Python -- how to force enumerate to start at 1 -- or workaround?

Question

I have a simple for loop:

for index, (key, value) in enumerate(self.addArgs["khzObj"].nodes.items()):

and I want to start a new wx horizontal boxsizer after every 3rd item to create a panel with 3 nodes each and going on for as many as are in nodes. The obvious solution is to do:

if index % 3 == 0: add a horizontal spacer

but the enumerate starts at 0, so 0 % 3 == 0 and it would start a new row right off the bat. I've tried doing:

if index == 0: index = index + 1

but of course that doesn't work because it creates a new var instead of changing the original -- so I get 1, 1, 2, 3, 4, etc and that won't work because I'll get 4 nodes before I hit a index % 3 == 0.

Any suggestions on how to do this? This isn't a big enumerate, usually only about 10-15 items. Thanks.

Read. The. Documentation. `pydoc enumerate` is less effort than asking here... — Has QUIT--Anony-Mousse, Mar 09 '15 at 22:48
A similar question has already been answered in http://stackoverflow.com/questions/3303608/how-to-enumerate-a-range-of-numbers-starting-at-1 — prabhugs, Mar 24 '16 at 09:31

score 19 · Accepted Answer · answered Mar 07 '15 at 23:44

19

Since Python 2.6, enumerate() takes an optional start parameter to indicate where to start the enumeration. See the documentation for enumerate.

answered Mar 07 '15 at 23:44

Daniel Pryden

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1

I didn't know that, guess you learn something new every day. Good job. – Malik Brahimi Mar 07 '15 at 23:47
1

Before that was added, `itertools.izip(itertools.count(1), L)` would have been the way to do it. – Dan D. Mar 07 '15 at 23:47
1

`＋1` This is actually pretty handy to know, but the op could also just use `index % 3 == 2` – Mar 08 '15 at 00:36
Using enumerate will still loop through all of the elements, it just changes which position it starts at. – Speerian Jan 13 '18 at 00:40

score 3 · Answer 2 · answered Mar 08 '15 at 00:31

3

You are going to hate this answer for how obvious it is, but you could just do:

if index % 3 == 2: add a horizontal spacer

This adds the spacer after the 2nd element (which is actually the third), and every third element after it.

answered Mar 08 '15 at 00:31

I didn't notice that -- fancy. I marked Daniel's as correct, though, because it's a bit more explicit and it helps me in other situations that aren't so neat and tidy ;) – linus72982 Mar 08 '15 at 02:21
His solution is certainly nifty. – Mar 08 '15 at 02:23

Python -- how to force enumerate to start at 1 -- or workaround?

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