How To Sort Tuple Element First On The Basis Of Key And Then On The Basis Of Value

How to sort a tuple of elements in python, first on the basis of value and then on the basis of key. Consider the program in which I am taking input from the user as a string. I wa

Solution 1:

Sort a list of tuples, the first value in descending order (reverse=True) and the second value in ascending order (reverse=False, by default). Here is a MWE.

lists = [(2, 'c'), (2, 'a'), (3, 'b')]

result = sorted(lists, key=lambda x: (-x[0], x[1])) # -x[0] represents descending orderprint(result)
# Output
[(3, 'b'), (2, 'a'), (2, 'c')]

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It is straightforward to use collections.Counter to count each letter's frequency in a string.

import collections

s = 'bcabcab'# If you don't care the order, just use `most_common`#most_common = collections.Counter(s).most_common(3)

char_and_frequency = collections.Counter(s)
result = sorted(char_and_frequency.items(), key=lambda x:(-x[1], x[0]))[:3]    # sorted by x[1] in descending order, x[0] in ascending orderprint(result)
# Output
[('b', 3), ('a', 2), ('c', 2)]

Solution 2:

As a general solution for "sort by value descending then by key ascending" you can use itertools.groupby:

import itertools
from operator import itemgetter

defsort_by_value_then_key(count_dict):
    first_level_sorted = sorted(count_dict.items(), key=itemgetter(1), reverse=True)
    for _, group in itertools.groupby(first_level_sorted,itemgetter(1)):
        for pair insorted(group):
            yield pair


strr=list("aabbccdef")
count=dict()

#store the count of each character in dictionaryfor i inrange(len(strr)):
    count[strr[i]]=count.get(strr[i],0)+1


temp = list(sort_by_value_then_key(count))

for (letter, freq) in temp[:3]:
    print letter, freq

Output:

a2b2
c 2

since you are using integers @sparkandshine's solution of sorting by the negative number will probably be easier to work with but this is much more generalized.