How to get max value and min value of a given list into tuple form with out using built in max and min function from python

Question

I am new to python.I have list of values like data = [100,50,3,205,10] and i am trying to write single function which will get a min and max value into tuple without using min ,max function

Here is my code and when i execute this i am getting ma,mi values (3,3) - which are incorrect.

passing data into findmax function

def findmax(data):
    a = data
    ma = a[0]
    mi = a[len(a)-1]
    k = 0
    while k != len(a):
        for i in range(0,len(a)):
            k = i+1
            if ma > a[i]:
                ma = a[i]
            continue
        for l in range(0,len(a)):
            if mi > a[l]:
                mi = a[l]
     print (ma,mi)

print(findmax([100,50,3,205,10]))

Answer 1

This might be helpfull. There's no need to loop over the input data twice.

data = [100,50,3,205,10] 

def minmax(data):
    _min = data[0]
    _max = data[0]
    for i in range(len(data)):
        if _min > data[i]:
            _min = data[i]
        if _max < data[i]:
            _max = data[i]
    return (_min, _max)

Answer 2

def minmax(ls):
    if len(ls) == 0:
        return None, None # or some default val
    mini = maxi = ls[0]
    for val in ls[1:]:
        if val < mini:
            mini = val
        elif val > maxi:
            maxi = val
    return mini, maxi

This handles the case where input list length is 0. Also, I think having if-else-if makes more sense because if the value is smaller than minimum then there's no use performing the next check.

Answer 3

You should be able to go over the list only once and save min and max values.

data = [100,50,3,205,10]

def findmax(data):
    a = data
    ma = a[0]
    mi = a[0]

    for item in data:
        if item > ma:
            ma = item

        if item < mi:
            mi = item

    return (ma,mi)

print(findmax(data))

Answer 4

Here is an interesting way to do this using a bithack however any performance improvements you would see from something like this are so insignificant that the compromise between readability and performance is pointless.

def findminmax(data):
    mi = ma = data[0]
    for value in data:
        mi = value if mi ^ ((value^mi) & -(value < mi))
        ma = value if value ^ ((ma^value) & -(ma < value))
    return mi, ma

Answer 5

Here is a different approach to single pass finding of min and max, by reading values two at a time. Instead of performing 2 comparisons per item, it performs 3 for every 2 items. This also uses an iterator to move through the input, and zip_longest to go through the list in pairs:

from itertools import zip_longest

def minmax(data):
    it = iter(data)
    _min = _max = next(it)
    for a, b in zip_longest(it, it, fillvalue=_min):
        if a > b:
            # swap items to make a <= b
            a, b = b, a
        if  a < _min:
            _min = a
        if b > _max:
            _max = b
    return _min, _max

When the list elements are ints or floats, this improvement is not going to be measurable unless the lists are extremely long. But if the list items are complicated user objects, where > and < operations might involve some non-trivial amount of work, this might be worth pursuing.

Answer 6

Hello looping if you have a lot of elements in your list might take a toll on your computing time. I suggest do it in pandas instead

import pandas as pd

Then create a DataFrame using your data:

df = pd.DataFrame({'Values':data})

Using this code you'll get the max value: print(df.loc[df['Values'].idxmax()]

You'll get:

205

Hope this helps :))

Answer 7

You could do it in a single line, albeit very inefficiently, with a sort and list comprehension

minmax = [ (s[0],s[-1]) for s in [sorted(data)] ]

or, more seriously, you should take into account the possibility of being given an empty list.

minValue,maxValue = (data[:1] or [None])*2
for value in data[1:]:
    minValue = value if value < minValue else minValue
    maxValue = value if value > maxValue else maxValue