Changing a list item without changing original values

Question

I'm doing some practice problems and one program is asking me to change numbers in the list to 99, only if the number previously is even. I'm running into the trouble that when I change the number to 99, when it moves on the the next number, it checks to see if 99 is even, which I don't want, I want it to check the original value I had there, not what I changed it to .

d = [9, 8, 2, 15, 6, 33, 10, 4]
i = 1
while i < len(d) - 1 :
    if d[i-1]%2 == 0 :
        d[i] = 99
    i = i + 1

print d

returns [9, 8, 99, 15, 6, 99, 10, 4]

I want it to return [9,8,99,99,6,99,10,99]

How would I add 99 into the original list without changing its original value if that makes any sense? List methods like pop or insert can not be used.

Be very careful about your wording (I understand English might not be your first language). You said `"if the number previously is even`". I think you mean "if *the previous number* was even". There's a big difference. — Jonathon Reinhart, Oct 23 '13 at 07:19
and how would i do the "was" part in python. do i make like a temporary values assigned to the previous value or something? — Zach Santiago, Oct 23 '13 at 07:22
possible duplicate of [Unexpected IndexError while removing list items](http://stackoverflow.com/questions/19312021/unexpected-indexerror-while-removing-list-items) — aIKid, Oct 23 '13 at 07:50

score 5 · Answer 1 · answered Oct 23 '13 at 07:34

5

Try this.

d = [9, 8, 2, 15, 6, 33, 10, 4]
for i in reversed(xrange(1, len(d))):
  d[i] = 99 if d[i - 1] % 2 == 0 else d[i]

answered Oct 23 '13 at 07:34

moofins

150
2
4

oleg · Answer 2 · 2013-10-23T07:34:15.773

I would advice to iterate descending:

d = [9, 8, 2, 15, 6, 33, 10, 4]
for x in xrange(len(d)-1, 0, -1):
    if d[x - 1] % 2 == 0:
        d[x] = 99
print d

In this case next iteration will operate not changed values.

Or You can create new list

or You can add variable for previous value and use it in if statement

d = [9, 8, 2, 15, 6, 33, 10, 4]
previous = d[0]
for i, x in enumerate(d[1:]):
    if previous % 2 == 0 :
        previous = d[i]
        d[i] = 99
    else:
        previous = d[i]

print d

score 0 · Answer 3 · answered Oct 23 '13 at 07:18

0

search the list backwards

see last and before last value (n-1, n-2)
change the last value if needed
move to previous values (n-2,n-3)
repeat until whole list is updated

answered Oct 23 '13 at 07:18

Spektre

49,595
11
110
380

score 0 · Answer 4 · answered Oct 23 '13 at 07:55

0

Try this:

d = [9, 8, 2, 15, 6, 33, 10, 4]

prev=d[0]
for i,j in enumerate(d):
        if prev%2==0:
                prev=j
                d[i]=99
        else:
                prev=j

print d

answered Oct 23 '13 at 07:55

Darknight

1,132
2
13
31

Changing a list item without changing original values

4 Answers4