Simple definition

Developers
1

Functions.dyn (9.3 KB)
2019-03-18_20h47_51 2019-03-18_20h50_42

Hello Dynos,

I try to define Funktions (def), I have agiain some basic issuses:

I try to give strings a value f.e. “B” has 10 or “A” has 5.

For now I have no syntex-error, but the result is 0 :slight_smile:

KR

Andreas

2019-03-19_08h17_03
2019-03-19_08h17_03.png1175×191 23.3 KB

1 Like
2

you aren’t returning anything in your function.

Plus

you are defining “count” as variabel of your function but in your function you are using “letter”

So with count = 0 you are saying:
Dynamo run me the score function (score(0)).

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3

A few things. What @3Pinter said as well as that you are not adding count to anything so it will only return the final outcome once you do iterate through the letters, which is 0. Also, because count is inside the function, it becomes a local variable. Basically that means the variable only exists within the function and has no meaning outside of it. It is also disconnected from the original count outside of the function.

I think this is more of what you are going for:
pythoncountletters

letters = IN[0]
count = 0

def score(l):
    if l == "B":
        return 10
    elif l == "A":
        return 5
    else: return 0

for letter in letters:
    count += score(letter)

OUT = count
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4

@kennyb6

LOL! EXACTLY as I was gonna suggest, including the letter/letters :slight_smile:

letters = IN[0]
count = 0

def score(letter):
	if letter == "B":
		count = 10
	elif letter == "A":
		count = 5
	else:
		count = 0
	return count

for letter in letters
	count += score(letter)

OUT = count
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5

Ok and for collecting the results I use not count.append(letter) so I use an Operator +=

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6

Haha that’s great. I actually wanted to suggest using a dictionary for this as it would be more organized:

letters = IN[0]
points = {"A" : 5, "B" : 10}
count = 0

def score(l):
    if l in points:
        return points[l]
    else: return 0

for letter in letters:
    count += score(letter)

OUT = count
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7

And how to count separatly? Item by item f.e. 5,5,5,5,10,10,10 count[letters]

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8

If you want to add to a list, first you have to initialize a list and then you can use .append() to append the result of the function.

letters = IN[0]
points = {"A" : 5, "B" : 10}
count = [] # initializing an empty list so you can append to it later

def score(l):
    if l in points:
        return points[l]
    else: return 0

for letter in letters:
    count.append(score(letter))

OUT = count

or shorter version:

letters = IN[0]
points = {"A" : 5, "B" : 10}

def score(l):
    if l in points:
        return points[l]
    else: return 0

count = [score(letter) for letter in letters]

OUT = count

pythoncountletterslist

2 Likes