如何遍历python数组并将元素与前一个元素进行比较？

Question

I'm having a bit of trouble with an array in python. I want to loop through it, and compare element n to element n-1. For example:

 [(11, 11), (11, 10), (11, 9), (11, 8), (11, 7), (11, 6), (11, 5),
  (11, 4), (10, 4), (9, 4), (8, 4), (8, 5), (7, 5), (6, 5), (5, 5),
  (4, 5), (3, 5), (3, 4), (3, 3), (2, 3), (1, 3), (1, 2), (1, 1), (1, 0)]

Using the above array, I want to apply the following moves/logic:

• 0,1 = right

• 1,0 = down

• -1,0 = up

• 0,-1 = left

So if the element of the array we are looking at's first value is less than the previous I want to print up.

So the result for the array above would be (assuming the start is always 0,0)

[Start, down, right, right, right, down, down, right, right, down, 
down, down, down, down, left, down, down, down, right, right, right, 
right, right, right, right] 

It's a tricky one to explain so apologies if this is a bit confusing. Also the element will never go diagonal so it will never got from (1,1) to (2,2) one one of the 2 sub-elements will change at any given time.

Answer 1

Considering that your array of coordinates is called coords , you could do:

steps = [(x2-x1, y2-y1) for ((x1, y1), (x2, y2)) in  zip(coords, coords[1:])]

Explanation:

• coords[1:] represents all coordinate pairs starting with the second one
• zip(coords, coords[1:]) makes pairs of each coordinate with the one before
• for an array of n corrdinate pairs, zip will output n-1 steps, since zip 's output length is equal to the length of the shorter one of its arguments. So, due to the second list being shorter, the first list will only be enumerated excluding its last element (thanks Ev. Kounis for suggesting this clarification)

EDIT: to make a list of strings representing the steps made, you can make a dictionary of possible movements:

coords = [(11, 11), (11, 10), (11, 9), (11, 8), (11, 7), (11, 6),
          (11, 5), (11, 4), (10, 4), (9, 4), (8, 4), (8, 5), (7, 5),
          (6, 5), (5, 5), (4, 5), (3, 5), (3, 4), (3, 3), (2, 3),
          (1, 3), (1, 2), (1, 1), (1, 0)]

movements = {
        (0, 1):  'right',
        (1, 0):  'down',
        (-1, 0): 'up',
        (0, -1): 'left'
        }

steps = [movements[(x2-x1, y2-y1)]
         for ((x1, y1), (x2, y2)) in zip(coords, coords[1:])]

Answer 2

You could always just access index i-1 for the previous element starting at i=1 :

from operator import sub

lst = [(11, 11), (11, 10), (11, 9), (11, 8), (11, 7), (11, 6), (11, 5), (11, 4), (10, 4), (9, 4), (8, 4), (8, 5), (7, 5), (6, 5), (5, 5), (4, 5), (3, 5), (3, 4), (3, 3), (2, 3), (1, 3), (1, 2), (1, 1), (1, 0)]

d = {(0, 1):'right', (1, 0): 'down', (-1, 0): 'up', (0, -1): 'left'}

result = [d[tuple(map(sub, lst[i], lst[i-1]))] for i in range(1, len(lst))]

print(result)
# ['left', 'left', 'left', 'left', 'left', 'left', 'left', 'up', 'up', 'up', 'right', 'up', 'up', 'up', 'up', 'up', 'left', 'left', 'up', 'up', 'left', 'left', 'left']