how to sort by length of string followed by alphabetical order?

Question

I'm currently new to python and got stuck at this question, can't seem to find the proper answer.

question:Given a list of words, return a list with the same words in order of length (longest to shortest), the second sort criteria should be alphabetical. Hint: you need think of two functions.

This is what I have so far:

def bylength(word1,word2):
    return len(word2)-len(word1)

def sortlist(a):
    a.sort(cmp=bylength)
    return a

it sorts by length but I don't know how to apply the second criteria to this sort, which is by alphabetical descending.

http://stackoverflow.com/questions/4655591/python-sort-list - looks like a homework of a large classroom... — eumiro, Jan 11 '11 at 15:56

Jochen Ritzel · Accepted Answer · 2011-01-11T15:58:32.323

176

You can do it in two steps like this:

the_list.sort() # sorts normally by alphabetical order
the_list.sort(key=len, reverse=True) # sorts by descending length

Python's sort is stable, which means that sorting the list by length leaves the elements in alphabetical order when the length is equal.

You can also do it like this:

the_list.sort(key=lambda item: (-len(item), item))

Generally you never need cmp, it was even removed in Python3. key is much easier to use.

edited Jan 11 '11 at 15:58

answered Jan 11 '11 at 15:53

Jochen Ritzel

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the lambda solution is awe-some! – dmeu Mar 18 '16 at 16:31
The second solution was a bit of art and I didn't know key works that way! – Masked Man Apr 30 '19 at 15:45
1

@jochen When I do this I get an `'str' object has no attribute 'sort'` error because I'm not using a list. It's a string from opening the csv. Any suggestions? – Edison May 13 '20 at 15:46
@Edison old question, but you could split at commas/line separators to create a list and then apply this solution – samm82 Dec 17 '20 at 23:54
just wondering why you are returning a tuple in the lambda function when you could only return `-len(item)` – Anthony Awuley Mar 06 '22 at 15:58

score 6 · Answer 2 · edited Mar 31 '17 at 00:53

6

n = ['aaa', 'bbb', 'ccc', 'dddd', 'dddl', 'yyyyy']

for i in reversed(sorted(n, key=len)):
    print i

yyyyy dddl dddd ccc bbb aaa

for i in sorted(n, key=len, reverse=True):
     print i

yyyyy dddd dddl aaa bbb ccc

edited Mar 31 '17 at 00:53

Ninjakannon

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answered Apr 15 '14 at 11:35

Arindam Roychowdhury

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This only works if the array is already sorted. Only `sorted( sorted( iterable ), key=len )` gives the correct answer always. – user Jun 16 '18 at 18:46
@user I just tried with unsorted array . Getting the same (correct) result . Can you please provide your input array ? – Arindam Roychowdhury Jun 18 '18 at 07:36
`sorted( ['bb', 'b', 'aa', 'a'], key=len, reverse=True)` produces `['bb', 'aa', 'b', 'a']`, but it should be `['aa', 'bb', 'a', 'b']` as get from `sorted( sorted( ['bb', 'b', 'aa', 'a'] ), key=len, reverse=True)` – user Jun 18 '18 at 13:48
This is simply sorting by string length in descending order. – frederick99 Sep 21 '21 at 23:09

score 3 · Answer 3 · answered Jun 15 '15 at 10:34

-Sort your list by alpha order, then by length.

See the following exmple:

>>> coursesList = ["chemistry","physics","mathematics","art"]
>>> sorted(coursesList,key=len)
['art', 'physics', 'chemistry', 'mathematics']
>>> coursesList.append("mopsosa")
>>> sorted(coursesList,key=len)
['art', 'physics', 'mopsosa', 'chemistry', 'mathematics']
>>> coursesList.sort()
>>> sorted(coursesList,key=len)
['art', 'mopsosa', 'physics', 'chemistry', 'mathematics']

score 1 · Answer 4 · answered Oct 22 '20 at 13:53

1

First sort by Alphabet and then sort by Length.

Here is a working example

mylist.sort()
mylist = sorted(mylist, key=len, reverse=False)

# Print the items on individual line
for i in mylist:
    print(i)

answered Oct 22 '20 at 13:53

Naveed Abbas

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score -1 · Answer 5 · answered May 23 '18 at 01:10

Although Jochen Ritzel said you don't need cmp, this is actually a great use case for it! Using cmp you can sort by length and then alphabetically at the same time in half the time sorting twice would take!

def cmp_func(a, b):
    # sort by length and then alphabetically in lowercase
    if len(a) == len(b):
        return cmp(a, b)
    return cmp(len(a), len(b))

sorted_the_way_you_want = sorted(the_list, cmp=cmp_func)

Example:

>>> the_list = ['B', 'BB', 'AA', 'A', 'Z', 'C', 'D']
>>> sorted(the_list, cmp=cmp_func)
['A', 'B', 'C', 'D', 'Z', 'AA', 'BB']

Note, if your list is a mix of upper and lower case replace cmp(a, b) with cmp(a.lower(), b.lower()) as python sorts 'a' > 'Z'.

In python3 you'd need to be sorting objects with __lt__ style comparison functions defined or functools.cmp_to_key() which does that for you.

not very interesting as `cmp` doesn't exist in python 3 anymore (you said it yourself) — Jean-François Fabre, Dec 04 '18 at 21:04

score -1 · Answer 6 · answered May 13 '22 at 03:10

-1

def cmp_func(a, b): # sort by length and then alphabetically in lowercase if len(a) == len(b): return cmp(a, b) return cmp(len(a), len(b))

sorted_the_way_you_want = sorted(the_list, cmp=cmp_func)

answered May 13 '22 at 03:10

Wendell Blanco

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You need to sourround your code with ``` or prepend 4 spaces to put it in code blocks. – ljmc May 13 '22 at 13:20
1

As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – ljmc May 13 '22 at 13:20

how to sort by length of string followed by alphabetical order?

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