How to find nth smallest value in python dictionary-CodePudding

If I have a Python dictionary, how do I get the key to the entry which contains the nth minimum value?

Given the input:

input = {'Apple':1, 'Banana':3, 'Orange':9,
 'Watermelon':5, 'Mango': 4, 'Peach': 4,'Blueberry':10}

def return_smallest_key(input, n):

if n = 3 
It should return 'Mango', 'Peach'

Also, if someone could introduce two different solutions,

Solution one: n = 6, return Blueberry (since Mango and Peach both equal to 4, we count them in the rank as 1,2,3,3,4,5,6)

Solution two: n = 6 return Orange (Count Mango and Peach in the rank as 1,2,3,3,5,6,7)

CodePudding user response：

Solution one:

def return_smallest_key(dictionary, n):
    ser = pd.Series(dictionary).rank(method='dense')
    return list(ser[ser==float(n)].index)

return_smallest_key(input, 6)

Solution two:

def return_smallest_key(dictionary, n):
    ser = pd.Series(dictionary).rank(method='first')
    return list(ser[ser==float(n)].index)

return_smallest_key(input, 6)

CodePudding user response：

Approach

Create sorted key, value pairs of dictionary items
Group pairs by value
Find rank by counting items from lowest value group until we get the desired rank

Note: input should never be used as a global variable names since it conflicts with builtin function

Code

from itertools import groupby

# input values (shouldn't use input as variable name in Python)
data = {'Apple':1, 
        'Banana':3, 
        'Orange':9, 
        'Watermelon':5, 
        'Mango': 4, 
        'Peach': 4,
        'Blueberry':10}

def solution_one(d, rank):
    
    ' Sort dictionary value, creating a list of tuples of key, value pairs '
    values = sorted(d.items(), key = lambda kv: kv[1])
   
    # Group tuples by value
    # Determine rank by counting pairs until we reach desired rank
    count = 0
    for k, v in groupby(values, key = lambda kv: kv[1]):
        v = list(v)
        if count   1 == rank:
            return [x[0] for x in v]  # List of values at this rank
        elif count > rank:
            return    # Will not have any values that match rank
        
        count  = 1               # update count of tuples encountered
        
def solution_two(d, rank):
    
    ' Sort dictionary value, creating a list of tuples of key, value pairs '
    values = sorted(d.items(), key = lambda kv: kv[1])
   
    # Group tuples by value
    # Determine rank by counting pairs until we reach desired rank
    count = 0
    for k, v in groupby(values, key = lambda kv: kv[1]):
        v = list(v)
        if count   1 == rank:
            return [x[0] for x in v]  # List of values at this rank
        elif count > rank:
            return    # Will not have any values that match rank
        
        count  = len(v)               # update count of tuples encountered

Test

Test solutions 1 & 2 by displaying ranks 1 through 8

print('# Ranks for Solution 1')
for r in range(1, 9):
    print(r, solution_one(data, r))
    
print()
print('# Ranks for Solution 2')
for r in range(1, 9):
    print(r, solution_two(data, r))

Output

# Ranks for Solution 1
1 ['Apple']
2 ['Banana']
3 ['Mango', 'Peach']
4 ['Watermelon']
5 ['Orange']
6 ['Blueberry']
7 None
8 None

# Ranks for Solution 2
1 ['Apple']
2 ['Banana']
3 ['Mango', 'Peach']
4 None
5 ['Watermelon']
6 ['Orange']
7 ['Blueberry']
8 None