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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++]

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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.

python reinventing-the-wheel

share|improve this question

edited 5 hours ago

200_success

135k2121 gold badges173173 silver badges443443 bronze badges

asked 8 hours ago

jeremy radcliffjeremy radcliff

25022 silver badges99 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  7 hours ago
  

  
  
  
  

add a comment | 

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.

python reinventing-the-wheel

share|improve this question

edited 5 hours ago

200_success

135k2121 gold badges173173 silver badges443443 bronze badges

asked 8 hours ago

jeremy radcliffjeremy radcliff

25022 silver badges99 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

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.

python reinventing-the-wheel

share|improve this question

edited 5 hours ago

200_success

135k2121 gold badges173173 silver badges443443 bronze badges

asked 8 hours ago

jeremy radcliffjeremy radcliff

25022 silver badges99 bronze badges

$endgroup$

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

share|improve this question

edited 5 hours ago

200_success

135k2121 gold badges173173 silver badges443443 bronze badges

asked 8 hours ago

jeremy radcliffjeremy radcliff

25022 silver badges99 bronze badges

share|improve this question

edited 5 hours ago

200_success

135k2121 gold badges173173 silver badges443443 bronze badges

asked 8 hours ago

jeremy radcliffjeremy radcliff

25022 silver badges99 bronze badges

edited 5 hours ago

200_success

135k2121 gold badges173173 silver badges443443 bronze badges

edited 5 hours ago

200_success

135k2121 gold badges173173 silver badges443443 bronze badges

asked 8 hours ago

jeremy radcliffjeremy radcliff

25022 silver badges99 bronze badges

asked 8 hours ago

jeremy radcliffjeremy radcliff

25022 silver badges99 bronze badges

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

edited 5 hours ago

answered 7 hours ago

RootTwoRootTwo

1,22933 silver badges66 bronze badges

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

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

answered 7 hours ago

dfhwzedfhwze

8,44111 gold badge1818 silver badges4949 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  7 hours ago
  

  
  
  
  

add a comment | 

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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

edited 5 hours ago

answered 7 hours ago

RootTwoRootTwo

1,22933 silver badges66 bronze badges

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

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

edited 5 hours ago

answered 7 hours ago

RootTwoRootTwo

1,22933 silver badges66 bronze badges

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 5 hours ago

answered 7 hours ago

RootTwoRootTwo

1,22933 silver badges66 bronze badges

$endgroup$

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

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

edited 5 hours ago

answered 7 hours ago

RootTwoRootTwo

1,22933 silver badges66 bronze badges

answered 7 hours ago

RootTwoRootTwo

1,22933 silver badges66 bronze badges

answered 7 hours ago

RootTwoRootTwo

1,22933 silver badges66 bronze badges

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

answered 7 hours ago

dfhwzedfhwze

8,44111 gold badge1818 silver badges4949 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  7 hours ago
  

  
  
  
  

add a comment | 

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

answered 7 hours ago

dfhwzedfhwze

8,44111 gold badge1818 silver badges4949 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

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

answered 7 hours ago

dfhwzedfhwze

8,44111 gold badge1818 silver badges4949 bronze badges

$endgroup$

share|improve this answer

answered 7 hours ago

dfhwzedfhwze

8,44111 gold badge1818 silver badges4949 bronze badges

answered 7 hours ago

dfhwzedfhwze

8,44111 gold badge1818 silver badges4949 bronze badges

answered 7 hours ago

dfhwzedfhwze

8,44111 gold badge1818 silver badges4949 bronze badges