Comparing every element in a list with the next one

Question

I have a list of integers a and I want to iterate through every element of it, and if the element is smaller or the same size as the next element in the list (the upcoming index-neighbour), I want to convert the element into a string and concatenate "000" onto it. But if it's larger than the next element, I want to concatenate "111" onto it.

I use Python 3.7.3.

Here's what I've already tried:

a = [41433, 23947, 10128, 89128, 29523, 47106]

for I in a:
      if a[I] <= a[I+1]:
             a[I] = str(a[I] + "000")
      else:
             a[I] = str(a[I] + "111")

I've actuallly tried a lot more than that, but nothing worked. When I run the code, I always get "IndexError: list index out of range".

I'm quite new in Python, anyone know a solution?

Given your description, all elements but the last one are converted to strings. Correct? So, for i in range(len(a)-1): a[i] = str(a[i]) + ('000' if a[i] <= a[i+1] else '111') — tonypdmtr
– tonypdmtr, Commented Jul 19, 2019 at 21:52
So what is your expected output, and what is the treatment for the last item in the list where it can't be compared to the one after it? — Alexander
– Alexander, Commented Jul 19, 2019 at 21:57

OneCricketeer · Accepted Answer · 2019-07-19 21:42:20Z

4

Your I value isn't an index, it's an element in a (41433, 23947 ect), so a[41433] doesn't exist.

Try using a range

for i in range(0, len(a)-1):
  if a[i] <= a[i+1]:
    a[i] = str(a[i]) + "000"
  else:
    a[i] = str(a[i]) + "111"

edited Jul 19, 2019 at 21:42

answered Jul 19, 2019 at 21:34

OneCricketeer

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Comments

chryss · Accepted Answer · 2019-07-19 22:14:35Z

0

I would use itertools to create two iterators for your list, and zip them together so that you get pairs of successive values:

from itertools import tee

i1, i2 = tee(a)
next(i2, None)    # advances i2 by one step
pairs = zip(i1, i2)
a = [str(elem1)+'000' 
        if elem1 <= elem2 else str(elem1)+'111' 
        for elem1, elem2 in pairs]

answered Jul 19, 2019 at 22:14

chryss

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Comments

garyboland · Accepted Answer · 2019-07-19 22:14:50Z

enumerate is perfect for this use case. it will also increase readability:

a = [41433, 23947, 10128, 89128, 29523, 47106]                  

for index, elem in enumerate(a[:-1]):                           
    if a[index] <= a[index+1]:                                  
        a[index] = str(a[index]) + "000"                     
    else:                                                       
        a[index] = str(a[index]) + "111"                     
print(a)

['41433111', '23947111', '10128000', '89128111', '29523000', 47106]

SkippyElvis · Accepted Answer · 2019-08-16 19:03:12Z

0

A pythonic way would be to use enumerate, which generates tuples of (index, list[index).

for idx, val in enumerate(a[:-1]):
    if val <= a[idx+1]:
        a[idx] = str(a[idx]) + "000"
    else:
        a[idx] = str(a[idx]) + "111"

Notes
Loop until second-to-last item by using a[:-1] so that you don't run into index out of bounds errors.

edited Aug 16, 2019 at 19:03

answered Jul 19, 2019 at 22:13

SkippyElvis

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