# Python: Are Terms in this List Monotonic?

Gold Member
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 :

Python:
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]

Last edited by a moderator:

2021 Award
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
Gold Member
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.

2021 Award
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.

Vanadium 50, Wrichik Basu and WWGD
Gold Member
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.

2021 Award
But you must not multiply zeros, you must skip them, since they can allow the sign of the difference to flip.

WWGD
Mentor
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.

anorlunda
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.

pasmith
Gold Member
Python:
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)

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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.

Gold Member
• If a2=a1 then not monotonic.
No, we are not looking for strict monotonicity.

2021 Award
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.

Gold Member
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.

Gold Member
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.

Gold Member
Now even I am not strictly non-confused.
Ok, my bad, let's do just non decreasing/increasing for both.

Gold Member
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?

2021 Award
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
Gold Member
Deleted: going back to reread the OP!

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Gold Member
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!

Python:
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 dont 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.

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Gold Member
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 :

Python:
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.

Python:
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'

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WWGD
2021 Award
Here is a non-python fast algorithm that uses one loop, written in VB pseudocode.
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

Try this. I think its pretty clear and fast.

Python:
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
Gold Member
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

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

Gold Member
Python:
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?

Gold Member
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 dont know...

Staff Emeritus
Gold Member
2021 Award
.
I dont 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?

2021 Award
Sorting a list is O(n log n). Checking a list once is O(n).
Avoid sorting long lists, twice.
Python:
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 ) )

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Arman777
Gold Member
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)##.

2021 Award
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.

Gold Member
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.

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

Gold Member
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.

Only in Python 2: it does not exist in Python 3
Spot the person with the elderly python install. In that case, try
Python:
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.