stocks that have gone up x of the last y days

[Scotty123]

TOS code to find stocks that have gone up x of the last y days (e.g stocks that have gone up 3 of the last 5 days)

Hi, I believe I have seen this code before but cannot find it again.

I would like to have a scan script that would find stocks that were up, say 3 of the last 5 days, and are up 20% during that period.

Can anyone help me with that code?

Thanks in advance.

Scotty

 

[BenTen]

I believe TOS already has some sort of pre-made scanners for this.

Maybe you can sort by gap-up stocks on the Daily chart?

 

[Scotty123]

I believe TOS already has some sort of pre-made scanners for this.

Maybe you can sort by gap-up stocks on the Daily chart?

Thanks BenTen. So you can't think of any script that will allow you to search in the last X periods for symbols that are up Y times (where X would be 5 and Y would be 3 in the example I gave)?

I'm pretty sure I saw some code like that. If I find it, I will post it here.

Scotty

 

[mashume]

Adjust as necessary. I've set it to look at Previous Close vs This Close to determine whether it went up. You might prefer Open vs Close. As you see fit.

def lookback = 5;
def up_days = 3;
def is_up = if close > close[1] then 1 else 0;
plot eureka = if sum(is_up, lookback) == up_days then 1 else double.nan;

Note that this finds symbols which are up EXACTLY 3 of the last 5 days. You can replace == with >= if you want 3, 4, or 5 of the last 5 days. All we're really doing is creating an array with 1s and 0s for each period (Day?): a 1 if the day is up and a 0 if the day is not up. Then we sum the array over the last n days (lookback) and decide if it is equal to the number of up days you're looking for.

hope that helps.
-mashume

 

[Scotty123]

Thanks Mashume!

Question: if today's close is below yesterday's, but the previous 4 days were up, will the symbol be plotted, or does today's close need to be up?

Wondering if I would I need to change the is_up variable assignment as follows which would check that at least 1 of the last 3 days were up (which is a requirement for 3 of the last 5 days to be up):

def is_up = if close > close[1] or close[1] > close[2] or close[2] > close[3] then 1 else 0;

Also, is there a difference in output between these statements?

plot eureka = if sum(is_up, lookback) == up_days then 1 else double. Nan;
plot eureka = if sum(is_up, lookback) == up_days then 1 else 0;

 

[mashume]

Thanks Mashume!

Question: if today's close is below yesterday's, but the previous 4 days were up, will the symbol be plotted, or does today's close need to be up?

Wondering if I would I need to change the is_up variable assignment as follows which would check that at least 1 of the last 3 days were up (which is a requirement for 3 of the last 5 days to be up):



Also, is there a difference in output between these statements?

All the code looks at is whether x of the last n days were up days (close > previous close). It does not look at order or anything else. You don't need to do the long string of OR conditions. The sum does that for you pretty well.

If you want to add a condition that at least one of the last 3 days is up, I would do it this way:

def eureka = if sum(is_up, lookback) == up_days AND sum(is_up, 3) >= 1 then 1 else double.nan;

As always, adjust the logic to suit what you need. I chose >= for the second condition.

While your logic statement is essentially equivalent, it is a bit harder to understand at a glance and certainly more difficult to debug. I can't comment on whether it is faster or not.

You can even get into fun things like "Are there three up days in the 5 days preceding the most recent five days (not counting them)?" like this:

def eureka = if sum(is_up[5], lookback) == up_days then 1 else double.nan;

and other such fun things with sums of arrays. Note that it may actually be is_up[6]... I'm just typing at the browser not in the ThinkScript editor.

As to the difference between the two statements at the end of your post, if you are using it in a scanner, I prefer to have eureka return no value if there is not a match and a value if there is -- that's the logic behind the 1 vs double.nan. I really don't do scans very often and don't fully understand whether any numeric result would cause the inclusion of a symbol in the returns list or not. I do know that 1 vs NaN works as I expect it to.

Happy Trading, and if you have more questions do come back. We'll see what we can figure out
-mashume

 

[hballa]

Hi @mashume, can you please share scanner link?