(place of [thing] in [list v]::list)

I don't know why this block doesn't exist yet. It's MUCH easier than making a bunch of blocks to scan through a list.

“Oh what if there isn't any of it in the list”
Two solutions.

if <[vegetables v] contains [apple ]> then
    do whatever to (place of [apple] in [vegetables v]::list)
end
or
< (place of [apple] in [vegetables v]::list) = [0]>

I'm not a programming language programmer so I don't know if you can or can't do this, but I think you'd have a block like this by now.
THE END

add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
set [index v] to (place of [thing] in [list v] ::list)

now what

Zro716 wrote:

add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
set [index v] to (place of [thing] in [list v] ::list)

now what

Average them - 2.5

jk

add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
set [index v] ((1st v) occurrence of [thing] in [list v] :: list) :: variables
set [index v] ((last v) occurrence of [thing] in [list v] :: list) :: variables

I've actually wanted this

Support, it would be useful for indexing lists.

Zro716 wrote:

add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
set [index v] to (place of [thing] in [list v] ::list)

now what

Hmmmmm
I guess the first one, unless you use the block ChocolatePi made
What other problems could that come across? I know you guys love finding those in your free time

Support. This would make creating “project cookies” (data that is specific to a user for that project, e.g. highscores) much easier.

Zro716 wrote:

add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
set [index v] to (place of [thing] in [list v] ::list)

now what

It would return the first index where it was found. I mean, that's how things like JavaScript arrays work.

Support for this

MegaApuTurkUltra wrote:

Zro716 wrote:

add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
set [index v] to (place of [thing] in [list v] ::list)

now what

It would return the first index where it was found. I mean, that's how things like JavaScript arrays work.

Support for this

So how about

(first occurence of  [thing] in [list v] :: list)

To avoid confusing people?

Cyoce wrote:

MegaApuTurkUltra wrote:

Zro716 wrote:

add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
set [index v] to (place of [thing] in [list v] ::list)

now what

It would return the first index where it was found. I mean, that's how things like JavaScript arrays work.

Support for this

So how about

first occurence of  [thing] in [list v] :: list

To avoid confusing people?

what about mine?

ChocolatePi wrote:

add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
set [index v] ((1st v) occurrence of [thing] in [list v] :: list) :: variables
set [index v] ((last v) occurrence of [thing] in [list v] :: list) :: variables

I've actually wanted this

ChocolatePi wrote:

Cyoce wrote:

MegaApuTurkUltra wrote:

Zro716 wrote:

add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
set [index v] to (place of [thing] in [list v] ::list)

now what

It would return the first index where it was found. I mean, that's how things like JavaScript arrays work.

Support for this

So how about

(first occurence of  [thing] in [list v] :: list)

To avoid confusing people?

what about mine?

ChocolatePi wrote:

add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
set [index v] ((1st v) occurrence of [thing] in [list v] :: list) :: variables
set [index v] ((last v) occurrence of [thing] in [list v] :: list) :: variables

I've actually wanted this

I like yours better, but my suggestion was just to clarify the block he was creating.
I would prefer yours, but if the block were to simply return the first one, it should be named as such.

Zro716 wrote:

add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
set [index v] to (place of [thing] in [list v] ::list)

now what

1.

Or zero if you do python/other languages that I don't know about

Alberknyis wrote:

Zro716 wrote:

add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
set [index v] to (place of [thing] in [list v] ::list)

now what

1.

Or zero if you do python/other languages that I don't know about

You meant bad languages.

Support!

Support

ChocolatePi wrote:

add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
set [index v] ((1st v) occurrence of [thing] in [list v] :: list) :: variables
set [index v] ((last v) occurrence of [thing] in [list v] :: list) :: variables

I've actually wanted this

Support this!

theonlygusti wrote:

Alberknyis wrote:

Zro716 wrote:

add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
add [thing] to [list v]
set [index v] to (place of [thing] in [list v] ::list)

now what

1.

Or zero if you do python/other languages that I don't know about

You meant bad languages.

No, I meant other languages.

Support for

((1st v) occurence of [thing] in [list v] ::list)
((last v) occurence of [thing] in [list v] ::list)

Support

Support! In fact, I've been REALLY wishing there was a

delete all occurrences of [thing] in [list v]::list

for a while.
But I agree with rollercoasterfan's idea:

rollercoasterfan wrote:

Support for

((1st v) occurence of [thing] in [list v] ::list)
((last v) occurence of [thing] in [list v] ::list)

since it wouldn't work if there was more than one occurence.
This is workaroundable, however.

(occurence (2 v) of [thing] in [list v] ::list)

would be workaroundable as follows.
For example,

say (join [The second occurrence of the string "thing" in the list is item ] (join (occurrence (2 v) of [thing] in [list v]::list) [.])

would be

set [item v] to [1]
set [occurrences v] to [0]
repeat until <(occurrences) = [2]>
if <(item (item) of [list v]) = [thing]> then
change [occurrences v] by [1]    
end
change [item v] by [1]    
end
say (join [The second occurrence of the string "thing" in the list is item ] (join ((item) - (1)) [.])

But that workaround is not that easy, so I still support.