CODEWITHSUNDEEP
pythonif-statement
Glenn
Glenn

Reputation: 69

if statement with list contains

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

Upvotes: 1

Views: 1269

Answers (2)

Martijn Pieters
Martijn Pieters

Reputation: 1124818

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)

Upvotes: 5

Henry Keiter
Henry Keiter

Reputation: 17188

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:

Upvotes: 1

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