列表中相邻元素之间的差异越来越大

Question

Write a function expanding(l) that takes as input a list of integer l and returns True if the absolute difference between each adjacent pair of elements strictly increases.

I tried to execute this code but this isn't returning correct value for some lists.

def expanding(l):
 for i in range(0,len(l)-3):
  if (abs(l[i+2]-l[i+1])>abs(l[i+1]-l[i])):
   Answer=True
  else:
   Answer=False
 return Answer

expanding([1,3,7,2,-3]) should be False but the output is True .

Answer 1

Use a temporary variable to store the difference, and exit once you reach a non-increasing difference.

def expanding(l):

    dif = abs(l[1] - l[0])

    for i in range(1, len(l)-1):
        temp = abs(l[i+1] - l[i])

        # Non-increasing difference, return
        if temp < dif:
            return False
        else:
            dif = temp

    # All differences are increasing
    return True

Answer 2

Numpy is your friend:

import numpy as np

x=np.array([1,3,7,2,-3])
(np.diff(abs(x[1:] - x[:-1])) > 0).all()  # returns False

If you're fancy a function, I would do like this:

def expanding(l):
    # convert list to  np.array if a list was passed
    if isinstance(l, list):
        l = np.array(l)

    return (np.diff(abs(l[1:] - l[:-1])) > 0).all()

Answer 3

Yet another solution using iterators:

from itertools import tee, islice, starmap
from operator import lt, sub

def pairwise(x):
    a, b = tee(x, 2)
    return zip(a, islice(b, 1, None))

a = [1,3,7,2,-3]

pairs = pairwise(a) # this will be (a[0], a[1]), (a[1], a[2]), ...

# The next will produce the result abs(a[1]-a[0]), abs(a[2]-a[1]), ...
differences = map(abs, starmap(sub, pairs))

# This will be abs(a[2]-a[1])>abs(a[1]-a[0]), abs(a[3]-a[2])>abs(a[2]-a[1]), ...
cmp = starmap(lt, pairwise(differences))

# Differences strictly increases if all items in cmp are evaluated to True...
result = all(cmp)
print(result)

The output for such input is False

Answer 4

I would recommend writing the condition in an iterator that you then pass to any so to make sure that on the first occurrence of a non-expanding difference the iteration stops and False is returned.

Here is how this would look:

def expanding(l):
    return not any((abs(l[i-1]-l[i])>=abs(l[i]-l[i+1]) for i in range(1,len(l)-1)))

Answer 5

Your logic is a little bit wrong. Once you know that one item is out of order you should answer False for the whole list.

def expanding(l):
 for i in range(0,len(l)-3):
  if (abs(l[i+2]-l[i+1])<=abs(l[i+1]-l[i])):
   return False 
 return True

Answer 6

def expanding(l):
 for i in range(0,len(l)-2):
  if (abs(l[i+2]-l[i+1])>abs(l[i+1]-l[i])):
   Answer=True
  else:
   Answer=False
   return Answer
 return Answer

expanding([1,3,7,2,-3]) 

As soon as you know that one item is out of order you should answer False for the whole list without waiting,so that other pairs not change the answer.

Also note the change from range(0,len(l)-3) to range(0,len(l)-2) . The original implementation was missing the last pair of list elements.

Answer 7

Code Written By Debkanta Mondal

Python3

def expanding(l):
dif=[]
for x,y in zip(l,l[1:]):
    diff=abs(y-x)
    dif.append(diff)
return all(i<j for i,j in zip(dif,dif[1:]))

print(expanding([1,3,7,2,-3]) )