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Reimplementation of min() in Python


Reimplementation of C++ vectorSize improvements for cat reimplementationReimplementation of a Hashtable and hash functionReimplementation of Underscore _.throttleReimplementation of Laravel's config systemReimplementation of Python's dequeOptimisation suggestions for Min HeapMin/Max Heap implementation in PythonReimplementation of call() method in JavaScriptMin Heap implementation [C++]






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








1












$begingroup$


I'm reimplementing the min() function as an exercise (EDIT: not all the functionality of the python std library function, just the minimum of a list of numbers). Here is my code:



def my_min(num_list):
minimum = num_list[0]
for num in num_list[1:]:
if num < minimum:
minimum = num
return minimum


My question is: How bad is num_list[1:] in the for loop? And are there any other optimizations I could make to the code?



My intention by truncating the list is to avoid comparing the list's first element to itself. While insignificant in terms of wasted time and resources, I just find it lacking elegance.










share|improve this question











$endgroup$













  • $begingroup$
    what happens if num_list has only 1 element?
    $endgroup$
    – dfhwze
    8 hours ago










  • $begingroup$
    @dfhwze, I should check for the list having no elements anyway, so I could check for its having only 1 element and return that element
    $endgroup$
    – jeremy radcliff
    8 hours ago










  • $begingroup$
    If the purpose is to comply to the specification of min, you are far away from home: programiz.com/python-programming/methods/built-in/min.
    $endgroup$
    – dfhwze
    8 hours ago






  • 1




    $begingroup$
    @dfhwze, good point, I edited my question
    $endgroup$
    – jeremy radcliff
    8 hours ago

















1












$begingroup$


I'm reimplementing the min() function as an exercise (EDIT: not all the functionality of the python std library function, just the minimum of a list of numbers). Here is my code:



def my_min(num_list):
minimum = num_list[0]
for num in num_list[1:]:
if num < minimum:
minimum = num
return minimum


My question is: How bad is num_list[1:] in the for loop? And are there any other optimizations I could make to the code?



My intention by truncating the list is to avoid comparing the list's first element to itself. While insignificant in terms of wasted time and resources, I just find it lacking elegance.










share|improve this question











$endgroup$













  • $begingroup$
    what happens if num_list has only 1 element?
    $endgroup$
    – dfhwze
    8 hours ago










  • $begingroup$
    @dfhwze, I should check for the list having no elements anyway, so I could check for its having only 1 element and return that element
    $endgroup$
    – jeremy radcliff
    8 hours ago










  • $begingroup$
    If the purpose is to comply to the specification of min, you are far away from home: programiz.com/python-programming/methods/built-in/min.
    $endgroup$
    – dfhwze
    8 hours ago






  • 1




    $begingroup$
    @dfhwze, good point, I edited my question
    $endgroup$
    – jeremy radcliff
    8 hours ago













1












1








1





$begingroup$


I'm reimplementing the min() function as an exercise (EDIT: not all the functionality of the python std library function, just the minimum of a list of numbers). Here is my code:



def my_min(num_list):
minimum = num_list[0]
for num in num_list[1:]:
if num < minimum:
minimum = num
return minimum


My question is: How bad is num_list[1:] in the for loop? And are there any other optimizations I could make to the code?



My intention by truncating the list is to avoid comparing the list's first element to itself. While insignificant in terms of wasted time and resources, I just find it lacking elegance.










share|improve this question











$endgroup$




I'm reimplementing the min() function as an exercise (EDIT: not all the functionality of the python std library function, just the minimum of a list of numbers). Here is my code:



def my_min(num_list):
minimum = num_list[0]
for num in num_list[1:]:
if num < minimum:
minimum = num
return minimum


My question is: How bad is num_list[1:] in the for loop? And are there any other optimizations I could make to the code?



My intention by truncating the list is to avoid comparing the list's first element to itself. While insignificant in terms of wasted time and resources, I just find it lacking elegance.







python reinventing-the-wheel






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 5 hours ago









200_success

135k21 gold badges173 silver badges443 bronze badges




135k21 gold badges173 silver badges443 bronze badges










asked 8 hours ago









jeremy radcliffjeremy radcliff

2502 silver badges9 bronze badges




2502 silver badges9 bronze badges














  • $begingroup$
    what happens if num_list has only 1 element?
    $endgroup$
    – dfhwze
    8 hours ago










  • $begingroup$
    @dfhwze, I should check for the list having no elements anyway, so I could check for its having only 1 element and return that element
    $endgroup$
    – jeremy radcliff
    8 hours ago










  • $begingroup$
    If the purpose is to comply to the specification of min, you are far away from home: programiz.com/python-programming/methods/built-in/min.
    $endgroup$
    – dfhwze
    8 hours ago






  • 1




    $begingroup$
    @dfhwze, good point, I edited my question
    $endgroup$
    – jeremy radcliff
    8 hours ago
















  • $begingroup$
    what happens if num_list has only 1 element?
    $endgroup$
    – dfhwze
    8 hours ago










  • $begingroup$
    @dfhwze, I should check for the list having no elements anyway, so I could check for its having only 1 element and return that element
    $endgroup$
    – jeremy radcliff
    8 hours ago










  • $begingroup$
    If the purpose is to comply to the specification of min, you are far away from home: programiz.com/python-programming/methods/built-in/min.
    $endgroup$
    – dfhwze
    8 hours ago






  • 1




    $begingroup$
    @dfhwze, good point, I edited my question
    $endgroup$
    – jeremy radcliff
    8 hours ago















$begingroup$
what happens if num_list has only 1 element?
$endgroup$
– dfhwze
8 hours ago




$begingroup$
what happens if num_list has only 1 element?
$endgroup$
– dfhwze
8 hours ago












$begingroup$
@dfhwze, I should check for the list having no elements anyway, so I could check for its having only 1 element and return that element
$endgroup$
– jeremy radcliff
8 hours ago




$begingroup$
@dfhwze, I should check for the list having no elements anyway, so I could check for its having only 1 element and return that element
$endgroup$
– jeremy radcliff
8 hours ago












$begingroup$
If the purpose is to comply to the specification of min, you are far away from home: programiz.com/python-programming/methods/built-in/min.
$endgroup$
– dfhwze
8 hours ago




$begingroup$
If the purpose is to comply to the specification of min, you are far away from home: programiz.com/python-programming/methods/built-in/min.
$endgroup$
– dfhwze
8 hours ago




1




1




$begingroup$
@dfhwze, good point, I edited my question
$endgroup$
– jeremy radcliff
8 hours ago




$begingroup$
@dfhwze, good point, I edited my question
$endgroup$
– jeremy radcliff
8 hours ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

iterators



You can use an iterator



def my_min(num_list):
# empty lists
if not num_list:
raise ValueError('Empty list')

list_iter = iter(num_list)
minimum = next(list_iter)
for num in list_iter:
if num < minimum:
minimum = num
return minimum


In response to Mathias' comment, here is a version that works with an iterable:



def my_min(seq):
seq = iter(seq)

try:
minimum = next(seq)

for num in seq:
if num < minimum:
minimum = num

return minimum

except StopIteration as e:
pass

raise ValueError('Empty list')





share|improve this answer











$endgroup$










  • 1




    $begingroup$
    This is the right approach to avoid duplicating the whole list in memory. If only you took the special case of the empty parameter around the next call, you could handle all kind of iterables, not only lists.
    $endgroup$
    – Mathias Ettinger
    6 hours ago










  • $begingroup$
    @MathiasEttinger good idea. Added it to the answer.
    $endgroup$
    – RootTwo
    5 hours ago


















1












$begingroup$

Review




I should check for the list having no elements anyway, so I could check for its having only 1 element and return that element




  • You have implemented a simple function, so it shouldn't have been that hard to provide a couple of unit tests. You would have immediately found bugs on the most obvious edge cases as (1) empty list and self-created edge case (2) single item.


While insignificant in terms of wasted time and resources, I just
find it lacking elegance.




  • What you gain in elegance is lost by the edge case guards you'd have to build in to fix the bugs.





share|improve this answer









$endgroup$














  • $begingroup$
    To your second point, are you saying there's no way to get the best of both worlds and therefore comparing the first element to itself is the way to go in terms of readability of code, etc...?
    $endgroup$
    – jeremy radcliff
    8 hours ago













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

iterators



You can use an iterator



def my_min(num_list):
# empty lists
if not num_list:
raise ValueError('Empty list')

list_iter = iter(num_list)
minimum = next(list_iter)
for num in list_iter:
if num < minimum:
minimum = num
return minimum


In response to Mathias' comment, here is a version that works with an iterable:



def my_min(seq):
seq = iter(seq)

try:
minimum = next(seq)

for num in seq:
if num < minimum:
minimum = num

return minimum

except StopIteration as e:
pass

raise ValueError('Empty list')





share|improve this answer











$endgroup$










  • 1




    $begingroup$
    This is the right approach to avoid duplicating the whole list in memory. If only you took the special case of the empty parameter around the next call, you could handle all kind of iterables, not only lists.
    $endgroup$
    – Mathias Ettinger
    6 hours ago










  • $begingroup$
    @MathiasEttinger good idea. Added it to the answer.
    $endgroup$
    – RootTwo
    5 hours ago















3












$begingroup$

iterators



You can use an iterator



def my_min(num_list):
# empty lists
if not num_list:
raise ValueError('Empty list')

list_iter = iter(num_list)
minimum = next(list_iter)
for num in list_iter:
if num < minimum:
minimum = num
return minimum


In response to Mathias' comment, here is a version that works with an iterable:



def my_min(seq):
seq = iter(seq)

try:
minimum = next(seq)

for num in seq:
if num < minimum:
minimum = num

return minimum

except StopIteration as e:
pass

raise ValueError('Empty list')





share|improve this answer











$endgroup$










  • 1




    $begingroup$
    This is the right approach to avoid duplicating the whole list in memory. If only you took the special case of the empty parameter around the next call, you could handle all kind of iterables, not only lists.
    $endgroup$
    – Mathias Ettinger
    6 hours ago










  • $begingroup$
    @MathiasEttinger good idea. Added it to the answer.
    $endgroup$
    – RootTwo
    5 hours ago













3












3








3





$begingroup$

iterators



You can use an iterator



def my_min(num_list):
# empty lists
if not num_list:
raise ValueError('Empty list')

list_iter = iter(num_list)
minimum = next(list_iter)
for num in list_iter:
if num < minimum:
minimum = num
return minimum


In response to Mathias' comment, here is a version that works with an iterable:



def my_min(seq):
seq = iter(seq)

try:
minimum = next(seq)

for num in seq:
if num < minimum:
minimum = num

return minimum

except StopIteration as e:
pass

raise ValueError('Empty list')





share|improve this answer











$endgroup$



iterators



You can use an iterator



def my_min(num_list):
# empty lists
if not num_list:
raise ValueError('Empty list')

list_iter = iter(num_list)
minimum = next(list_iter)
for num in list_iter:
if num < minimum:
minimum = num
return minimum


In response to Mathias' comment, here is a version that works with an iterable:



def my_min(seq):
seq = iter(seq)

try:
minimum = next(seq)

for num in seq:
if num < minimum:
minimum = num

return minimum

except StopIteration as e:
pass

raise ValueError('Empty list')






share|improve this answer














share|improve this answer



share|improve this answer








edited 5 hours ago

























answered 8 hours ago









RootTwoRootTwo

1,2293 silver badges6 bronze badges




1,2293 silver badges6 bronze badges










  • 1




    $begingroup$
    This is the right approach to avoid duplicating the whole list in memory. If only you took the special case of the empty parameter around the next call, you could handle all kind of iterables, not only lists.
    $endgroup$
    – Mathias Ettinger
    6 hours ago










  • $begingroup$
    @MathiasEttinger good idea. Added it to the answer.
    $endgroup$
    – RootTwo
    5 hours ago












  • 1




    $begingroup$
    This is the right approach to avoid duplicating the whole list in memory. If only you took the special case of the empty parameter around the next call, you could handle all kind of iterables, not only lists.
    $endgroup$
    – Mathias Ettinger
    6 hours ago










  • $begingroup$
    @MathiasEttinger good idea. Added it to the answer.
    $endgroup$
    – RootTwo
    5 hours ago







1




1




$begingroup$
This is the right approach to avoid duplicating the whole list in memory. If only you took the special case of the empty parameter around the next call, you could handle all kind of iterables, not only lists.
$endgroup$
– Mathias Ettinger
6 hours ago




$begingroup$
This is the right approach to avoid duplicating the whole list in memory. If only you took the special case of the empty parameter around the next call, you could handle all kind of iterables, not only lists.
$endgroup$
– Mathias Ettinger
6 hours ago












$begingroup$
@MathiasEttinger good idea. Added it to the answer.
$endgroup$
– RootTwo
5 hours ago




$begingroup$
@MathiasEttinger good idea. Added it to the answer.
$endgroup$
– RootTwo
5 hours ago













1












$begingroup$

Review




I should check for the list having no elements anyway, so I could check for its having only 1 element and return that element




  • You have implemented a simple function, so it shouldn't have been that hard to provide a couple of unit tests. You would have immediately found bugs on the most obvious edge cases as (1) empty list and self-created edge case (2) single item.


While insignificant in terms of wasted time and resources, I just
find it lacking elegance.




  • What you gain in elegance is lost by the edge case guards you'd have to build in to fix the bugs.





share|improve this answer









$endgroup$














  • $begingroup$
    To your second point, are you saying there's no way to get the best of both worlds and therefore comparing the first element to itself is the way to go in terms of readability of code, etc...?
    $endgroup$
    – jeremy radcliff
    8 hours ago















1












$begingroup$

Review




I should check for the list having no elements anyway, so I could check for its having only 1 element and return that element




  • You have implemented a simple function, so it shouldn't have been that hard to provide a couple of unit tests. You would have immediately found bugs on the most obvious edge cases as (1) empty list and self-created edge case (2) single item.


While insignificant in terms of wasted time and resources, I just
find it lacking elegance.




  • What you gain in elegance is lost by the edge case guards you'd have to build in to fix the bugs.





share|improve this answer









$endgroup$














  • $begingroup$
    To your second point, are you saying there's no way to get the best of both worlds and therefore comparing the first element to itself is the way to go in terms of readability of code, etc...?
    $endgroup$
    – jeremy radcliff
    8 hours ago













1












1








1





$begingroup$

Review




I should check for the list having no elements anyway, so I could check for its having only 1 element and return that element




  • You have implemented a simple function, so it shouldn't have been that hard to provide a couple of unit tests. You would have immediately found bugs on the most obvious edge cases as (1) empty list and self-created edge case (2) single item.


While insignificant in terms of wasted time and resources, I just
find it lacking elegance.




  • What you gain in elegance is lost by the edge case guards you'd have to build in to fix the bugs.





share|improve this answer









$endgroup$



Review




I should check for the list having no elements anyway, so I could check for its having only 1 element and return that element




  • You have implemented a simple function, so it shouldn't have been that hard to provide a couple of unit tests. You would have immediately found bugs on the most obvious edge cases as (1) empty list and self-created edge case (2) single item.


While insignificant in terms of wasted time and resources, I just
find it lacking elegance.




  • What you gain in elegance is lost by the edge case guards you'd have to build in to fix the bugs.






share|improve this answer












share|improve this answer



share|improve this answer










answered 8 hours ago









dfhwzedfhwze

8,4411 gold badge18 silver badges49 bronze badges




8,4411 gold badge18 silver badges49 bronze badges














  • $begingroup$
    To your second point, are you saying there's no way to get the best of both worlds and therefore comparing the first element to itself is the way to go in terms of readability of code, etc...?
    $endgroup$
    – jeremy radcliff
    8 hours ago
















  • $begingroup$
    To your second point, are you saying there's no way to get the best of both worlds and therefore comparing the first element to itself is the way to go in terms of readability of code, etc...?
    $endgroup$
    – jeremy radcliff
    8 hours ago















$begingroup$
To your second point, are you saying there's no way to get the best of both worlds and therefore comparing the first element to itself is the way to go in terms of readability of code, etc...?
$endgroup$
– jeremy radcliff
8 hours ago




$begingroup$
To your second point, are you saying there's no way to get the best of both worlds and therefore comparing the first element to itself is the way to go in terms of readability of code, etc...?
$endgroup$
– jeremy radcliff
8 hours ago

















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Smell Mother Skizze Discussion Tachometer Jar Alligator Star 끌다 자세 의문 과학적t Barbaric The round system critiques the connection. Definition: A wind instrument of music in use among the Spaniards Nasty Level 이상 분노 금년 월급 근교 Cloth Owner Permissible Shock Purring Parched Raise 오전 장면 햄 서투르다 The smash instructs the squeamish instrument. Large Nosy Nalpure Chalk Travel Crayon Bite your tongue The Hulk 신호 대사 사과하다 The work boosts the knowledgeable size. Steeplump Level Wooden Shake Teaching Jump 이제 복도 접다 공중전화 부지런하다 Rub Average Ruthless Busyglide Glost oven Didelphia Control A fly on the wall Jaws 지하철 거