Python：找出整数列表是否一致

Question

I am trying to find out whether a list of integers is coherent or 'at one stretch', meaning that the difference between two neighboring elements must be exactly one and that the numbers must be increasing monotonically. I found a neat approach where we can group by the number in the list minus the position of the element in the list -- this difference changes when the numbers are not coherent. Obviously, there should be exactly one group when the sequence does not contain gaps or repetitions.

Test:

>>> l1 = [1, 2, 3, 4, 5, 6]
>>> l2 = [1, 2, 3, 4, 5, 7]
>>> l3 = [1, 2, 3, 4, 5, 5]
>>> l4 = [1, 2, 3, 4, 5, 4]
>>> l5 = [6, 5, 4, 3, 2, 1]
>>> def is_coherent(seq):
...     return len(list(g for _, g in itertools.groupby(enumerate(seq), lambda (i,e): i-e))) == 1
... 
>>> is_coherent(l1)
True
>>> is_coherent(l2)
False
>>> is_coherent(l3)
False
>>> is_coherent(l4)
False
>>> is_coherent(l5)
False

It works well, but I personally find that this solution is a bit too convoluted in view of the simplicity of the problem. Can you come up with a clearer way to achieve the same without significantly increasing the code length?

Edit: summary of answers

From the answers given below, the solution

def is_coherent(seq):
    return seq == range(seq[0], seq[-1]+1)

clearly wins. For small lists (10^3 elements), it is on the order of 10 times faster than the groupby approach and (on my machine) still four times faster than the next best approach (using izip_longest ). It has the worst scaling behavior, but even for a large list with 10^8 elements it is still two times faster than the next best approach, which again is the izip_longest -based solution.

Relevant timing information obtained with timeit :

Testing is_coherent_groupby...
   small/large/larger/verylarge duration: 8.27 s, 20.23 s, 20.22 s, 20.76 s
   largest/smallest = 2.51
Testing is_coherent_npdiff...
   small/large/larger/verylarge duration: 7.05 s, 15.81 s, 16.16 s, 15.94 s
   largest/smallest = 2.26
Testing is_coherent_zip...
   small/large/larger/verylarge duration: 5.74 s, 20.54 s, 21.69 s, 24.62 s
   largest/smallest = 4.28
Testing is_coherent_izip_longest...
   small/large/larger/verylarge duration: 4.20 s, 10.81 s, 10.76 s, 10.81 s
   largest/smallest = 2.58
Testing is_coherent_all_xrange...
   small/large/larger/verylarge duration: 6.52 s, 17.06 s, 17.44 s, 17.30 s
   largest/smallest = 2.65
Testing is_coherent_range...
   small/large/larger/verylarge duration: 0.96 s, 4.14 s, 4.48 s, 4.48 s
   largest/smallest = 4.66

Testing code:

import itertools
import numpy as np
import timeit


setup = """
import numpy as np
def is_coherent_groupby(seq):
    return len(list(g for _, g in itertools.groupby(enumerate(seq), lambda (i,e): i-e))) == 1

def is_coherent_npdiff(x):
    return all(np.diff(x) == 1)

def is_coherent_zip(seq):
    return all(x==y+1 for x, y in zip(seq[1:], seq))

def is_coherent_izip_longest(l):
    return all(a==b for a, b in itertools.izip_longest(l, xrange(l[0], l[-1]+1)))

def is_coherent_all_xrange(l):
    return all(l[i] + 1 == l[i+1] for i in xrange(len(l)-1))

def is_coherent_range(seq):
    return seq == range(seq[0], seq[-1]+1)


small_list = range(10**3)
large_list = range(10**6)
larger_list = range(10**7)
very_large_list = range(10**8)
"""


fs = [
    'is_coherent_groupby',
    'is_coherent_npdiff',
    'is_coherent_zip',
    'is_coherent_izip_longest',
    'is_coherent_all_xrange',
    'is_coherent_range'
    ]


for n in fs:
    print "Testing %s..." % n
    t1 = timeit.timeit(
        '%s(small_list)' % n, 
        setup,
        number=40000
        )      
    t2 = timeit.timeit(
        '%s(large_list)' % n, 
        setup,
        number=100
        )     
    t3 = timeit.timeit(
        '%s(larger_list)' % n, 
        setup,
        number=10
        )
    t4 =  timeit.timeit(
        '%s(very_large_list)' % n, 
        setup,
        number=1
        )
    print "   small/large/larger/verylarge duration: %.2f s, %.2f s, %.2f s, %.2f s" % (t1, t2, t3, t4)
    print "   largest/smallest = %.2f" % (t4/t1)

Test machine:

• Linux 3.2.0 (Ubuntu 12.04)
• Python 2.7.3 (gcc 4.1.2)
• numpy 1.6.2 built with Intel compiler
• CPU: E5-2650 @ 2.00GHz
• 24 GB of memory

Answer 1

how bout

sorted_list = sorted(my_list)
return sorted_list == range(sorted_list[0],sorted_list[-1]+1)

or if its only coherent if it is already sorted

return my_list == range(my_list[0],my_list[-1]+1)

if you are using python 3 you will need list(range(...))

Answer 2

Unless I'm overlooking something in your examples, this simpler solution is actually shorter.

>>> l1 = [1, 2, 3, 4, 5, 6]
>>> l2 = [1, 2, 3, 4, 5, 7]
>>> l3 = [1, 2, 3, 4, 5, 5]
>>> l4 = [1, 2, 3, 4, 5, 4]
>>> l5 = [6, 5, 4, 3, 2, 1]
>>> 
>>> def is_coherent(seq):
...     return seq == range(seq[0], seq[0]+len(seq), 1)
... 
>>> is_coherent(l1)
True
>>> is_coherent(l2)
False
>>> is_coherent(l3)
False
>>> is_coherent(l4)
False
>>> is_coherent(l5)
False
>>> 

The results of some basic performance tests seem to indicate that this method is significantly quicker (I've added your example as is_coherent2 ):

Carl > python -m timeit -s 'from t import is_coherent, l1' 'is_coherent(l1)'
1000000 loops, best of 3: 0.782 usec per loop
Carl > python -m timeit -s 'from t import is_coherent, l3' 'is_coherent(l3)'
1000000 loops, best of 3: 0.796 usec per loop
Carl > python -m timeit -s 'from t import is_coherent2, l1' 'is_coherent2(l1)'
100000 loops, best of 3: 4.54 usec per loop
Carl > python -m timeit -s 'from t import is_coherent2, l3' 'is_coherent2(l3)'
100000 loops, best of 3: 4.93 usec per loop

Answer 3

If you're looking for a numpy solution:

import numpy as np

def is_coherent(x):
    return all(np.diff(x) == 1)

is_coherent(np.array([1,2,3,4,5]))
Out[39]: True

is_coherent(np.array([1,2,3,4,8]))
Out[40]: False

Answer 4

def is_coherent(seq):
    return all(x==y+1 for x, y in zip(seq[1:], seq))

Answer 5

This short circuits and does not create an extra list making it useful for testing very large lists.

def is_coherent(l):
    return all(a==b for a, b in izip_longest(l, xrange(l[0], l[-1]+1)))

Or

def is_coherent(l):
    return all(l[i] + 1 == l[i+1] for i in xrange(len(l)-1))

Answer 6

i dont know python but i know its functional so heres a small loop function that will do it if you change the syntax for correct python.

PSEUDO CODE

def is_coherent(seq):
     for x in xrange(1, len(seq)-1):
        if (seq[x+1]-seq[x] != 1)  return false;   
     return true