JS Problems geeksforgeeks Exercises and Solutions for Beginners

How to Get the Key with the Maximum Value in Python [All Method]️

The step-by-step guide on this page will show you How to Get the Key with the Maximum Value in Python. Error found! Why this could be happening? know and Learn everything.
Question: What is the best solution for this problem? Answer: This blog code can help you solve errors How to Get the Key with the Maximum Value in Python. Question:”What should you do if you run into code errors?” Answer:”By following this blog, you can find a solution.”

Sometimes, when we’re counting occurrences of an item, we need to obtain the item with the highest frequency.

Another way to express this is: how can we get the key with the maximum value in a dictionary?

Well. There are many ways to do this:

Given a dictionary structured as follows, we’ll walk through six methods to obtain the key with the maximum value.

d = {
  'corgi': 2,
  'shih tzu': 5,
  'pug': 16,
  'husky': 3

Method 1

To understand this method, it’s important to know that you can obtain the keys and values of a dictionary using d.keys() and d.values(). They will return an iterable list of keys and values, respectively.

arr.index(elem) will provide the index of elem in array arr.

This method involves converting all the keys and values into separate lists, finding the maximum value, obtaining its index, and returning the corresponding value in the keys list.

def method1():
  k = list(d.keys())
  v = list(d.values())
  return k[v.index(max(v))]

Method 2

Similar to d.keys() and d.values(), d.items() will return both the keys and values in a list of tuples.

Additionally, max(d) will provide us with the maximum key of dictionary d.

In this method, we reverse the original dictionary, call max() to get the maximum key in the new dictionary, and then look up the value associated with that key.

def method2():
  d_new = { v: k for k,v in d.items() }
  return d_new[max(d_new)]

Method 3

This third method makes use of lambda functions and the filter() method.

Similar to how we can use def to define a function, lambda can be used to define an inline, anonymous function.

filter() allows us to select only elements in a list that follow some condition. We use a lambda function to specify this condition. As an example, the following lambda function will return only the objects whose first element is equal to 5.

lambda t: t[0] == 5

Be sure to read up on lambda functions here.

This method uses the lambda function and filter() to retrieve only the sets of key-value pairs whose value is equivalent to the maximum value in the entire dictionary. It is then converted to a list and indexed to obtain just the key.

def method3():
  return list(filter(lambda t: t[1] == max(d.values()), d.items()))[0][0]   

Method 4

In this method, we utilize the max() function again to easily obtain the maximum of any given collection.

However, this time, we use the key parameter, which defines the basis of comparison.

In this case, we are evaluating the maximum in d.items() based on the second element, or the first index. The second element in each d.items() object is the value in the dictionary. As a result, we are comparing the value field, giving us the maximum value.

To do this, we will use the built-in operator module. You can read up on operators and itemgetters here.

import operator
def method4():
  return max(d.items(), key = operator.itemgetter(1))[0]

Method 5

The fifth and sixth methods are essentially the same as the fourth, but they take a different route to define the key.

In this method, we use the built-in get method for dictionaries, which returns the value for each key in the dictionary.

def method5():
  return max(d, key = d.get)

Method 6

We can create the same functionality using a lambda function.

Each key will simply be outputting its respective value in the dictionary. This is the same idea as the previous method.

def method6():
  return max(d, key = lambda k: d[k])  

Which Method is the Fastest?

There are all these different methods, but which one should you use?

I used Python’s timeit module to time each method separately. This module allows us to record the time it takes to execute any number of lines in a given program one million times.

I created the dictionary in test_dict and surrounded it in quotes. timeit will only execute lines inside quotes (“…execute me…”) as shown below. To execute multiple lines, we can surround the lines in three quotes (“””…execute me…”””).

test_dict = "d = {'corgi': 2, 'pug': 5, 'shih tzu': 16, 'husky': 3}"

method1 = """
v = list(d.values())
k = list(d.keys())
method2 = """
d_new = { v: k for k,v in d.items() }
method3 = """
list(filter(lambda t: t[1] == max(d.values()), d.items()))[0][0]
method4 = """
import operator
max(d.items(), key = operator.itemgetter(1))[0]
method5 = "max(d, key = d.get)"
method6 = "max(d, key = lambda k: d[k])"

Since test_dict is input as the setup parameter, the time it takes to run test_dict will not be included in our timing.

Only the code inside method# (method1, method2, etc.) will be run, as indicated by the stmt (statement) parameter.

import timeit
# 4.646706566985813
print(timeit.timeit(stmt = method1, setup = test_dict))
# 4.319815654002014
print(timeit.timeit(stmt = method2, setup = test_dict))
# 11.278023117003613
print(timeit.timeit(stmt = method3, setup = test_dict))
# 5.527265636992524
print(timeit.timeit(stmt = method4, setup = test_dict))
# 2.376368289013044
print(timeit.timeit(stmt = method5, setup = test_dict))
# 3.923244642006466
print(timeit.timeit(stmt = method6, setup = test_dict))

Over one million iterations, Method 5 runs faster than every other method, with Method 6 coming in second.

def method5():
 return max(d, key = d.get)

This is the fastest and probably the cleanest way to obtain the key with the maximum value in a dictionary.

Now you learned, How you can use & How to Get the Key with the Maximum Value in Python.
Final Note: Try to Avoid this type of mistake(error) in future!

Related Articles

Leave a Reply

Your email address will not be published.

Back to top button