Python, next element to array

Question

I need to access the next element of an array to compare it with the previous one, try to do it by indexes however the index is out of range

lista =  [1,2,2,3,4,5,5,6,7,8]
for i in range(len(lista)):
if lista[i]==lista[i+1]: print("same number")

you should use len(lista)-1 since the last element has no next — Abdeslem SMAHI
– Abdeslem SMAHI, Commented Mar 10, 2020 at 1:04
What is the maximum value that i will be assigned here? If it is out of range of lista, then perhaps change the loop so that it never has a value greater than the largest index. — Thomas Matecki
– Thomas Matecki, Commented Mar 10, 2020 at 1:06

Mark · Accepted Answer · 2020-03-10 01:05:41Z

4

You can zip the list with itself offset by one and avoid the indices altogether:

lista =  [1,2,2,3,4,5,5,6,7,8]
for a, b in zip(lista, lista[1:]):
    if a == b: 
        print("same number", a, b)

Prints:

same number 2 2
same number 5 5

answered Mar 10, 2020 at 1:05

Mark

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Comments

Masoud Khani · Accepted Answer · 2020-03-10 01:43:35Z

0

You can simply check the size of the list before doing the process. check the code bellow

for i in range(len(lista)):
if i == len(lista)-1:
    print("end Process")
else:
    a = lista[i]
    b = lista[i+1]
    if a == b:
        print( str(a) + " and " + str(b) + " are the same number ")

or using len(lista)-1 in your for loop

for i in range(len(lista)-1):
        a = lista[i]
        b = lista[i+1]
        if a == b:
            print( str(a) + " and " + str(b) + " are the same number ")

answered Mar 10, 2020 at 1:43

Masoud Khani

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