args and **kwargsHow do I remove a particular element from an array in JavaScript?
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
New contributor
|
show 1 more comment
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
New contributor
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 namei
both for indicesi in range(5)
then also for items/values:for i in item_no
. Better to dofor no in item_no
– smci
30 mins ago
|
show 1 more comment
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
New contributor
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
python arrays python-3.x
New contributor
New contributor
edited 27 mins ago
smci
15.4k677108
15.4k677108
New contributor
asked 17 hours ago
user11206537user11206537
7512
7512
New contributor
New contributor
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 namei
both for indicesi in range(5)
then also for items/values:for i in item_no
. Better to dofor no in item_no
– smci
30 mins ago
|
show 1 more comment
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 namei
both for indicesi in range(5)
then also for items/values:for i in item_no
. Better to dofor no in item_no
– smci
30 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
30 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
30 mins ago
|
show 1 more comment
5 Answers
5
active
oldest
votes
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.
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
17 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 than0
.
– Eric Duminil
16 hours ago
2
@user11206537 To get the index as well, you can modify the code slightly and useenumerate
onitem_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
|
show 4 more comments
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]
One last question, how do you find the index of the result in item_no?
– user11206537
16 hours ago
add a comment |
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
.
2
note that your solution callsmax
a total oflen(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
16 hours ago
@Ev.Kounis I assume the Python interpreter optimizes this, no?
– user1717828
8 hours ago
add a comment |
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)]
How do you find the index of the result in item_no?
– user11206537
16 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
13 hours ago
add a comment |
Count the occurrence of max number
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]
How do you find the index of the result in item_no?
– user11206537
16 hours ago
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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
17 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 than0
.
– Eric Duminil
16 hours ago
2
@user11206537 To get the index as well, you can modify the code slightly and useenumerate
onitem_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
|
show 4 more comments
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.
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
17 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 than0
.
– Eric Duminil
16 hours ago
2
@user11206537 To get the index as well, you can modify the code slightly and useenumerate
onitem_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
|
show 4 more comments
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.
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.
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
17 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 than0
.
– Eric Duminil
16 hours ago
2
@user11206537 To get the index as well, you can modify the code slightly and useenumerate
onitem_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
|
show 4 more comments
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
17 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 than0
.
– Eric Duminil
16 hours ago
2
@user11206537 To get the index as well, you can modify the code slightly and useenumerate
onitem_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
17 hours ago
@user11206537 I kinda saw this coming. Knowing what a better way would be is still of some value. Have fun!
– Ev. Kounis
17 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
16 hours ago
You might want to start with a smaller number than
0
.– Eric Duminil
16 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
|
show 4 more comments
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]
One last question, how do you find the index of the result in item_no?
– user11206537
16 hours ago
add a comment |
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]
One last question, how do you find the index of the result in item_no?
– user11206537
16 hours ago
add a comment |
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]
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]
edited 13 hours ago
answered 17 hours ago
AllanAllan
7,91731534
7,91731534
One last question, how do you find the index of the result in item_no?
– user11206537
16 hours ago
add a comment |
One last question, how do you find the index of the result in item_no?
– user11206537
16 hours ago
One last question, how do you find the index of the result in item_no?
– user11206537
16 hours ago
One last question, how do you find the index of the result in item_no?
– user11206537
16 hours ago
add a comment |
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
.
2
note that your solution callsmax
a total oflen(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
16 hours ago
@Ev.Kounis I assume the Python interpreter optimizes this, no?
– user1717828
8 hours ago
add a comment |
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
.
2
note that your solution callsmax
a total oflen(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
16 hours ago
@Ev.Kounis I assume the Python interpreter optimizes this, no?
– user1717828
8 hours ago
add a comment |
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
.
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
.
edited 15 hours ago
answered 17 hours ago
DaweoDaweo
1,03525
1,03525
2
note that your solution callsmax
a total oflen(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
16 hours ago
@Ev.Kounis I assume the Python interpreter optimizes this, no?
– user1717828
8 hours ago
add a comment |
2
note that your solution callsmax
a total oflen(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
16 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
16 hours ago
How do you find the index of the result in item_no?
– user11206537
16 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
add a comment |
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)]
How do you find the index of the result in item_no?
– user11206537
16 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
13 hours ago
add a comment |
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)]
How do you find the index of the result in item_no?
– user11206537
16 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
13 hours ago
add a comment |
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)]
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)]
edited 15 hours ago
376writer
132
132
answered 17 hours ago
Pradeep PandeyPradeep Pandey
15417
15417
How do you find the index of the result in item_no?
– user11206537
16 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
13 hours ago
add a comment |
How do you find the index of the result in item_no?
– user11206537
16 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
13 hours ago
How do you find the index of the result in item_no?
– user11206537
16 hours ago
How do you find the index of the result in item_no?
– user11206537
16 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
13 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
13 hours ago
add a comment |
Count the occurrence of max number
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]
How do you find the index of the result in item_no?
– user11206537
16 hours ago
add a comment |
Count the occurrence of max number
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]
How do you find the index of the result in item_no?
– user11206537
16 hours ago
add a comment |
Count the occurrence of max number
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]
Count the occurrence of max number
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]
answered 17 hours ago
DirtyBitDirtyBit
9,04821540
9,04821540
How do you find the index of the result in item_no?
– user11206537
16 hours ago
add a comment |
How do you find the index of the result in item_no?
– user11206537
16 hours ago
How do you find the index of the result in item_no?
– user11206537
16 hours ago
How do you find the index of the result in item_no?
– user11206537
16 hours ago
add a comment |
user11206537 is a new contributor. Be nice, and check out our Code of Conduct.
user11206537 is a new contributor. Be nice, and check out our Code of Conduct.
user11206537 is a new contributor. Be nice, and check out our Code of Conduct.
user11206537 is a new contributor. Be nice, and check out our Code of Conduct.
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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 indicesi in range(5)
then also for items/values:for i in item_no
. Better to dofor no in item_no
– smci
30 mins ago