if statement with list contains

Question

I have a list which contains the numbers (lijstkleur) 1,4,6,7. I also have a range from 0 till 8. Now i have te following code:

for x in range(0, len(L), 1):
 if x in lijstkleur == True:
 self.label = Label(self.frame, text=string[x], fg="yellow", bg="red")
 self.label.pack(side=LEFT)
 else:
 self.label = Label(self.frame, text=string[x], fg="white", bg="red")
 self.label.pack(side=LEFT)

but all the numbers become white, what is wrong with this if statement

Answer 1

No need to use == True:

if x in lijstkleur:

The expression x in lijstkleur==True is interpreted as (x in lijstkleur) and (lijstkleur == True); a list is never equal to boolean True and thus you end up testing something and False, guaranteed to be False instead. This is called comparison chaining, making expressions like 10 < a < 20 possible.

You can simplify your range() call to just len(L):

for x in range(len(L)):

and there is no need to repeat the .pack() call:

if x in lijstkleur:
 self.label=Label(self.frame,text=string[x],fg="yellow",bg="red")
else:
 self.label=Label(self.frame,text=string[x],fg="white",bg="red")
self.label.pack(side=LEFT)

• hughdbrown: And no need for for x in range(0,len(L),1) when for x in range(len(L)) will do. Or for x, s in zip(L, string). And self.label.pack(side=LEFT) can be hoisted ... There's a lot here.

Answer 2

Your conditional isn't evaluating the way you think it is. It's doing this:

if (x in lijstkleur) and (lijstkleur==True):

The result of lijstkleur==True is always False, since a list is never equal to a boolean, so the conditional always returns False. What you want is this:

if x in lijstkleur:

• Karoly Horvath: lijstkleur==True is True.. or False.. neither of them are iterable. Why don't you get any error message?
• Henry Keiter: @KarolyHorvath Whoops; skipped a piece. Fixed.
• Karoly Horvath: Ahh.. I forgot that operators can be on both side. Thx!