How can I get the minimum and the maximum element of a list in python

Question

If a have a list like:

l = [1,2,3,4,5]

and I want to have at the end

min = 1   
max = 5

WITHOUT

Answer 1

If you only require one looping through the list, you could use reduce a (not so) creative way. The helper function could have been reduced as a lambda but I don't do it for sake of readability:

>>> def f(solutions, item):
...     return (item if item < solutions[0] else solutions[0],
...             item if item > solutions[1] else solutions[1])
... 

>>> L = [i for i in range(5)]
>>> functools.reduce(f, L, (L[0],L[0]))
(0, 4)

Answer 2

For finding max:

print reduce(lambda x,y: x if x>y else y, map(int,raw_input().split()))

For finding min:

print reduce(lambda x,y: x if x<y else y, map(int,raw_input().split()))

Answer 3

If you are trying to avoid using two loops, hoping a single loop will be faster, you need to reconsider. Calling two O(N) functions still gives you a O(N) algorithm, all you do is double the constant per-iteration cost. A single Python loop with comparisons can't do better than O(N) either (unless your data is already sorted), and interpreting bytecode for each iteration has a sizeable constant cost too. Which approach has the higher constant cost can only be determined by timing your runs.

To do this in a single loop, iterate over the list and test each item against the minimum and maximum found so far. float('inf') and float('-inf') (infinity and negative infinity) are good starting points to simplify the logic:

minimum = float('inf')
maximum = float('-inf')
for item in l:
    if item < minimum:
        minimum = item
    if item > maximum:
        maximum = item

Alternatively, start with the first element and only loop over the rest. Turn the list into an iterable first, store the first element as the result-to-date, and then loop over the rest:

iterl = iter(l)
minimum = maximum = next(iterl)
for item in iterl:
    if item < minimum:
        minimum = item
    if item > maximum:
        maximum = item

Don't use sorting. Python's Tim Sort implementation is a O(N log N) algorithm, which can be expected to be slower than a straight-up O(N) approach.

Timing comparisons with a larger, random list:

>>> from random import shuffle
>>> l = list(range(1000))
>>> shuffle(l)
>>> from timeit import timeit
>>> def straight_min_max(l):
...     return min(l), max(l)
... 
>>> def sorted_min_max(l):
...     s = sorted(l)
...     return s[0], s[-1]
... 
>>> def looping(l):
...     l = iter(l)
...     min = max = next(l)
...     for i in l:
...         if i < min: min = i
...         if i > max: max = i
...     return min, max
... 
>>> timeit('f(l)', 'from __main__ import straight_min_max as f, l', number=10000)
0.5266690254211426
>>> timeit('f(l)', 'from __main__ import sorted_min_max as f, l', number=10000)
2.162343978881836
>>> timeit('f(l)', 'from __main__ import looping as f, l', number=10000)
1.1799919605255127

So even for lists of 1000 elements, the min() and max() functions are fastest. Sorting is slowest here. The sorting version can be faster if you allow for in-place sorting, but then you'd need to generate a new random list for each timed run as well.

Moving to a million items (and only 10 tests per timed run), we see:

>>> l = list(range(1000000))
>>> shuffle(l)
>>> timeit('f(l)', 'from __main__ import straight_min_max as f, l', number=10)
1.6176080703735352
>>> timeit('f(l)', 'from __main__ import sorted_min_max as f, l', number=10)
6.310506105422974
>>> timeit('f(l)', 'from __main__ import looping as f, l', number=10)
1.7502741813659668

Last but not least, using a million items and l.sort() instead of sorted():

>>> def sort_min_max(l):
...     l.sort()
...     return l[0], l[-1]
... 
>>> timeit('f(l[:])', 'from __main__ import straight_min_max as f, l', number=10)
1.8858389854431152
>>> timeit('f(l[:])', 'from __main__ import sort_min_max as f, l', number=10)
8.408858060836792
>>> timeit('f(l[:])', 'from __main__ import looping as f, l', number=10)
2.003532886505127

Note the l[:]; we give each test run a copy of the list.

Conclusion: even for large lists, you are better off using the min() and max() functions anyway, it is hard to beat the low per-iteration cost of a good C loop. But if you have to forgo those functions, the straight loop is the next better option.

Answer 4

The fastest approach I can think of would be to sort the original list and then pick the first and last elements. This avoids looping multiple times, but it does destroy the original structure of your list. This can be solved by simply copying the list and sorting only the copied list. I was curious if this was slower than just using max() and min() with this quick example script:

import time

l = [1,2,4,5,3]

print "Run 1"
t1 = time.time()
print "Min =", min(l)
print "Max =", max(l)
print "time =", time.time() - t1
print ""
print "l =", l
print ""


l = [1,2,4,5,3]
l1 = list(l)

print "Run 2"
t1 = time.time()
l1.sort()
print "Min =", l1[0]
print "Max =", l1[-1]
print "time =", time.time() - t1
print ""
print "l =", l
print "l1 =", l1
print ""


l = [1,2,4,5,3]

print "Run 3"
minimum = float('inf')
maximum = float('-inf')
for item in l:
    if item < minimum:
        minimum = item
    if item > maximum:
        maximum = item
print "Min =", minimum
print "Max =", maximum
print "time =", time.time() - t1
print ""
print "l =", l

Surprisingly, the second approach is faster by about 10ms on my computer. Not sure how effective this would be with very large list, but this approach is faster for at least the example list you provided.

I added @Martijn Pieters's simple loop algorithm to my timing script. (As timing would be the only important parameter worth exploring in this question.) My results are:

Run 1: 0.0199999809265s
Run 2: 0.00999999046326s
Run 3: 0.0299999713898s

---

Edit: Inclusion of timeit module for timing.

import timeit
from random import shuffle

l = range(10000)
shuffle(l)

def Run_1():
    #print "Min =", min(l)
    #print "Max =", max(l)
    return min(l), max(l)

def Run_2():
    l1 = list(l)
    l1.sort()
    #print "Min =", l1[0]
    #print "Max =", l1[-1]
    return l1[0], l1[-1]


def Run_3():
    minimum = float('inf')
    maximum = float('-inf')
    for item in l:
        if item < minimum:
            minimum = item
        if item > maximum:
            maximum = item
    #print "Min =", minimum
    #print "Max =", maximum
    return minimum, maximum


if __name__ == '__main__':
    num_runs = 10000
    print "Run 1"
    run1 = timeit.Timer(Run_1)
    time_run1 = run1.repeat(3, num_runs)
    print ""
    print "Run 2"
    run2 = timeit.Timer(Run_2)
    time_run2 = run2.repeat(3,num_runs)
    print ""
    print "Run 3"
    run3 = timeit.Timer(Run_3)
    time_run3 = run3.repeat(3,num_runs)
    print ""

    print "Run 1"
    for each_time in time_run1:
        print "time =", each_time
    print ""
    print "Run 2"
    for each_time in time_run2:
        print "time =", each_time
    print ""
    print "Run 3"
    for each_time in time_run3:
        print "time =", each_time
    print ""

My results are:

Run 1
time = 3.42100585452
time = 3.39309908229
time = 3.47903182233

Run 2
time = 26.5261287922
time = 26.2023346397
time = 26.7324208568

Run 3
time = 3.29800945144
time = 3.25067545773
time = 3.29783778232

sort algorithm is very slow for large arrays.

Answer 5

>>> L = [1,2,3,4,5]
>>> reduce(lambda x, y: x if x<y else y, L)
1
>>> reduce(lambda x, y: x if x>y else y, L)
5

Another way

>>> it = iter(L)
>>> mn = mx = next(it)
>>> for i in it:
...  if i<mn:mn=i
...  if i>mx:mx=i
... 
>>> mn
1
>>> mx
5

Answer 6

Homework questions with weird restrictions demand cheat answers

>>> l = [1,2,3,4,5]
>>> sorted(l)[::len(l)-1]
[1, 5]