Searching for an element in a list with a loop in Python

Question

I've been having some trouble finding if an element is already stored in a list in a Python code. I've used two scripts. The second one I wrote turned out to work, but the first one didn't and I don't understand why (and it's bothering me). The list I have is called grains and looks like this :
[[a, b], [c, d], [e, f], ...].

Here is the first piece of code, not working :

for i in range(0, len(grains) - 1) :
    if grains[i] == [x,y] :
        return
    else :
        grains.append([x,y])
        grains_restants = grains_restants - 1
        fourmi_chargee = False
        return

And here is the second piece of code, this one works :

if not([x,y] in grains) :
    grains.append([x,y])
    grains_restants = grains_restants - 1
    fourmi_chargee = False
    return

So I tried to understand why the first piece of code doesn't work, but I gave up. Do you know why ?

what is the value of a, b in grains? what is the value of x and y? — shahkalpesh
– shahkalpesh, Commented Apr 17, 2017 at 8:34
range's parameters are inclusive, exclusive - that means it will include the first number you give it, and go all the way up to but not including, the second number. Hence, you never actually access the last element in your for loop. — deepbrook
– deepbrook, Commented Apr 17, 2017 at 8:34
Because in your first piece of code it'll only compare grains[0] == [x, y], and then the for loop will return either it's True or False. — shizhz
– shizhz, Commented Apr 17, 2017 at 8:35
also, if your list example is correct, there shouldn't be an element [x,y] in your list. Instead, there should be [w, x] and ['y, z]. — deepbrook
– deepbrook, Commented Apr 17, 2017 at 8:36

shizhz · Accepted Answer · 2017-04-17 09:30:14Z

2

Your first piece of code will not loop list grains as your expectation, it'll actually only compare grains[0] == [x, y] and the for loop will exit due to you have return in either if and else branch, while the second implementation will check whether [x, y] is in the whole list grains or not.

You should NOT use code like this but the for loop code is supposed to be:

for i in grains:
    if i == [x,y] :
        break
else:
    grains.append([x,y])
    grains_restants = grains_restants - 1
    fourmi_chargee = False

edited Apr 17, 2017 at 9:30

answered Apr 17, 2017 at 8:47

shizhz

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masnun · Accepted Answer · 2017-04-17 08:38:08Z

for i in range(0, len(grains) - 1) :
    if grains[i] == [x,y] :
        return
    else :
        grains.append([x,y])
        grains_restants = grains_restants - 1
        fourmi_chargee = False
        return

You are matching every element in the list against [x,y] but you really want to check if [x,y] exists already. That's what the 2nd code snippet does using the in operator.

Now why does the first one not work? You didn't mention what errors / unexpected scenarios you were getting but I can see one. Let's assume that the [x,y] element is in index 3 (4th item). But you're comparing it to 0,1,2 => all of them won't match and add a duplicate item and then return. That's really not what you want.

To check if the element exists, using the in operator is the best idea. You could also use list.count to see if any element exists in the list (and how many times it appears in that list).

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