Python Python: Are Terms in this List Monotonic?

[WWGD]

TL;DR

Given a list of Real numbers in Python , how do we tell if its terms are monotonic,i.e., if they are in strictly non-decreasing or strictly non-increasing order. I know how to do either one; either strictly increasing,strictly decreasing or both.

Given a list [ a,b,c,..., z] in Python, write a program to determine if the terms are monotonic or not. I can do either , and combine theminto a long, clunky program, but trying to see if I can find a succint way of doing both. I have heard that using ' Nums' may do it,but it seems too high-level for my taste.

For, say, non-decreasing we can do something like :

List=L=[ a1, a2,...,an]
for i in in range(length(L)):
if L[i+1]-L[i] >=0:
return(' Yes,list is monotonic non-decreasing')
else:
print('No,not monotonic' non-decreasing)

Similar for non-increasing.
And maybe I can do a kludge to join the code for both.
Someone suggested I look up 'Nums'.I did and it looks a bit high-level ( i.e., black-boxed').
Any idea to cover both cases in a single program?[/i][/code]

 

[Baluncore]

Compute a list of the n-1 differences.
Check if the sign of the differences ever changes.
Is a zero real positive or negative?

 

[WWGD]

Compute a list of the n-1 differences.
Check if the sign of the differences ever changes.
Integers or reals? Is zero positive or negative?

Thanks: Yes, what I do is I loop from 1 to length of the list and test wether (i+i)st -ith item is larger or equal to zero if list is L, then check if L[i+1]-L >=0. But this works for either monotone decreasing or monotone increasing and I am not sure how to integrate both into a single program/algorithm.

 

[Baluncore]

You only need detect the first change of sign.
Find the first non-zero difference, then compare that with all following differences. If you multiply two adjacent differences and the result is < zero then the list fails the test.

 

[WWGD]

You only need detect the first change of sign.
Find the first non-zero difference, then compare that with all following differences. If you multiply two adjacent differences and the result is < zero then the list fails the test.

Cool, so I can then iterate over all products LxL[i+1] and if I detect one negative, I return a no, other
ways I conclude it's monotone. Thanks, nice job.

 

[Baluncore]

But you must not multiply zeros, you must skip them, since they can allow the sign of the difference to flip.

 

[Mark44]

Is a zero real positive or negative?

It could be either. The IEEE 754 specification of floating point numbers has separate representations for -0 and +0. Most programming languages these days adhere to this specification as far as floating point reals are concerned.

 

[Ibix]

Loops in python tend to be quite slow. I'd suspect it would be faster (possibly a lot faster) to use list comprehensions and the cmp() and any() or all() functions to get the signs of the differences and a summary of those signs.

 

[pbuk]

How's it going? Does your answer look anything like mine?

from functools import reduce

# myList = [ ... ]

isIncreasing, isDecreasing, _ = reduce(
    lambda s, el: [s[0] and el >= s[2], s[1] and el <= s[2], el],
    myList,
    [True, True, myList[0]]
)

# Show whether increasing, decreasing, constant (T, T) or none (F, F).
print(isIncreasing, isDecreasing)

# Just show True iff loosely monotonic (including constant).
print(isIncreasing or isDecreasing)

 

[Svein]

Just check the first two:

• If a2<a1 check if the rest is descending.
• If a2=a1 then not monotonic.
• If a2>a1 check if the rest is ascending.

 

[pbuk]

• If a2=a1 then not monotonic.

No, we are not looking for strict monotonicity.

 

[Baluncore]

No, we are not looking for strict monotonicity.

Summary:: Given a list of Real numbers in Python , how do we tell if its terms are monotonic,i.e., if they are in strictly non-decreasing or strictly non-increasing order. I know how to do either one; either strictly increasing,strictly decreasing or both.

Now even I am not strictly non-confused.

 

[pbuk]

Now even I am not strictly non-confused.

To be fair Mathworld, as well as Wikipedia use similar confusing definitions. I was working on the basis that the use of the >= test rather than > in posts #1 and #3 made it clear that we were not looking for strictness.

 

[WWGD]

But you must not multiply zeros, you must skip them, since they can allow the sign of the difference to flip.

I thought of recursively applying the min to the list: If L[1] is the min, L[2] is the min of the remaining terms, etc.

 

[WWGD]

Now even I am not strictly non-confused.

Ok, my bad, let's do just non decreasing/increasing for both.

 

[pbuk]

Ok, my bad, let's do just non decreasing/increasing for both.

That's still confusing: 'strictly increasing' is a lot easier to parse than 'non decreasing', and 'strictly monotonic' is certainly easier than 'non-(decreasing/increasing)'.

Have you got anything working yet - I could see a few problems in your first post?

 

[Baluncore]

I was working on the basis that the use of the >= test rather than > in posts #1 and #3 made it clear that we were not looking for strictness.

We are working with real numbers, so I assumed that duplicates were as acceptable as a one LSB difference in any floating point representation. The critical thing being that it does not change direction.

 

[pbuk]

Deleted: going back to reread the OP!

 

[pbuk]

Given a list of Real numbers in Python

There is no such thing as a list of real numbers in Python: we can have a list of floats though, is this what you mean?

how do we tell if its terms are monotonic

By this I assume you don't mean strictly monotonic...

,i.e., if they are in strictly non-decreasing or strictly non-increasing order.

... and this confirms that you don't mean strictly monotonic (strictly non-decreasing is the same as non-strictly increasing etc.)

I know how to do either one; either strictly increasing,strictly decreasing or both.

But now you are testing for strict monotonicity which is not what you want!

if L[i+1]-L[i] >=0:

But this test does not give you strictly increasing because equal values pass!

Ok, my bad, let's do just non decreasing/increasing for both.

And now you are confirming you don't want strictness!

Probably best to leave this horrible confusion and start again if you are still interested. Note however that if this is another interview question the interviewer could be more interested in a discussion about the advantages and disadvantages of solving it with a loop rather than a reduce as I did, or a combination of list comprehensions as somebody else suggested rather than the actual coding of a working solution, which ought to be pretty easy.

 

[Arman777]

Summary:: Given a list of Real numbers in Python , how do we tell if its terms are monotonic,i.e., if they are in strictly non-decreasing or strictly non-increasing order. I know how to do either one; either strictly increasing,strictly decreasing or both.

Given a list [ a,b,c,..., z] in Python, write a program to determine if the terms are monotonic or not. I can do either , and combine theminto a long, clunky program, but trying to see if I can find a succint way of doing both. I have heard that using ' Nums' may do it,but it seems too high-level for my taste.

For, say, non-decreasing we can do something like :

List=L=[ a1, a2,...,an]
for i in in range(length(L)):
if L[i+1]-L[i] >=0:
return(' Yes,list is monotonic non-decreasing')
else:
print('No,not monotonic' non-decreasing)

Similar for non-increasing.
And maybe I can do a kludge to join the code for both.
Someone suggested I look up 'Nums'.I did and it looks a bit high-level ( i.e., black-boxed').
Any idea to cover both cases in a single program?[/i][/code]

Try this. I think its pretty clear and fast.

def is_monotonic(xlist):
    xlist_sorted_inc = sorted(xlist)
    xlist_sorted_dec = sorted(xlist)[::-1]
    if xlist == xlist_sorted_inc or  xlist == xlist_sorted_dec:
        return 'monotonic'
    return 'not monotonic'

 

[Baluncore]

Here is a non-python fast algorithm that uses one loop, written in VB pseudocode.
[CODE title="VB"]' test list of floats for monotonicity, duplicates are OK.

Const As Integer n = 5
Dim As Double a( 1 To n ) = { 1, 1, 2, 2, 1 } ' a list

Dim As Boolean mono_flag = True
Dim As Double diff, last_diff = 0
For i As Integer = 2 To n
diff = a( i - 1 ) - a( i )
If diff <> 0 Then ' not a duplicate
If ( diff * last_diff ) < 0 Then ' sign changed
mono_flag = False
Exit For ' may as well quit
End If
last_diff = diff
End If
Next i

' report result
If mono_flag Then
Print " monotonic pass "
Else
Print " fails test"
End If
[/CODE]

 

[Ibix]

Try this. I think its pretty clear and fast.

def is_monotonic(xlist):
    xlist_sorted_inc = sorted(xlist)
    xlist_sorted_dec = sorted(xlist)[::-1]
    if xlist == xlist_sorted_inc or  xlist == xlist_sorted_dec:
        return 'not monotonic'
    return 'monotonic'

Isn't the logic flipped here? If the list is equal to a sorted copy then it's monotonic.

 

[Arman777]

Isn't the logic flipped here? If the list is equal to a sorted copy then it's monotonic.

Yes, sorry. I kind of mixed the definition of 'monotonic'. I fixed it

 

[glappkaeft]

When you got it working for longer lists think about what should happen if the list has zero or one element.

 

[pbuk]

def is_monotonic(xlist):
    xlist_sorted_inc = sorted(xlist)
    xlist_sorted_dec = sorted(xlist)[::-1]
    if xlist == xlist_sorted_inc or  xlist == xlist_sorted_dec:
        return 'monotonic'
    return 'not monotonic'

Questions:

1. Do you know another way to do sorted(xlist)[::-1]?
2. If so, why did you choose that way?
3. You create the descending list immediately after creating the ascending list. Can you think of something else you could do first?
4. Is 'monotonic' the kind of return value you would normally expect from a function called is_monotonic()?
5. Can you think of a situation where sorting a list to see if it is monotonic might not be a good idea?

 

[Arman777]

Do you know another way to do sorted(xlist)[::-1]?

yes

If so, why did you choose that way?

theres no reason.

Is 'monotonic' the kind of return value you would normally expect from a function called is_monotonic()?

You can return True or False as well.

Can you think of a situation where sorting a list to see if it is monotonic might not be a good idea?

I don't know...

 

[Office_Shredder]

.
I don't know...

Efficiency is often overrated, but what is the asymptotic time it takes to run your algorithm, vs something that just goes through the list twice to check each form of monotonicity?

 

[Baluncore]

Sorting a list is O(n log n). Checking a list once is O(n).
Avoid sorting long lists, twice.

def is_monotonic( a: float ) -> bool:
    last_diff = 0.0
    for i in range( 1, len( a ) ):
        diff = a[ i-1 ] - a[ i ]
        if diff != 0.0:         # is not a duplicate
            if ( diff * last_diff ) < 0.0:
                return False    # diff changed sign
            last_diff = diff
    return True

a = [ 3, 2, 2, -1, -1 ]         # a list
print( a, is_monotonic( a ) )

 

[Arman777]

Efficiency is often overrated, but what is the asymptotic time it takes to run your algorithm, vs something that just goes through the list twice to check each form of monotonicity?

If I am not mistaken its in the order $O(n)$.

 

[Baluncore]

Efficiency is often overrated, but what is the asymptotic time it takes to run your algorithm, vs something that just goes through the list twice to check each form of monotonicity?

You only need to scan the list once if you test for a sign change of the first difference.
If you use sort you must do it twice, once for ascending and once for descending.
But with Python the aim is to write the minimum source code, in a way that can be quickly understood. Sorting twice will win under that measure.

A monotonicity test is not the type of algorithm I would expect to find inside a tight loop, or running 50,000 times each day. If you only run the test once each day, on a 100 element list, then efficiency is irrelevant.

 

[pbuk]

You only need to scan the list once if you test for a sign change of the first difference.
If you use sort you must do it twice, once for ascending and once for descending.

Yes, and even if the algorithm used is $O(n \log n)$ then in practice performance is much worse than $k n \log n$ because there is a step change in $k$ when the list no longer fits in processor cache, and again, dramitcally when the list no longer fits in main memory. There is also the memory usage to consider.

But with Python the aim is to write the minimum source code, in a way that can be quickly understood.

Even if "minimum source code" were a criterion (and I don't see that anywhere in the question) I don't think it would ever be the only criterion without any regard to performance.

Sorting twice will win under that measure.

But it will lose against an equally or even more concise implementation of a list traversal such as the one in #9.

A monotonicity test is not the type of algorithm I would expect to find inside a tight loop, or running 50,000 times each day. If you only run the test once each day, on a 100 element list, then efficiency is irrelevant.

No list length was specified in the op. Edit: I agree that premature optimization is to be avoided, but selecting an (at best) $O(n \log n)$ time $O(n)$ memory implementation when there is an at least equally trivial $O(n)$ time $O(1)$ memory one available is premature pessimization, and this is never good.

 

[Ibix]

Since we all seem to be posting solutions, here's what I had in mind:

MONOTONIC_INCREASING=1
MONOTONIC_DECREASING=-1
NOT_MONOTONIC=0

def isMonotonic(lst):
    ordered=[cmp(l1,l) for (l,l1) in zip(lst[:-1],lst[1:]) if cmp(l,l1)!=0]
    if len(ordered)==0 or all(o==1 for o in ordered):
        return MONOTONIC_INCREASING
    if all(o==-1 for o in ordered):
        return MONOTONIC_DECREASING
    return NOT_MONOTONIC

The cmp ("compare") function returns 1, 0, or -1 depending on whether the first argument is greater than, equal to, or less than the second, and it's used with zip to get a list of each element compared to the next, dropping equalities. Then it just checks that all differences were positive (or all elements were equal, arbitrarily) or all negative.

 

[pbuk]

The cmp ("compare") function returns 1, 0, or -1 depending on whether the first argument is greater than, equal to, or less than the second

Only in Python 2: it does not exist in Python 3, although it is easy enough to implement. Also I believe your algorithm checks strict monotonicity, the OP (or a later clarification) asked for a loose check.

 

[Ibix]

Only in Python 2: it does not exist in Python 3

Spot the person with the elderly python install. In that case, try

ordered=[(1 if l1>l else -1) for (l,l1) in zip(lst[:-1],lst[1:]) if l!=l1]

(I think I have the comparison the right way round).

Also I believe your algorithm checks strict monotonicity, the OP (or a later clarification) asked for a loose check.

I don't think so. It drops the cases where they are equal, so [1,2,2,3] should give [1,1,1], which I think matches the loose check criterion.

 

[pbuk]

It drops the cases where they are equal, so [1,2,2,3] should give [1,1,1], which I think matches the loose check criterion.

Ah yes, didn't see that (and I couldn't check because I am not running Py2 which is now 18 months past EOL!)

 

[Ibix]

I have python 3 installed, but typing python invokes python2. I should probably change the behaviour...

 

[pbuk]

@Ibix just realized your algorithm is $O(n)$ in memory against $O(1)$ for a list traversal (either via iteration per @Balancore or functools.reduce per my #7), but I won't hold that against you

 

[Ibix]

It actually seems to be quite a lot slower than Baluncore's approach as well, which surprises me since I understood python loops to be slow and list comprehensions to be better. Perhaps I need to do some reading.

 

[PeterDonis]

I understood python loops to be slow and list comprehensions to be better.

The difference is not so much loops vs. comprehensions but how much code written in Python (as opposed to just built-in functions or operations, which are really C code running inside the interpreter) is executed on each iteration.

However, @Baluncore's algorithm exits early if it spots a failure of monotonicity, whereas yours will traverse the entire list each time, even if a failure in monotonicity occurs early in the traversal. That's probably why yours is slower.

 

[Ibix]

However, @Baluncore's algorithm exits early if it spots a failure of monotonicity, whereas yours will traverse the entire list each time, even if a failure in monotonicity occurs early in the traversal. That's probably why yours is slower.

It actually seems to be random noise. I set up a version that runs the two algorithms a thousand times on range(1000) and range(1000)[::-1] and the average is pretty close for monotonic cases. Looks like my two-or-three-runs-and-eyeball-it approach wasn't entirely reliable from a statistical point of view. Baluncore's approach is, as you note, way faster for non-monotonic cases since it exits early.

 

[pbuk]

It actually seems to be quite a lot slower than Baluncore's approach as well

Bear in mind that you are doing FOUR list traversals (for...in, zip and two all's) instead of just one.

which surprises me since I understood python loops to be slow and list comprehensions to be better

This is a common belief, partly based on misunderstanding and partly based on early implementations of Python which were really slow.

The difference is not so much loops vs. comprehensions but how much code written in Python (as opposed to just built-in functions or operations, which are really C code running inside the interpreter) is executed on each iteration.

Exactly this. So for instance finding the dot product of two vectors represented as numpy.arrays is implemented entirely in C and runs much quicker than iterating over plain lists, but when as in this case you are executing the same code to compare elements of a list, it doesn't matter whether that list is traversed via a for...in comprehension, a for..next iteration or functools.reduce. And where it is possible that the iteration might terminate early, the for..next loop will always win because it is the only one you can break out of.

 

[Ibix]

This is a common belief [...] partly based on early implementations of Python which were really slow.

That'll be it. I learned python originally from an O'Reilly book well over fifteen years ago, so it's probably 20+ year old info now. How time flies. I guess I need to update more than my install.

 

[wle]

Succinct version, haven't tested performance:

from itertools import accumulate

def is_monotonic(lst):
    def all_eq(acc):
        return all(x == y for x, y in zip(lst, accumulate(lst, acc)))
    return all_eq(max) or all_eq(min)

accumulate is like reduce except it returns a generator of all intermediate values (and it can stop early).

It is possible to rearrange this so everything is done in "one loop" but it's hairier and doesn't really seem worth it in Python.

 

[Baluncore]

The Baluncore technique was to imagine the stream of sequential data arriving from secondary storage, then constructing the minimum systolic processor to filter that. My Python code indexes the list twice for each entry, which can be reduced by simply saving the previous data item. I did not test that because I don't have Python 3 running here, but that is OK because I don't have Python 2 either. Thanks for checking my code. I must admit that the more I see of Python, the more human, hilarious and Pythonesque I find it.