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 finding 2 max values in an array in python
#1
I have an array in Python and i want to find the two largest values in that array and their corresponding position in that array.
Can anyone please share the approach for it?
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#2
What have you tried? Show us your code in python tags.
micseydel and nilamo like this post
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#3
Also, what column in each row of the array do you want to compare (assuming it is 2 dimensional and not 3 or 4, etc.).
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#4
a = [1,2,3,4,5]

high1 = 0
high2 = 0

for item in a:
	if item > high1:
		high1 = item
		pos = a.index(item)
		
print str(high1)
print str(pos)

for item in a:
	if item > high2 and item < high1:
		high2 = item
		pos = a.index(item)
		
print str(high2)
print str(pos)
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#5
Assuming your array has only unique values:

max_values = sorted(your_array, reverse=True)[:2]
indices = [your_array.index(value) for value in max_values]
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#6
>>> def n_max(seq, n=2):
...   max_items = sorted(enumerate(seq), key=lambda x: x[1])[-1 * n:]
...   return max_items[::-1]
...
>>> items = [4, 5, 3, 2, 63, 0, 1, 5]
>>> n_max(items, 2)
[(4, 63), (7, 5)]
Here you go, pal. Hopefully this isn't homework, because an instructor wouldn't accept this lol.
So what we do here, is use enumerate() to create a list of index-value pairs, which we then sort by value, using the key argument of sorted(). Once the list is sorted by value, we simply grab the n items off the end of the list, since the largest values are at the end, and reverse it before returning it so the largest item is the first in the result set.
buran likes this post
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#7
(Oct-17-2018, 08:03 PM)nilamo Wrote:
>>> def n_max(seq, n=2):
...   max_items = sorted(enumerate(seq), key=lambda x: x[1])[-1 * n:]
...   return max_items[::-1]
...
>>> items = [4, 5, 3, 2, 63, 0, 1, 5]
>>> n_max(items, 2)
[(4, 63), (7, 5)]
....
from operator import itemgetter
def n_max(seq, n=2):
    return sorted(enumerate(seq), key=itemgetter(1), reverse=True)[:n]
Test everything in a Python shell (iPython, Azure Notebook, etc.)
  • Someone gave you an advice you liked? Test it - maybe the advice was actually bad.
  • Someone gave you an advice you think is bad? Test it before arguing - maybe it was good.
  • You posted a claim that something you did not test works? Be prepared to eat your hat.
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#8
I've tried this:

from operator import itemgetter

def n_max(seq, n=2):
    max_items = sorted(enumerate(seq), key=lambda x: x[1])[-1 * n:]
    return max_items[::-1]

def my_max(seq, n=2):
    max_values = sorted(seq, reverse=True)[:n]
    indices = [seq.index(value) for value in max_values]
    return list(zip(indices, max_values))

def n_max_2(seq, n=2):
    return sorted(enumerate(seq), key=itemgetter(1), reverse=True)[:n]


if __name__ == '__main__':
    import timeit
    print(timeit.timeit('n_max(seq)', setup='seq=[x for x in range(10000)]; from __main__ import n_max', number=1000))
    print(timeit.timeit('my_max(seq)', setup='seq=[x for x in range(10000)]; from __main__ import my_max', number=1000))
    print(timeit.timeit('n_max_2(seq)', setup='seq=[x for x in range(10000)]; from __main__ import n_max_2, itemgetter', number=1000))

and the results are:

2.84397915685567
0.6598219036493851
2.251084051447339
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#9
(Oct-17-2018, 08:56 PM)LeSchakal Wrote: and the results are:

2.84397915685567
0.6598219036493851
2.251084051447339

Ok, but because you use .index(), it's slower the more results you want (I believe O(n) is what it'd be called). For example, if you change it to be 1) an unsorted starting list (sorted starting list is cheating kind of), and b) more than just the max(2), you'll see a huge difference. Here's an example (I moved to a global seq so each has a fair starting point):
if __name__ == '__main__':
    import random, timeit
    seq = [x for x in range(10000)]
    random.shuffle(seq)

    print(timeit.timeit('n_max(seq, 100)', setup='from __main__ import seq, n_max', number=1000))
    print(timeit.timeit('my_max(seq, 100)', setup='from __main__ import seq, my_max', number=1000))
    print(timeit.timeit('n_max_2(seq, 100)', setup='from __main__ import seq, n_max_2, itemgetter', number=1000))
Output:
3.9099400770000003 10.747478319 3.6070847269999984
buran likes this post
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#10
(Oct-17-2018, 09:17 PM)nilamo Wrote: Ok, but because you use .index(), it's slower the more results you want (I believe O(n) is what it'd be called). For example, if you change it to be 1) an unsorted starting list (sorted starting list is cheating kind of), and b) more than just the max(2), you'll see a huge difference.

Thanks for the explanation.

I was afraid of this a little bit, but never thought it would happen so early.
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