通过 Counter Python 对元组列表进行排序

Question

I have read and tried to implement suggestions from around Stack Overflow.

In Python 3.6+ I have a list of tuples that looks something like this:

tuple_list=[(a=3,b=gt,c=434),(a=4,b=lodf,c=We),(a=3,b=gt,c=434)]

created by

for row in result:    
    tuple_list.append(var_tuple(row['d'], row['f'], row['q']))

I want to count the number of duplicates in the list and then sort the list so the number with the highest duplicates is at the top so I used

tuple_counter = collections.Counter(tuple(sorted(tup)) for tup in tuple_list)

But this returns in error because

TypeError: unorderable types: int() < str()

I've also tried this but it doesn't seem to sort by the highest counter.

tuple_counter = collections.Counter(tuple_list)
tuple_counter = sorted(tuple_counter, key=lambda x: x[1])

As well as this

tuple_counter = collections.Counter(tuple_list)
tuple_counter = tuple_counter.most_common()

Is there a better way to do this?

Answer 1

tuple contains different type s

tuple_counter = collections.Counter(tuple(sorted(tup)) for tup in tuple_list)

This line errors saying that int < str cannot be ordered. before any of this is evaluated, the generator expression must be, and sorted(tup) immediately breaks. Why? From the error, I am confident that tup contains both integers and strings. You can't sort integers and strings in the same list because you can't compare an integer and a string with < . If you have a method of comparing int s and str s, try sorted(tup, key = function) with your function to order int s and str s.

Since you want to count by the number of occurrences, try this:

sorted_tuples = sorted(tuple_list, key = tuple_list.count)

This sorts the tuples using the counter function of tuple_list as a key. If you want to sort descending, do sorted(tuple_list, key = tuple_list.count, reversed = True) .