Finding Median of List in Python

How can i find the median of a number from a list? Python

Here's a nice trick to avoid using if/else to handle the odd and even length cases separately: the indices in the middle are (len(nums) - 1) // 2 and len(nums) // 2. If the length is odd, then these indices are equal, so adding the values and dividing by 2 has no effect.

Note that you should do floor-division with the // operator to get an integer to use as an index.

def median(nums):
    nums = sorted(nums)
    middle1 = (len(nums) - 1) // 2
    middle2 = len(nums) // 2
    return (nums[middle1] + nums[middle2]) / 2

Examples:

>>> median([1, 2, 3, 4])
2.5
>>> median([1, 2, 3, 4, 5])
3.0

Python - median value of a list

You could also use NumPy's built-in functions, which could potentially be faster.

import numpy as np
def median(x):
    return np.median(np.array(x))

NumPy has a whole suite of array-based data analysis functions, such as Mean, Mode, Range, Standard Deviation and more: https://docs.scipy.org/doc/numpy/reference/routines.statistics.html

Hope this helps!

How do I find the median of a list in python

Hello User9123,

Main Use of the Medain() function

This module provides functions for calculating mathematical statistics of numeric (Real-valued) data.

Note Unless explicitly noted otherwise, these functions support int, float, decimal.Decimal and fractions.Fraction. Behaviour with other types (whether in the numeric tower or not) is currently unsupported. Mixed types are also undefined and implementation-dependent. If your input data consists of mixed types, you may be able to use map() to ensure a consistent result, e.g. map(float, input_data).

Your Problem

When you use this function for the string so it is not work properly some time so better i suggest you used custom function. I give the solution in my below code.

Explanation Function Median()

statistics.median(data)

Return the median (middle value) of numeric data, using the common “mean of middle two” method. If data is empty, StatisticsError is raised. data can be a sequence or iterator.

The median is a robust measure of central location, and is less affected by the presence of outliers in your data. When the number of data points is odd, the middle data point is returned:

Solution of Problem

If you use this function for the numeric value so try this below code,

def getMedian(numericValues):
  theValues = sorted(numericValues)

  if len(theValues) % 2 == 1:
    return theValues[(len(theValues)+1)/2-1]
  else:
    lower = theValues[len(theValues)/2-1]
    upper = theValues[len(theValues)/2]
    return (float(lower + upper)) / 2 

print getMedian([0,1,2,3,4,5]) # output = 2.5

If you are use median function for the string so try this below code,

def medianFind(mystring):
    return mystring[(len(mystring)-1)/2]  #len returns the length of the string

print medianFind("vmr")  #prints m

I hope my answer is helpful.

If any query so comment please.

Median of a List in Python

Exception is caused by / operator that returns float on Python 3.x. Use integer division // instead when calculating the index:

>>> 2 / 1
2.0
>>> 2 // 1
2

Finding median of a list using quick select and median-of-medians

    int finalMedian = quick_select(arrayofMedians, 0, numberOfGroups-1,rank);

Here rank is wrong - it is set to the middle of array A, but should rather be the middle of array arrayofMedians. Change to:

    int finalMedian = quick_select(arrayofMedians, 0, numberOfGroups-1, ctr/2);

Also in the function int median_of_medians(int A[], int p, int r) you're indexing array A always from index 0 on, while indexing should rather start at p. To correct this, insert A += p; near the start of the function.

Finding median of a list of list of different length

Without the overhead of numpy you could do this:

values = [
    [1, 1, 1, 18, 35, 35, 70, 133, 280],
    [1, 1, 1, 53, 90, 101, 130, 148, 178],
    [1, 1, 1, 18, 35, 133, 133, 164],
    [1, 1, 1, 18, 101, 108],
    [1, 1, 18, 36, 86, 118, 126]
]

def median(list_):
    m = len(list_) // 2
    list_ = sorted(list_) # to ensure compatibility with numpy.median
    if len(list_) % 2 == 0:
        return sum(list_[m-1:m+1]) / 2
    else:
        return float(list_[m])

for e in values:
    print(median(e))

Output: