如何比较列表中的元素

Question

I'm having some trouble with an algorithm (To find the max and min values), not the algorithm itself but the implementation, let me explain:

Let's say the list is n = [0,1,1,2,3,5,8,13,-1,99] ; len(n) = 10 Then globalMin, globalMax = n[0], n[0] #To skip 1 iteration from the list

And now what I have to do is compare by 'pairs', so since I already used n[0] I start comparing n[1] and n[2] to find the max and min values between those 2 and then compare it with y global min, max values, then n[3] and n[4] and compare the nm with my global values, then n[5] and n[6] .. until I have to compare n[9] and n[10], as you can see n[10] doesn't exist on my list. I thought I could solve this with list slicing using the next code:

 for i in range(1, len(n), 2):

    if n[i:i+1] > n[i+1:i+2]:
      minl, maxl = n[i+1:i+2], n[i:i+1] # minl = local min; maxl = local max
    else:
      maxl, minl = n[i+1:i+2], n[i:i+1]

But this wont work if my last element is only one (as on the example above) so, as you can guess, If the min or max is the last element on my list it will be ignored. I have been trying to fix this with index or list slicing but no luck at all, any suggestions? I have to do this in a 'Pythonic' way and making sure to make this as simple and short as possible without using imports. I have figured the rest of the algorithm which is based on the next image: Image

Answer 1

You can check the length of list first, if it's of odd length then you can append the last element at the end of the list. Appending is an O(1) operation, so, it won't affect the time complexity.

You can use a while loop:

In [78]: lis = [0,1,1,2,3,5,8,13,-1]

In [79]: if len(lis)%2 !=0 :  #if the list is of odd length then append the last item to it
    lis.append(lis[-1])
   ....:     

In [80]: i=0

In [81]: while i<len(lis)-1:
    if lis[i]>lis[i+1]:
        local_max,local_min=lis[i],lis[i+1]
    elif lis[i]<lis[i+1]:    
        local_max,local_min=lis[i+1],lis[i]
    else:
        local_max,local_min=lis[i],lis[i+1]
    print local_min,local_max
    i+=2
   ....:     
0 1
1 2
3 5
8 13
-1 -1

or you can use an iterator:

In [86]: it=iter(lis)

In [87]: lis = [0,1,1,2,3,5,8,13,-1]

In [88]: if len(lis)%2 !=0 :
    lis.append(lis[-1])
   ....:     

In [89]: it=iter(lis)

In [90]: for _ in xrange(len(lis)/2):
   ....:     a,b=next(it),next(it)
   ....:     if a>b:
   ....:         local_max,local_min=a,b
   ....:     elif a<b:    
   ....:         local_max,local_min=b,a
   ....:     else:    
   ....:         local_max,local_min=a,b
   ....:     print local_min,local_max    
   ....:     
0 1
1 2
3 5
8 13
-1 -1

Answer 2

I would argue this is the most Pythonic implementation of this:

from itertools import zip_longest

n = [0, 1, 1, 2, 3, 5, 8, 13, -1, 99]
global_min = global_max = n[0]

groups = [iter(n)] * 2
for a, b in zip_longest(*groups):
    if b is not None:
        if a > b:
            local_min, local_max = b, a
        else:
            local_min, local_max = a, b
    else:
        local_min = local_max = a
    if local_min < global_min:
        global_min = local_min
    if local_max > global_max:
        global_max = local_max

print(global_min, global_max)

We group the items by using itertools.zip_longest() ( itertools.izip_longest() in 2.x) and a list of the same iterator repeated. We then loop over this, which gives us the values in pairs. We then do a check, if the value is the last one, we assign it as the local_min and local_max , otherwise, we compare the two values and assign as appropriate.

We then compare (and potentially update) global_min and global_max .