排序功能的自定义键不起作用

Question

Given an array A[] of integers, A has to be sorted according to frequency of elements. If frequencies of two elements are same, then smaller number comes first.

I've tried to solve the problem using the following code using the sort function but my custom key for the sort() does not seem to work. Can someone tell me what's going on here?

'''a is the input array and n is it's sizeQ'''
def sortByFreq(a,n):
    def count(k):
        return n-a.count(k)
    a.sort(key=int)
    a.sort(key=count)
    a=list(map(str,a))
    answer=' '.join(a)
    print(answer)

For an input array [9,9,9,2,5], the code is supposed to print 9 9 9 2 5, (as the array already is), but it prints 2 5 9 9 9 instead. (The second time I call sort it does not seem to work)

Answer 1

The problem is that you just cannot use the original list inside a sort key, because sort uses an out of place location to compute the sort.

At some point, the original a is empty. Don't expect something like "all elements of a are in it" when calling the sort key.

In fact, if you print a in the count method, you'll get an empty list, everytime.

The workaround is to make a copy of the list and use it in the sort key (aside: no need to pass the size of the list, since len can compute that)

def sortByFreq(a):
    def count(k):
        return len(a_copy)-a_copy.count(k)

    a_copy = a.copy()  # create a copy of the list
    a.sort(key=count)
    a=list(map(str,a))
    answer=' '.join(a)
    print(answer)

an alternative is to pre-count elements with a counter (uses hashes, so faster). Note that you have to store the length of a too.

def sortByFreq(a):
    def count(k):
        return n-c[k]
    c = collections.Counter(a)
    n = len(a)
    a.sort(key=count)

finally, a shorter way would use reverse=True to simplify sort key, which we can turn to a lambda :

def sortByFreq(a):
    c = collections.Counter(a)
    a.sort(key=lambda k:c[k],reverse=True) 

Answer 2

def sort_by_frequency(sequence):
    """Sort sequence by frequency."""
    return ''.join(sorted([char for char in sequence], key=sequence.count, reverse=True))


if __name__ == '__main__':
    print(sort_by_frequency('92959'))