Sorting a list of numbers, each with a character label
$begingroup$
I am working on a project that requires me to modify an input list as follows:
INPUT: List of numbers, each tagged with a label i.e. [[2, 'a'], [3, 'b'], [1, 'c'], ...]
DESIRED OUTPUT: Sort the list and grab the elements with the highest numerical component, store them in another list, and maybe rerun the function on the new list. This is an order determination scheme based on players rolling dice.
I have created the following code in Python 3.7 and it works just fine for me. I am curious to see if there are more clever ways to arrive at the same output. For some reason this took me about two hours to come up with (I am learning from scratch).
list = [[1,'a'],[3,'b'],[7,'c'],[6,'d'],[4,'e'],[1,'f'],[7,'g'],[4,'h'],[6,'i']]
def comp(list):
new_list =
list.sort(reverse = True)
for y in range(1, len(list)):
if list[0][0] > list[y][0]:
new_list.append(list[y])
top_rolls = [x for x in list if x not in new_list]
return(top_rolls)
python python-3.x sorting
$endgroup$
add a comment |
$begingroup$
I am working on a project that requires me to modify an input list as follows:
INPUT: List of numbers, each tagged with a label i.e. [[2, 'a'], [3, 'b'], [1, 'c'], ...]
DESIRED OUTPUT: Sort the list and grab the elements with the highest numerical component, store them in another list, and maybe rerun the function on the new list. This is an order determination scheme based on players rolling dice.
I have created the following code in Python 3.7 and it works just fine for me. I am curious to see if there are more clever ways to arrive at the same output. For some reason this took me about two hours to come up with (I am learning from scratch).
list = [[1,'a'],[3,'b'],[7,'c'],[6,'d'],[4,'e'],[1,'f'],[7,'g'],[4,'h'],[6,'i']]
def comp(list):
new_list =
list.sort(reverse = True)
for y in range(1, len(list)):
if list[0][0] > list[y][0]:
new_list.append(list[y])
top_rolls = [x for x in list if x not in new_list]
return(top_rolls)
python python-3.x sorting
$endgroup$
1
$begingroup$
I'd like to point out that "maybe" has no place in a proper requirements statement.
$endgroup$
– Reinderien
Dec 21 '18 at 15:50
$begingroup$
Thanks Reinderien, I was simply writing shorthand instead of saying that I would like to rerun the function if the new list contains more than one element. I understand the logic behind your statement.
$endgroup$
– JacobCheverie
Dec 21 '18 at 16:53
add a comment |
$begingroup$
I am working on a project that requires me to modify an input list as follows:
INPUT: List of numbers, each tagged with a label i.e. [[2, 'a'], [3, 'b'], [1, 'c'], ...]
DESIRED OUTPUT: Sort the list and grab the elements with the highest numerical component, store them in another list, and maybe rerun the function on the new list. This is an order determination scheme based on players rolling dice.
I have created the following code in Python 3.7 and it works just fine for me. I am curious to see if there are more clever ways to arrive at the same output. For some reason this took me about two hours to come up with (I am learning from scratch).
list = [[1,'a'],[3,'b'],[7,'c'],[6,'d'],[4,'e'],[1,'f'],[7,'g'],[4,'h'],[6,'i']]
def comp(list):
new_list =
list.sort(reverse = True)
for y in range(1, len(list)):
if list[0][0] > list[y][0]:
new_list.append(list[y])
top_rolls = [x for x in list if x not in new_list]
return(top_rolls)
python python-3.x sorting
$endgroup$
I am working on a project that requires me to modify an input list as follows:
INPUT: List of numbers, each tagged with a label i.e. [[2, 'a'], [3, 'b'], [1, 'c'], ...]
DESIRED OUTPUT: Sort the list and grab the elements with the highest numerical component, store them in another list, and maybe rerun the function on the new list. This is an order determination scheme based on players rolling dice.
I have created the following code in Python 3.7 and it works just fine for me. I am curious to see if there are more clever ways to arrive at the same output. For some reason this took me about two hours to come up with (I am learning from scratch).
list = [[1,'a'],[3,'b'],[7,'c'],[6,'d'],[4,'e'],[1,'f'],[7,'g'],[4,'h'],[6,'i']]
def comp(list):
new_list =
list.sort(reverse = True)
for y in range(1, len(list)):
if list[0][0] > list[y][0]:
new_list.append(list[y])
top_rolls = [x for x in list if x not in new_list]
return(top_rolls)
python python-3.x sorting
python python-3.x sorting
edited Dec 21 '18 at 13:51
JacobCheverie
asked Dec 21 '18 at 13:07
JacobCheverieJacobCheverie
1608
1608
1
$begingroup$
I'd like to point out that "maybe" has no place in a proper requirements statement.
$endgroup$
– Reinderien
Dec 21 '18 at 15:50
$begingroup$
Thanks Reinderien, I was simply writing shorthand instead of saying that I would like to rerun the function if the new list contains more than one element. I understand the logic behind your statement.
$endgroup$
– JacobCheverie
Dec 21 '18 at 16:53
add a comment |
1
$begingroup$
I'd like to point out that "maybe" has no place in a proper requirements statement.
$endgroup$
– Reinderien
Dec 21 '18 at 15:50
$begingroup$
Thanks Reinderien, I was simply writing shorthand instead of saying that I would like to rerun the function if the new list contains more than one element. I understand the logic behind your statement.
$endgroup$
– JacobCheverie
Dec 21 '18 at 16:53
1
1
$begingroup$
I'd like to point out that "maybe" has no place in a proper requirements statement.
$endgroup$
– Reinderien
Dec 21 '18 at 15:50
$begingroup$
I'd like to point out that "maybe" has no place in a proper requirements statement.
$endgroup$
– Reinderien
Dec 21 '18 at 15:50
$begingroup$
Thanks Reinderien, I was simply writing shorthand instead of saying that I would like to rerun the function if the new list contains more than one element. I understand the logic behind your statement.
$endgroup$
– JacobCheverie
Dec 21 '18 at 16:53
$begingroup$
Thanks Reinderien, I was simply writing shorthand instead of saying that I would like to rerun the function if the new list contains more than one element. I understand the logic behind your statement.
$endgroup$
– JacobCheverie
Dec 21 '18 at 16:53
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Names
Using the name list
for a variable is legal but usually avoided because it hides the list
builtin. The name lst
would be a better candidate.
The function name comp
could be improved as well. Maybe get_best_elements
(even though I am not fully convinced).
Tests
Before going further, it may be wise to add tests so that you can easily be sure that we do not break anything.
A proper solution would involve a unit-test framework but for the time being, we can go for the simple:
TESTS = (
(, ),
([[1,'a']], [[1, 'a']]),
([[1,'a'], [1,'a'], [1,'a'], [1,'a']], [[1,'a'], [1,'a'], [1,'a'], [1,'a']]),
([[1,'a'], [1,'b'], [1,'c'], [1,'d']], [[1,'d'], [1,'c'], [1,'b'], [1,'a']]),
([[1,'a'],[3,'b'],[7,'c'],[6,'d'],[4,'e'],[1,'f'],[7,'g'],[4,'h'],[6,'i']], [[7, 'g'], [7, 'c']]),
([[1,'a'],[3,'b'],[7,'c'],[6,'d'],[4,'e'],[1,'f'],[4,'h'],[6,'i']], [[7, 'c']]),
)
for test_input, expected_output in TESTS:
output = get_best_elements(test_input)
assert output == expected_output
print("OK")
Style
Parenthesis are not required in the return statements.
Also, the temporary variable is not really required.
List comprehension
The new_list
could be defined with a list comprehension.
def get_best_elements(lst):
lst.sort(reverse = True)
new_lst = [lst[y] for y in range(1, len(lst)) if lst[0][0] > lst[y][0]]
return [x for x in lst if x not in new_lst]
Loop like a native
I highly recommend Ned Batchelder's talk "Loop like a native" about iterators. One of the most simple take away is that whenever you're doing range(len(lst))
, you can probably do things in a better way: more concise, clearer and more efficient.
In our case, we'd have something like:
def get_best_elements(lst):
lst.sort(reverse = True)
new_lst = [e for e in lst if e[0] < lst[0][0]]
return [x for x in lst if x not in new_lst]
Another approach
The last list comprehension can be quite expensive because for each element x
, you may perform a look-up in the list. This can lead to a O(n²)
behavior.
Instead of filtering out elements that do not correspond to the biggest number, we could just keep the one that do correspond to the biggest number.
def get_best_elements(lst):
if not lst:
return
lst.sort(reverse = True)
big_n = lst[0][0]
return [x for x in lst if x[0] == big_n]
Sort a smaller numbers of elements
You could use max
to get big_n
.
Also, we could perform the sorting on the filtered list so that we have fewer elements to sort:
def get_best_elements(lst):
if not lst:
return
big_n = max(lst)[0]
return list(sorted((x for x in lst if x[0] == big_n), reverse=True))
$endgroup$
add a comment |
$begingroup$
You could simplify by directly making a list with the highest scores instead of creating and subtracting another list containing what you don't want :
def comp(rolls):
top_rolls =
rolls.sort(reverse=True)
for y in range(len(rolls)):
if rolls[0][0] == rolls[y][0]:
top_rolls.append(rolls[y])
return top_rolls
Once you've done that it becomes easy to fit it in a list comprehension:
def comp(list):
rolls.sort(reverse=True)
top_rolls = [rolls[y] for y in range(len(rolls)) if rolls[0][0] == rolls[y][0]]
return top_rolls
Or even shorter:
def comp(list):
list.sort(reverse=True)
return [rolls[y] for y in range(len(rolls)) if rolls[0][0] == rolls[y][0]]
$endgroup$
$begingroup$
That's great, thank you. I don't know why I created a list of things that I don't want just to subtract it from the overall list. Your method is much more concise.
$endgroup$
– JacobCheverie
Dec 21 '18 at 16:55
$begingroup$
Don't name an argumentlist
. That will shadow the built-in type also calledlist
.
$endgroup$
– Reinderien
Dec 21 '18 at 18:00
$begingroup$
I am renaming everything after I am finished, I am more concerned about the appropriate syntax for now.
$endgroup$
– JacobCheverie
Dec 21 '18 at 18:45
add a comment |
$begingroup$
Based on your stated use case:
I would like to rerun the function if the new list contains more than one element.
you don't even need to return a list; just return the highest element's roll and name, which can be unpacked by the caller as a 2-tuple:
def comp(rolls):
return max(rolls)
That said, you haven't explicitly stated how to resolve ties with rolls of the same value. That will affect this solution.
$endgroup$
$begingroup$
The max function on it's own will only return one players roll, so if there is a tie it will not be handled. Working with a simplification similar to one that @Comte_Zero provided, I am handling the case of a tie separately. I am sure that code could use some enhancing as well, I am just trying to get the basic form nailed down and make edits later.
$endgroup$
– JacobCheverie
Dec 21 '18 at 18:47
$begingroup$
I'd propose that tie resolution should be built into this function.
$endgroup$
– Reinderien
Dec 21 '18 at 19:12
add a comment |
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3 Answers
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Names
Using the name list
for a variable is legal but usually avoided because it hides the list
builtin. The name lst
would be a better candidate.
The function name comp
could be improved as well. Maybe get_best_elements
(even though I am not fully convinced).
Tests
Before going further, it may be wise to add tests so that you can easily be sure that we do not break anything.
A proper solution would involve a unit-test framework but for the time being, we can go for the simple:
TESTS = (
(, ),
([[1,'a']], [[1, 'a']]),
([[1,'a'], [1,'a'], [1,'a'], [1,'a']], [[1,'a'], [1,'a'], [1,'a'], [1,'a']]),
([[1,'a'], [1,'b'], [1,'c'], [1,'d']], [[1,'d'], [1,'c'], [1,'b'], [1,'a']]),
([[1,'a'],[3,'b'],[7,'c'],[6,'d'],[4,'e'],[1,'f'],[7,'g'],[4,'h'],[6,'i']], [[7, 'g'], [7, 'c']]),
([[1,'a'],[3,'b'],[7,'c'],[6,'d'],[4,'e'],[1,'f'],[4,'h'],[6,'i']], [[7, 'c']]),
)
for test_input, expected_output in TESTS:
output = get_best_elements(test_input)
assert output == expected_output
print("OK")
Style
Parenthesis are not required in the return statements.
Also, the temporary variable is not really required.
List comprehension
The new_list
could be defined with a list comprehension.
def get_best_elements(lst):
lst.sort(reverse = True)
new_lst = [lst[y] for y in range(1, len(lst)) if lst[0][0] > lst[y][0]]
return [x for x in lst if x not in new_lst]
Loop like a native
I highly recommend Ned Batchelder's talk "Loop like a native" about iterators. One of the most simple take away is that whenever you're doing range(len(lst))
, you can probably do things in a better way: more concise, clearer and more efficient.
In our case, we'd have something like:
def get_best_elements(lst):
lst.sort(reverse = True)
new_lst = [e for e in lst if e[0] < lst[0][0]]
return [x for x in lst if x not in new_lst]
Another approach
The last list comprehension can be quite expensive because for each element x
, you may perform a look-up in the list. This can lead to a O(n²)
behavior.
Instead of filtering out elements that do not correspond to the biggest number, we could just keep the one that do correspond to the biggest number.
def get_best_elements(lst):
if not lst:
return
lst.sort(reverse = True)
big_n = lst[0][0]
return [x for x in lst if x[0] == big_n]
Sort a smaller numbers of elements
You could use max
to get big_n
.
Also, we could perform the sorting on the filtered list so that we have fewer elements to sort:
def get_best_elements(lst):
if not lst:
return
big_n = max(lst)[0]
return list(sorted((x for x in lst if x[0] == big_n), reverse=True))
$endgroup$
add a comment |
$begingroup$
Names
Using the name list
for a variable is legal but usually avoided because it hides the list
builtin. The name lst
would be a better candidate.
The function name comp
could be improved as well. Maybe get_best_elements
(even though I am not fully convinced).
Tests
Before going further, it may be wise to add tests so that you can easily be sure that we do not break anything.
A proper solution would involve a unit-test framework but for the time being, we can go for the simple:
TESTS = (
(, ),
([[1,'a']], [[1, 'a']]),
([[1,'a'], [1,'a'], [1,'a'], [1,'a']], [[1,'a'], [1,'a'], [1,'a'], [1,'a']]),
([[1,'a'], [1,'b'], [1,'c'], [1,'d']], [[1,'d'], [1,'c'], [1,'b'], [1,'a']]),
([[1,'a'],[3,'b'],[7,'c'],[6,'d'],[4,'e'],[1,'f'],[7,'g'],[4,'h'],[6,'i']], [[7, 'g'], [7, 'c']]),
([[1,'a'],[3,'b'],[7,'c'],[6,'d'],[4,'e'],[1,'f'],[4,'h'],[6,'i']], [[7, 'c']]),
)
for test_input, expected_output in TESTS:
output = get_best_elements(test_input)
assert output == expected_output
print("OK")
Style
Parenthesis are not required in the return statements.
Also, the temporary variable is not really required.
List comprehension
The new_list
could be defined with a list comprehension.
def get_best_elements(lst):
lst.sort(reverse = True)
new_lst = [lst[y] for y in range(1, len(lst)) if lst[0][0] > lst[y][0]]
return [x for x in lst if x not in new_lst]
Loop like a native
I highly recommend Ned Batchelder's talk "Loop like a native" about iterators. One of the most simple take away is that whenever you're doing range(len(lst))
, you can probably do things in a better way: more concise, clearer and more efficient.
In our case, we'd have something like:
def get_best_elements(lst):
lst.sort(reverse = True)
new_lst = [e for e in lst if e[0] < lst[0][0]]
return [x for x in lst if x not in new_lst]
Another approach
The last list comprehension can be quite expensive because for each element x
, you may perform a look-up in the list. This can lead to a O(n²)
behavior.
Instead of filtering out elements that do not correspond to the biggest number, we could just keep the one that do correspond to the biggest number.
def get_best_elements(lst):
if not lst:
return
lst.sort(reverse = True)
big_n = lst[0][0]
return [x for x in lst if x[0] == big_n]
Sort a smaller numbers of elements
You could use max
to get big_n
.
Also, we could perform the sorting on the filtered list so that we have fewer elements to sort:
def get_best_elements(lst):
if not lst:
return
big_n = max(lst)[0]
return list(sorted((x for x in lst if x[0] == big_n), reverse=True))
$endgroup$
add a comment |
$begingroup$
Names
Using the name list
for a variable is legal but usually avoided because it hides the list
builtin. The name lst
would be a better candidate.
The function name comp
could be improved as well. Maybe get_best_elements
(even though I am not fully convinced).
Tests
Before going further, it may be wise to add tests so that you can easily be sure that we do not break anything.
A proper solution would involve a unit-test framework but for the time being, we can go for the simple:
TESTS = (
(, ),
([[1,'a']], [[1, 'a']]),
([[1,'a'], [1,'a'], [1,'a'], [1,'a']], [[1,'a'], [1,'a'], [1,'a'], [1,'a']]),
([[1,'a'], [1,'b'], [1,'c'], [1,'d']], [[1,'d'], [1,'c'], [1,'b'], [1,'a']]),
([[1,'a'],[3,'b'],[7,'c'],[6,'d'],[4,'e'],[1,'f'],[7,'g'],[4,'h'],[6,'i']], [[7, 'g'], [7, 'c']]),
([[1,'a'],[3,'b'],[7,'c'],[6,'d'],[4,'e'],[1,'f'],[4,'h'],[6,'i']], [[7, 'c']]),
)
for test_input, expected_output in TESTS:
output = get_best_elements(test_input)
assert output == expected_output
print("OK")
Style
Parenthesis are not required in the return statements.
Also, the temporary variable is not really required.
List comprehension
The new_list
could be defined with a list comprehension.
def get_best_elements(lst):
lst.sort(reverse = True)
new_lst = [lst[y] for y in range(1, len(lst)) if lst[0][0] > lst[y][0]]
return [x for x in lst if x not in new_lst]
Loop like a native
I highly recommend Ned Batchelder's talk "Loop like a native" about iterators. One of the most simple take away is that whenever you're doing range(len(lst))
, you can probably do things in a better way: more concise, clearer and more efficient.
In our case, we'd have something like:
def get_best_elements(lst):
lst.sort(reverse = True)
new_lst = [e for e in lst if e[0] < lst[0][0]]
return [x for x in lst if x not in new_lst]
Another approach
The last list comprehension can be quite expensive because for each element x
, you may perform a look-up in the list. This can lead to a O(n²)
behavior.
Instead of filtering out elements that do not correspond to the biggest number, we could just keep the one that do correspond to the biggest number.
def get_best_elements(lst):
if not lst:
return
lst.sort(reverse = True)
big_n = lst[0][0]
return [x for x in lst if x[0] == big_n]
Sort a smaller numbers of elements
You could use max
to get big_n
.
Also, we could perform the sorting on the filtered list so that we have fewer elements to sort:
def get_best_elements(lst):
if not lst:
return
big_n = max(lst)[0]
return list(sorted((x for x in lst if x[0] == big_n), reverse=True))
$endgroup$
Names
Using the name list
for a variable is legal but usually avoided because it hides the list
builtin. The name lst
would be a better candidate.
The function name comp
could be improved as well. Maybe get_best_elements
(even though I am not fully convinced).
Tests
Before going further, it may be wise to add tests so that you can easily be sure that we do not break anything.
A proper solution would involve a unit-test framework but for the time being, we can go for the simple:
TESTS = (
(, ),
([[1,'a']], [[1, 'a']]),
([[1,'a'], [1,'a'], [1,'a'], [1,'a']], [[1,'a'], [1,'a'], [1,'a'], [1,'a']]),
([[1,'a'], [1,'b'], [1,'c'], [1,'d']], [[1,'d'], [1,'c'], [1,'b'], [1,'a']]),
([[1,'a'],[3,'b'],[7,'c'],[6,'d'],[4,'e'],[1,'f'],[7,'g'],[4,'h'],[6,'i']], [[7, 'g'], [7, 'c']]),
([[1,'a'],[3,'b'],[7,'c'],[6,'d'],[4,'e'],[1,'f'],[4,'h'],[6,'i']], [[7, 'c']]),
)
for test_input, expected_output in TESTS:
output = get_best_elements(test_input)
assert output == expected_output
print("OK")
Style
Parenthesis are not required in the return statements.
Also, the temporary variable is not really required.
List comprehension
The new_list
could be defined with a list comprehension.
def get_best_elements(lst):
lst.sort(reverse = True)
new_lst = [lst[y] for y in range(1, len(lst)) if lst[0][0] > lst[y][0]]
return [x for x in lst if x not in new_lst]
Loop like a native
I highly recommend Ned Batchelder's talk "Loop like a native" about iterators. One of the most simple take away is that whenever you're doing range(len(lst))
, you can probably do things in a better way: more concise, clearer and more efficient.
In our case, we'd have something like:
def get_best_elements(lst):
lst.sort(reverse = True)
new_lst = [e for e in lst if e[0] < lst[0][0]]
return [x for x in lst if x not in new_lst]
Another approach
The last list comprehension can be quite expensive because for each element x
, you may perform a look-up in the list. This can lead to a O(n²)
behavior.
Instead of filtering out elements that do not correspond to the biggest number, we could just keep the one that do correspond to the biggest number.
def get_best_elements(lst):
if not lst:
return
lst.sort(reverse = True)
big_n = lst[0][0]
return [x for x in lst if x[0] == big_n]
Sort a smaller numbers of elements
You could use max
to get big_n
.
Also, we could perform the sorting on the filtered list so that we have fewer elements to sort:
def get_best_elements(lst):
if not lst:
return
big_n = max(lst)[0]
return list(sorted((x for x in lst if x[0] == big_n), reverse=True))
answered Dec 21 '18 at 16:09
JosayJosay
25.7k14087
25.7k14087
add a comment |
add a comment |
$begingroup$
You could simplify by directly making a list with the highest scores instead of creating and subtracting another list containing what you don't want :
def comp(rolls):
top_rolls =
rolls.sort(reverse=True)
for y in range(len(rolls)):
if rolls[0][0] == rolls[y][0]:
top_rolls.append(rolls[y])
return top_rolls
Once you've done that it becomes easy to fit it in a list comprehension:
def comp(list):
rolls.sort(reverse=True)
top_rolls = [rolls[y] for y in range(len(rolls)) if rolls[0][0] == rolls[y][0]]
return top_rolls
Or even shorter:
def comp(list):
list.sort(reverse=True)
return [rolls[y] for y in range(len(rolls)) if rolls[0][0] == rolls[y][0]]
$endgroup$
$begingroup$
That's great, thank you. I don't know why I created a list of things that I don't want just to subtract it from the overall list. Your method is much more concise.
$endgroup$
– JacobCheverie
Dec 21 '18 at 16:55
$begingroup$
Don't name an argumentlist
. That will shadow the built-in type also calledlist
.
$endgroup$
– Reinderien
Dec 21 '18 at 18:00
$begingroup$
I am renaming everything after I am finished, I am more concerned about the appropriate syntax for now.
$endgroup$
– JacobCheverie
Dec 21 '18 at 18:45
add a comment |
$begingroup$
You could simplify by directly making a list with the highest scores instead of creating and subtracting another list containing what you don't want :
def comp(rolls):
top_rolls =
rolls.sort(reverse=True)
for y in range(len(rolls)):
if rolls[0][0] == rolls[y][0]:
top_rolls.append(rolls[y])
return top_rolls
Once you've done that it becomes easy to fit it in a list comprehension:
def comp(list):
rolls.sort(reverse=True)
top_rolls = [rolls[y] for y in range(len(rolls)) if rolls[0][0] == rolls[y][0]]
return top_rolls
Or even shorter:
def comp(list):
list.sort(reverse=True)
return [rolls[y] for y in range(len(rolls)) if rolls[0][0] == rolls[y][0]]
$endgroup$
$begingroup$
That's great, thank you. I don't know why I created a list of things that I don't want just to subtract it from the overall list. Your method is much more concise.
$endgroup$
– JacobCheverie
Dec 21 '18 at 16:55
$begingroup$
Don't name an argumentlist
. That will shadow the built-in type also calledlist
.
$endgroup$
– Reinderien
Dec 21 '18 at 18:00
$begingroup$
I am renaming everything after I am finished, I am more concerned about the appropriate syntax for now.
$endgroup$
– JacobCheverie
Dec 21 '18 at 18:45
add a comment |
$begingroup$
You could simplify by directly making a list with the highest scores instead of creating and subtracting another list containing what you don't want :
def comp(rolls):
top_rolls =
rolls.sort(reverse=True)
for y in range(len(rolls)):
if rolls[0][0] == rolls[y][0]:
top_rolls.append(rolls[y])
return top_rolls
Once you've done that it becomes easy to fit it in a list comprehension:
def comp(list):
rolls.sort(reverse=True)
top_rolls = [rolls[y] for y in range(len(rolls)) if rolls[0][0] == rolls[y][0]]
return top_rolls
Or even shorter:
def comp(list):
list.sort(reverse=True)
return [rolls[y] for y in range(len(rolls)) if rolls[0][0] == rolls[y][0]]
$endgroup$
You could simplify by directly making a list with the highest scores instead of creating and subtracting another list containing what you don't want :
def comp(rolls):
top_rolls =
rolls.sort(reverse=True)
for y in range(len(rolls)):
if rolls[0][0] == rolls[y][0]:
top_rolls.append(rolls[y])
return top_rolls
Once you've done that it becomes easy to fit it in a list comprehension:
def comp(list):
rolls.sort(reverse=True)
top_rolls = [rolls[y] for y in range(len(rolls)) if rolls[0][0] == rolls[y][0]]
return top_rolls
Or even shorter:
def comp(list):
list.sort(reverse=True)
return [rolls[y] for y in range(len(rolls)) if rolls[0][0] == rolls[y][0]]
edited Dec 25 '18 at 16:46
answered Dec 21 '18 at 14:56
Comte_ZeroComte_Zero
13712
13712
$begingroup$
That's great, thank you. I don't know why I created a list of things that I don't want just to subtract it from the overall list. Your method is much more concise.
$endgroup$
– JacobCheverie
Dec 21 '18 at 16:55
$begingroup$
Don't name an argumentlist
. That will shadow the built-in type also calledlist
.
$endgroup$
– Reinderien
Dec 21 '18 at 18:00
$begingroup$
I am renaming everything after I am finished, I am more concerned about the appropriate syntax for now.
$endgroup$
– JacobCheverie
Dec 21 '18 at 18:45
add a comment |
$begingroup$
That's great, thank you. I don't know why I created a list of things that I don't want just to subtract it from the overall list. Your method is much more concise.
$endgroup$
– JacobCheverie
Dec 21 '18 at 16:55
$begingroup$
Don't name an argumentlist
. That will shadow the built-in type also calledlist
.
$endgroup$
– Reinderien
Dec 21 '18 at 18:00
$begingroup$
I am renaming everything after I am finished, I am more concerned about the appropriate syntax for now.
$endgroup$
– JacobCheverie
Dec 21 '18 at 18:45
$begingroup$
That's great, thank you. I don't know why I created a list of things that I don't want just to subtract it from the overall list. Your method is much more concise.
$endgroup$
– JacobCheverie
Dec 21 '18 at 16:55
$begingroup$
That's great, thank you. I don't know why I created a list of things that I don't want just to subtract it from the overall list. Your method is much more concise.
$endgroup$
– JacobCheverie
Dec 21 '18 at 16:55
$begingroup$
Don't name an argument
list
. That will shadow the built-in type also called list
.$endgroup$
– Reinderien
Dec 21 '18 at 18:00
$begingroup$
Don't name an argument
list
. That will shadow the built-in type also called list
.$endgroup$
– Reinderien
Dec 21 '18 at 18:00
$begingroup$
I am renaming everything after I am finished, I am more concerned about the appropriate syntax for now.
$endgroup$
– JacobCheverie
Dec 21 '18 at 18:45
$begingroup$
I am renaming everything after I am finished, I am more concerned about the appropriate syntax for now.
$endgroup$
– JacobCheverie
Dec 21 '18 at 18:45
add a comment |
$begingroup$
Based on your stated use case:
I would like to rerun the function if the new list contains more than one element.
you don't even need to return a list; just return the highest element's roll and name, which can be unpacked by the caller as a 2-tuple:
def comp(rolls):
return max(rolls)
That said, you haven't explicitly stated how to resolve ties with rolls of the same value. That will affect this solution.
$endgroup$
$begingroup$
The max function on it's own will only return one players roll, so if there is a tie it will not be handled. Working with a simplification similar to one that @Comte_Zero provided, I am handling the case of a tie separately. I am sure that code could use some enhancing as well, I am just trying to get the basic form nailed down and make edits later.
$endgroup$
– JacobCheverie
Dec 21 '18 at 18:47
$begingroup$
I'd propose that tie resolution should be built into this function.
$endgroup$
– Reinderien
Dec 21 '18 at 19:12
add a comment |
$begingroup$
Based on your stated use case:
I would like to rerun the function if the new list contains more than one element.
you don't even need to return a list; just return the highest element's roll and name, which can be unpacked by the caller as a 2-tuple:
def comp(rolls):
return max(rolls)
That said, you haven't explicitly stated how to resolve ties with rolls of the same value. That will affect this solution.
$endgroup$
$begingroup$
The max function on it's own will only return one players roll, so if there is a tie it will not be handled. Working with a simplification similar to one that @Comte_Zero provided, I am handling the case of a tie separately. I am sure that code could use some enhancing as well, I am just trying to get the basic form nailed down and make edits later.
$endgroup$
– JacobCheverie
Dec 21 '18 at 18:47
$begingroup$
I'd propose that tie resolution should be built into this function.
$endgroup$
– Reinderien
Dec 21 '18 at 19:12
add a comment |
$begingroup$
Based on your stated use case:
I would like to rerun the function if the new list contains more than one element.
you don't even need to return a list; just return the highest element's roll and name, which can be unpacked by the caller as a 2-tuple:
def comp(rolls):
return max(rolls)
That said, you haven't explicitly stated how to resolve ties with rolls of the same value. That will affect this solution.
$endgroup$
Based on your stated use case:
I would like to rerun the function if the new list contains more than one element.
you don't even need to return a list; just return the highest element's roll and name, which can be unpacked by the caller as a 2-tuple:
def comp(rolls):
return max(rolls)
That said, you haven't explicitly stated how to resolve ties with rolls of the same value. That will affect this solution.
answered Dec 21 '18 at 18:21
ReinderienReinderien
4,140822
4,140822
$begingroup$
The max function on it's own will only return one players roll, so if there is a tie it will not be handled. Working with a simplification similar to one that @Comte_Zero provided, I am handling the case of a tie separately. I am sure that code could use some enhancing as well, I am just trying to get the basic form nailed down and make edits later.
$endgroup$
– JacobCheverie
Dec 21 '18 at 18:47
$begingroup$
I'd propose that tie resolution should be built into this function.
$endgroup$
– Reinderien
Dec 21 '18 at 19:12
add a comment |
$begingroup$
The max function on it's own will only return one players roll, so if there is a tie it will not be handled. Working with a simplification similar to one that @Comte_Zero provided, I am handling the case of a tie separately. I am sure that code could use some enhancing as well, I am just trying to get the basic form nailed down and make edits later.
$endgroup$
– JacobCheverie
Dec 21 '18 at 18:47
$begingroup$
I'd propose that tie resolution should be built into this function.
$endgroup$
– Reinderien
Dec 21 '18 at 19:12
$begingroup$
The max function on it's own will only return one players roll, so if there is a tie it will not be handled. Working with a simplification similar to one that @Comte_Zero provided, I am handling the case of a tie separately. I am sure that code could use some enhancing as well, I am just trying to get the basic form nailed down and make edits later.
$endgroup$
– JacobCheverie
Dec 21 '18 at 18:47
$begingroup$
The max function on it's own will only return one players roll, so if there is a tie it will not be handled. Working with a simplification similar to one that @Comte_Zero provided, I am handling the case of a tie separately. I am sure that code could use some enhancing as well, I am just trying to get the basic form nailed down and make edits later.
$endgroup$
– JacobCheverie
Dec 21 '18 at 18:47
$begingroup$
I'd propose that tie resolution should be built into this function.
$endgroup$
– Reinderien
Dec 21 '18 at 19:12
$begingroup$
I'd propose that tie resolution should be built into this function.
$endgroup$
– Reinderien
Dec 21 '18 at 19:12
add a comment |
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$begingroup$
I'd like to point out that "maybe" has no place in a proper requirements statement.
$endgroup$
– Reinderien
Dec 21 '18 at 15:50
$begingroup$
Thanks Reinderien, I was simply writing shorthand instead of saying that I would like to rerun the function if the new list contains more than one element. I understand the logic behind your statement.
$endgroup$
– JacobCheverie
Dec 21 '18 at 16:53