multiple maximums in a list python

Question

Hey so being the NEWBY I am I was wondering how I could find multiple maximums (i.e. there is more than one maximum or item of the same length) and minimums(same situ as maximums) from a list. I've tried using the max function but yet it prints only one item, same with min. It is to be done for length of the string in the list (e.g. usinglen)!

this is the code i had so far

    def choice4(filelist):
       try:
           c4longest=max(filelist,key=len)
           print(c4longest,"is the longest item in the list")

The builtin will only give you the first maximum. If you want to return all of the maxima, you'll need to roll your own function for that (It should be pretty easy) — mgilson
– mgilson, Commented Jan 24, 2013 at 18:53

Steve Peak · Accepted Answer · 2013-01-24 19:14:19Z

5

Try this:

def choice4(filelist):
    mymax = max(map(len,filelist))
    return [a for a in filelist if len(a)==mymax]

a = ['joe','andy','mark','steve']
a.extend(a)
print choice4(a)

edited Jan 24, 2013 at 19:14

answered Jan 24, 2013 at 18:54

Steve Peak

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1 Comment

Steve Peak Over a year ago

Let me know if you would like a revision on my posting to better serve your question. Thanks!

Artjom B. · Accepted Answer · 2014-08-11 20:14:39Z

2

You could use sorting instead:

maxed = sorted(inputlist, key=lambda i: len(i), reverse=True)
allmax = list(takewhile(lambda e: len(e) == len(maxed[0]), maxed))

which takes O(n log n) time for the sort; but it's easy and short as the longest elements are all at the start for easy picking.

For a O(n) solution use a loop:

maxlist = []
maxlen = 0
for el in inputlist:
    l = len(el)
    if l > maxlen:
       maxlist = [el]
       maxlen = l
    elif l == maxlen:
       maxlist.append(el)

where maxlist is built and replaced as needed to hold only the longest elements:

>>> inputlist = 'And so we give a demo once more'.split()
>>> maxlist = []
>>> maxlen = 0
>>> for el in inputlist:
...     l = len(el)
...     if l > maxlen:
...        maxlist = [el]
...        maxlen = l
...     elif l == maxlen:
...        maxlist.append(el)
... 
>>> maxlist
['give', 'demo', 'once', 'more']

edited Aug 11, 2014 at 20:14

Artjom B.

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answered Jan 24, 2013 at 18:53

Martijn Pieters

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1 Comment

mgilson Over a year ago

This makes me cringe a bit from an algorithmic standpoint -- Why use an O(nlogn) algorithm when you only need to have O(n)? -- That said, for moderate sized lists, this is quick, easy and effective.

Lev Levitsky · Accepted Answer · 2013-01-24 18:53:18Z

0

In [1]: l = 'Is that what you mean'.split()

In [2]: [word for word in l if len(word) == max(map(len, l))]
Out[2]: ['that', 'what', 'mean']

answered Jan 24, 2013 at 18:53

Lev Levitsky

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2 Comments

mgilson Over a year ago

But the O(n**2) could easily be cut down to O(n) by computing the max only once ... (after all, the result will be the same every time)

Lev Levitsky Over a year ago

@MartijnPieters Well, yeah, but yours is O(n log n), and it really can be done in one loop. But meh.

ifmihai · Accepted Answer · 2015-01-15 20:25:53Z

0

use collections.Counter

from collections import Counter
d = Counter({k:v for k,v in enumerate(L)})
print d.most_common(5)  # to get top 5

answered Jan 15, 2015 at 20:25

ifmihai

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