Is there a pythonic way to check if a list is already sorted in ASC or DESC
listtimestamps = [1, 2, 3, 5, 6, 7]
something like isttimestamps.isSorted() that returns True or False.
I want to input a list of timestamps for some messages and check if the the transactions appeared in the correct order.
Definitely works in Python 3 and above for integers or strings:
def tail(t):
return t[:]
letters = ['a', 'b', 'c', 'd', 'e']
rest = tail(letters)
rest.sort()
if letters == rest:
print ('Given list is SORTED.')
else:
print ('List NOT Sorted.')
=====================================================================
Another way of finding if the given list is sorted or not
trees1 = list ([1, 4, 5, 3, 2])
trees2 = list (trees1)
trees2.sort()
if trees1 == trees2:
print ('trees1 is SORTED')
else:
print ('Not sorted')
Python 3.6.8
from more_itertools import pairwise
class AssertionHelper:
@classmethod
def is_ascending(cls, data: iter) -> bool:
for a, b in pairwise(data):
if a > b:
return False
return True
@classmethod
def is_descending(cls, data: iter) -> bool:
for a, b in pairwise(data):
if a < b:
return False
return True
@classmethod
def is_sorted(cls, data: iter) -> bool:
return cls.is_ascending(data) or cls.is_descending(data)
>>> AssertionHelper.is_descending((1, 2, 3, 4))
False
>>> AssertionHelper.is_ascending((1, 2, 3, 4))
True
>>> AssertionHelper.is_sorted((1, 2, 3, 4))
True
I'd do this (stealing from a lot of answers here [Aaron Sterling, Wai Yip Tung, sorta from Paul McGuire] and mostly Armin Ronacher):
from itertools import tee, izip
def pairwise(iterable):
a, b = tee(iterable)
next(b, None)
return izip(a, b)
def is_sorted(iterable, key=lambda a, b: a <= b):
return all(key(a, b) for a, b in pairwise(iterable))
One nice thing: you don't have to realize the second iterable for the series (unlike with a list slice).
As I don't see this option above I will add it to all the answers.
Let denote the list by l, then:
import numpy as np
# Trasform the list to a numpy array
x = np.array(l)
# check if ascendent sorted:
all(x[:-1] <= x[1:])
# check if descendent sorted:
all(x[:-1] >= x[1:])
How about this one ? Simple and straightforward.
def is_list_sorted(al):
llength =len(al)
for i in range (llength):
if (al[i-1] > al[i]):
print(al[i])
print(al[i+1])
print('Not sorted')
return -1
else :
print('sorted')
return true
I use this one-liner based on numpy.diff():
def issorted(x):
"""Check if x is sorted"""
return (numpy.diff(x) >= 0).all() # is diff between all consecutive entries >= 0?
I haven't really timed it against any other method, but I assume it's faster than any pure Python method, especially for large n, since the loop in numpy.diff (probably) runs directly in C (n-1 subtractions followed by n-1 comparisons).
However, you need to be careful if x is an unsigned int, which might cause silent integer underflow in numpy.diff(), resulting in a false positive. Here's a modified version:
def issorted(x):
"""Check if x is sorted"""
try:
if x.dtype.kind == 'u':
# x is unsigned int array, risk of int underflow in np.diff
x = numpy.int64(x)
except AttributeError:
pass # no dtype, not an array
return (numpy.diff(x) >= 0).all()
Although I don't think there is a guarantee for that the sorted built-in calls its cmp function with i+1, i, it does seem to do so for CPython.
So you could do something like:
def my_cmp(x, y):
cmpval = cmp(x, y)
if cmpval < 0:
raise ValueError
return cmpval
def is_sorted(lst):
try:
sorted(lst, cmp=my_cmp)
return True
except ValueError:
return False
print is_sorted([1,2,3,5,6,7])
print is_sorted([1,2,5,3,6,7])
Or this way (without if statements -> EAFP gone wrong? ;-) ):
def my_cmp(x, y):
assert(x >= y)
return -1
def is_sorted(lst):
try:
sorted(lst, cmp=my_cmp)
return True
except AssertionError:
return False
This is similar to the top answer, but I like it better because it avoids explicit indexing. Assuming your list has the name lst, you can generate
(item, next_item) tuples from your list with zip:
all(x <= y for x,y in zip(lst, lst[1:]))
In Python 3, zip already returns a generator, in Python 2 you can use itertools.izip for better memory efficiency.
Small demo:
>>> lst = [1, 2, 3, 4]
>>> zip(lst, lst[1:])
[(1, 2), (2, 3), (3, 4)]
>>> all(x <= y for x,y in zip(lst, lst[1:]))
True
>>>
>>> lst = [1, 2, 3, 2]
>>> zip(lst, lst[1:])
[(1, 2), (2, 3), (3, 2)]
>>> all(x <= y for x,y in zip(lst, lst[1:]))
False
The last one fails when the tuple (3, 2) is evaluated.
Bonus: checking finite (!) generators which cannot be indexed:
>>> def gen1():
... yield 1
... yield 2
... yield 3
... yield 4
...
>>> def gen2():
... yield 1
... yield 2
... yield 4
... yield 3
...
>>> g1_1 = gen1()
>>> g1_2 = gen1()
>>> next(g1_2)
1
>>> all(x <= y for x,y in zip(g1_1, g1_2))
True
>>>
>>> g2_1 = gen2()
>>> g2_2 = gen2()
>>> next(g2_2)
1
>>> all(x <= y for x,y in zip(g2_1, g2_2))
False
Make sure to use itertools.izip here if you are using Python 2, otherwise you would defeat the purpose of not having to create lists from the generators.
A beautiful way to implement this is to use the imap function from itertools:
from itertools import imap, tee
import operator
def is_sorted(iterable, compare=operator.le):
a, b = tee(iterable)
next(b, None)
return all(imap(compare, a, b))
This implementation is fast and works on any iterables.
This iterator form is 10-15% faster than using integer indexing:
# python2 only
if str is bytes:
from itertools import izip as zip
def is_sorted(l):
return all(a <= b for a, b in zip(l, l[1:]))
from functools import reduce
# myiterable can be of any iterable type (including list)
isSorted = reduce(lambda r, e: (r[0] and (r[1] or r[2] <= e), False, e), myiterable, (True, True, None))[0]
The derived reduction value is a 3-part tuple of (sortedSoFarFlag, firstTimeFlag, lastElementValue). It initially starts with (True, True, None), which is also used as the result for an empty list (regarded as sorted because there are no out-of-order elements). As it processes each element it calculates new values for the tuple (using previous tuple values with the next elementValue):
[0] (sortedSoFarFlag) evaluates true if: prev_0 is true and (prev_1 is true or prev_2 <= elementValue)
[1] (firstTimeFlag): False
[2] (lastElementValue): elementValue
The final result of the reduction is a tuple of:
[0]: True/False depending on whether the entire list was in sorted order
[1]: True/False depending on whether the list was empty
[2]: the last element value
The first value is the one we're interested in, so we use [0] to grab that from the reduce result.
If you want the fastest way for numpy arrays, use numba, which if you use conda should be already installed
The code will be fast because it will be compiled by numba
import numba
@numba.jit
def issorted(vec, ascending=True):
if len(vec) < 2:
return True
if ascending:
for i in range(1, len(vec)):
if vec[i-1] > vec[i]:
return False
return True
else:
for i in range(1, len(vec)):
if vec[i-1] < vec[i]:
return False
return True
and then:
>>> issorted(array([4,9,100]))
>>> True
Simplest way:
def isSorted(arr):
i = 1
while i < len(arr):
if(result[i] < result[i - 1]):
return False
i += 1
return True
if it's Sorted will print True else will print out False
def is_Sorted(lst):
if len(lst) == 1:
return True
return lst[0] <= lst[1] and is_Sorted(lst[1:])
any_list = [1,2,3,4]
print is_Sorted(any_list)
A solution using assignment expressions (added in Python 3.8):
def is_sorted(seq):
seq_iter = iter(seq)
cur = next(seq_iter, None)
return all((prev := cur) <= (cur := nxt) for nxt in seq_iter)
z = list(range(10))
print(z)
print(is_sorted(z))
import random
random.shuffle(z)
print(z)
print(is_sorted(z))
z = []
print(z)
print(is_sorted(z))
Gives:
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
True
[1, 7, 5, 9, 4, 0, 8, 3, 2, 6]
False
[]
True
As noted by @aaronsterling the following solution is the shortest and seems fastest when the array is sorted and not too small: def is_sorted(lst): return (sorted(lst) == lst)
If most of the time the array is not sorted, it would be desirable to use a solution that does not scan the entire array and returns False as soon as an unsorted prefix is discovered. Following is the fastest solution I could find, it is not particularly elegant:
def is_sorted(lst):
it = iter(lst)
try:
prev = it.next()
except StopIteration:
return True
for x in it:
if prev > x:
return False
prev = x
return True
Using Nathan Farrington's benchmark, this achieves better runtime than using sorted(lst) in all cases except when running on the large sorted list.
Here are the benchmark results on my computer.
sorted(lst)==lst solution
Second solution:
Not very Pythonic at all, but we need at least one reduce() answer, right?
def is_sorted(iterable):
prev_or_inf = lambda prev, i: i if prev <= i else float('inf')
return reduce(prev_or_inf, iterable, float('-inf')) < float('inf')
The accumulator variable simply stores that last-checked value, and if any value is smaller than the previous value, the accumulator is set to infinity (and thus will still be infinity at the end, since the 'previous value' will always be bigger than the current one).
more_itertools.is_sortedI'm not sure when this was added, but this hasn't been mentioned yet:
import more_itertools
ls = [1, 4, 2]
print(more_itertools.is_sorted(ls))
ls2 = ["ab", "c", "def"]
print(more_itertools.is_sorted(ls2, key=len))
SapphireSun is quite right. You can just use lst.sort(). Python's sort implementation (TimSort) check if the list is already sorted. If so sort() will completed in linear time. Sounds like a Pythonic way to ensure a list is sorted ;)
Just to add another way (even if it requires an additional module): iteration_utilities.all_monotone:
>>> from iteration_utilities import all_monotone
>>> listtimestamps = [1, 2, 3, 5, 6, 7]
>>> all_monotone(listtimestamps)
True
>>> all_monotone([1,2,1])
False
To check for DESC order:
>>> all_monotone(listtimestamps, decreasing=True)
False
>>> all_monotone([3,2,1], decreasing=True)
True
There is also a strict parameter if you need to check for strictly (if successive elements should not be equal) monotonic sequences.
It's not a problem in your case but if your sequences contains nan values then some methods will fail, for example with sorted:
def is_sorted_using_sorted(iterable):
return sorted(iterable) == iterable
>>> is_sorted_using_sorted([3, float('nan'), 1]) # definetly False, right?
True
>>> all_monotone([3, float('nan'), 1])
False
Note that iteration_utilities.all_monotone performs faster compared to the other solutions mentioned here especially for unsorted inputs (see benchmark).
With the upcoming Python 3.10 release schedule, the new pairwise function provides a way to slide through pairs of consecutive elements, and thus find if all of these pairs satisfy the same predicate of ordering:
from itertools import pairwise
all(x <= y for x, y in pairwise([1, 2, 3, 5, 6, 7]))
# True
The intermediate result of pairwise:
pairwise([1, 2, 3, 5, 6, 7])
# [(1, 2), (2, 3), (3, 5), (5, 6), (6, 7)]
from itertools import tee
def is_sorted(l):
l1, l2 = tee(l)
next(l2, None)
return all(a <= b for a, b in zip(l1, l2))
I ran a benchmark and . These benchmarks were run on a MacBook Pro 2010 13" (Core2 Duo 2.66GHz, 4GB 1067MHz DDR3 RAM, Mac OS X 10.6.5).sorted(lst, reverse=True) == lst was the fastest for long lists, and all(l[i] >= l[i+1] for i in xrange(len(l)-1)) was the fastest for short lists
UPDATE: I revised the script so that you can run it directly on your own system. The previous version had bugs. Also, I have added both sorted and unsorted inputs.
all(l[i] >= l[i+1] for i in xrange(len(l)-1))sorted(l, reverse=True) == lall(l[i] >= l[i+1] for i in xrange(len(l)-1))all(l[i] >= l[i+1] for i in xrange(len(l)-1))So in most cases there is a clear winner.
UPDATE: aaronsterling's answers (#6 and #7) are actually the fastest in all cases. #7 is the fastest because it doesn't have a layer of indirection to lookup the key.
#!/usr/bin/env python
import itertools
import time
def benchmark(f, *args):
t1 = time.time()
for i in xrange(1000000):
f(*args)
t2 = time.time()
return t2-t1
L1 = range(4, 0, -1)
L2 = range(100, 0, -1)
L3 = range(0, 4)
L4 = range(0, 100)
# 1.
def isNonIncreasing(l, key=lambda x,y: x >= y):
return all(key(l[i],l[i+1]) for i in xrange(len(l)-1))
print benchmark(isNonIncreasing, L1) # 2.47253704071
print benchmark(isNonIncreasing, L2) # 34.5398209095
print benchmark(isNonIncreasing, L3) # 2.1916718483
print benchmark(isNonIncreasing, L4) # 2.19576501846
# 2.
def isNonIncreasing(l):
return all(l[i] >= l[i+1] for i in xrange(len(l)-1))
print benchmark(isNonIncreasing, L1) # 1.86919999123
print benchmark(isNonIncreasing, L2) # 21.8603689671
print benchmark(isNonIncreasing, L3) # 1.95684289932
print benchmark(isNonIncreasing, L4) # 1.95272517204
# 3.
def isNonIncreasing(l, key=lambda x,y: x >= y):
return all(key(a,b) for (a,b) in itertools.izip(l[:-1],l[1:]))
print benchmark(isNonIncreasing, L1) # 2.65468883514
print benchmark(isNonIncreasing, L2) # 29.7504849434
print benchmark(isNonIncreasing, L3) # 2.78062295914
print benchmark(isNonIncreasing, L4) # 3.73436689377
# 4.
def isNonIncreasing(l):
return all(a >= b for (a,b) in itertools.izip(l[:-1],l[1:]))
print benchmark(isNonIncreasing, L1) # 2.06947803497
print benchmark(isNonIncreasing, L2) # 15.6351969242
print benchmark(isNonIncreasing, L3) # 2.45671010017
print benchmark(isNonIncreasing, L4) # 3.48461818695
# 5.
def isNonIncreasing(l):
return sorted(l, reverse=True) == l
print benchmark(isNonIncreasing, L1) # 2.01579380035
print benchmark(isNonIncreasing, L2) # 5.44593787193
print benchmark(isNonIncreasing, L3) # 2.01813793182
print benchmark(isNonIncreasing, L4) # 4.97615599632
# 6.
def isNonIncreasing(l, key=lambda x, y: x >= y):
for i, el in enumerate(l[1:]):
if key(el, l[i-1]):
return False
return True
print benchmark(isNonIncreasing, L1) # 1.06842684746
print benchmark(isNonIncreasing, L2) # 1.67291283607
print benchmark(isNonIncreasing, L3) # 1.39491200447
print benchmark(isNonIncreasing, L4) # 1.80557894707
# 7.
def isNonIncreasing(l):
for i, el in enumerate(l[1:]):
if el >= l[i-1]:
return False
return True
print benchmark(isNonIncreasing, L1) # 0.883186101913
print benchmark(isNonIncreasing, L2) # 1.42852401733
print benchmark(isNonIncreasing, L3) # 1.09229516983
print benchmark(isNonIncreasing, L4) # 1.59502696991
I would just use
if sorted(lst) == lst:
# code here
unless it's a very big list in which case you might want to create a custom function.
if you are just going to sort it if it's not sorted, then forget the check and sort it.
lst.sort()
and don't think about it too much.
if you want a custom function, you can do something like
def is_sorted(lst, key=lambda x: x):
for i, el in enumerate(lst[1:]):
if key(el) < key(lst[i]): # i is the index of the previous element
return False
return True
This will be O(n) if the list is already sorted though (and O(n) in a for loop at that!) so, unless you expect it to be not sorted (and fairly random) most of the time, I would, again, just sort the list.
Source: Stackoverflow.com