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I was trying to use a ternary to print out a string if something was true and something else if it wasn't (simple enough, right?). It gives no errors and doesn't seem like it would cause any trouble. However, when running...

print('[',dr+1,']\t',d[dr] if fullpath else '[',dr+1,']\t',d[dr].split("/")[len(d[dr].split("/"))-2], sep='')

...it prints out 

'[',dr+1,']\t'

twice. For example, it prints this:

[1] [1] Accoustic
[2] [2] Classical
[3] [3] Epic
[4] [4] Rock
[5] [5] Spoof
[6] [6] Techno & Electronic

If you want, in order to see what the variables are (which shouldn't matter since this is just a concept issue), you can see the rest of the code here

Thinking that it might be the variables, I tried printing the statements separately (no ternary) and they came out as I expected (i.e. without the odd duplication). Knowing that the variables were fine, I tried enclosing the strings with parenthesis (in an attempt to keep the strings from mixing) but, since python interprets them literally, they came out as arrays instead of strings. Also, + symbols (in lieu of commas) don't work because python can't combine the data type. I ended up changing the statement to

print('[',dr+1,']\t',d[dr]) if fullpath else print('[',dr+1,']\t',d[dr].split("/")[len(d[dr].split("/"))-2], sep='')

and it works fine.

Not so great coding aside, why is that first string getting printed twice?

Vreality
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2 Answers2

2

Allow me to break it down for you...

print(('['), (dr+1), (']\t'), (d[dr] if fullpath else '['), (dr+1), (']\t'), (d[dr].split("/")[len(d[dr].split("/"))-2]), sep='')

See the problem?

print(*(('[',dr+1,']\t', d[dr]) if fullpath else ('[',dr+1,']\t', d[dr].split("/")[len(d[dr].split("/"))-2])), sep='')
Ignacio Vazquez-Abrams
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  • Ah... Yes. I think. I was under the assumption that if and else somehow were not included in the groupings. – Vreality Dec 18 '12 at 01:03
2

I'm not sure why it wouldn't print twice:

print('[',dr+1,']\t',       #first one
      d[dr] if fullpath else '[',
      dr+1,']\t',           #second one (most of it anyway)
      d[dr].split("/")[len(d[dr].split("/"))-2],
      sep='')

Depending on fullpath, your statement could look like

print('[',dr+1,']\t',       #first one
      d[dr],dr+1,']\t',     #second one (more or less)
      d[dr].split("/")[len(d[dr].split("/"))-2], 
      sep='')

or it could look like:

print('[',dr+1,']\t',
      '[',dr+1,']\t',
      d[dr].split("/")[len(d[dr].split("/"))-2], 
      sep='')
mgilson
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  • Thank you, the other answer was easier to understand though. – Vreality Dec 18 '12 at 01:06
  • @Vreality2007 -- I'm glad you got it figured out. This one makes more sense to me, but the other is good too. I'll leave this for those who might come across this whose minds work like mine. :-) – mgilson Dec 18 '12 at 01:08
  • Good idea. :) Not that your answer is wrong, it's just that the grouping of the parenthesis helped me understand why it works like that (not just what it comes out to.) – Vreality Dec 18 '12 at 01:09