Is it possible to programmatically find ALL horizontal channels/floats at a certain time interval or at least formulate the ToR correctly?

 

Here's the problem, I know what should work, but in what way, I can't formulate it.

On a certain time interval, e.g. a week, I need to find ALL horizontal channels with a width of e.g. 50pp.

Additional signs of beginnings and endings of fluxes:

1. The price alternately crosses the boundaries of the channel (turns inside it twice), the trajectory of which, inside the channel, looks like v or ^ (two reversals inside the channel, alternately testing its boundaries).

2. If the price leaves the borders of the channel at the distance equal to its width (this can be variable, e.g. twice the channel width, three times, etc.), then we suggest that the flat has ceased to exist:).

From all the channels found (the flatlet), we read some information.

Help.

 

V or N is not yet a channel. W or M can already be conditionally considered a channel.

There are many channel indicators in codebase.

In addition, look at EAs working on the breakdown of the Asian session, they have a lot of these things.

 
sever29:

At a certain time interval, e.g. a week, you must programmatically find ALL horizontal channels with a width of e.g. 50pp.

You mean draw a system of levels?
 
gip:

1. V or N is not yet a channel. W or M can already be considered conditionally a channel.

2. There are many channel indicators in the codebase.

3. In addition, look at Expert Advisors that work on the breakdown of the Asian session, they have a lot of such things.

1. I have my own methods of working with youth ... :) and for me the flat is V or ^ (check the opposite). And the trajectory is W, not just a channel, but a flyaway (four flips):)

2. It's not the same... I don't need pictures, I need statistics, i.e. a script.

3. wrong...

 
Candid:
You mean draw a system of levels?


You don't have to draw it, you just... um... The script should identify these channels and read their number to start with.

Can you help? I'd estimate $50 for a professional scripting job.

 
sever29:


You don't need to draw, you need... um... The script needs to identify these channels and count their number to start with.

can you help me? I'll give you $50 for a professional scripting job.

)))))
 
sever29:

Here's the problem, I know what should work, but in what way, I can't formulate it.

At a certain time interval, e.g. a week, I need to programmatically find ALL horizontal channels with a width of, for example, 50 pips.

Additional attributes of the beginning and end of the channels existence:

1. The price alternately crosses the boundaries of the channel (turns inside it twice), the trajectory of which, inside the channel, looks like v or ^ (two reversals inside the channel, alternately testing its boundaries).

2. If the price leaves the borders of the channel at a distance equal to its width (this can be variable, e.g. twice the channel width or three times, etc.), we consider the channel to be non-existent:).

We read some information from all the channels found (flat).

Help.


that's what thinking about an avalanche leads to. :)

the original bet was 15 exactly trend. :) waiting for the resistance line.

 
sanyooooook:
)))))

Yeah, tell me about it and we'll have a laugh.
 
neama:


this is what thinking about an avalanche leads to. :)

the original bet was 15 exactly trending. :) waiting for the resistance line.


1961
sever29 31.08.2010 01:20 am correct | delete

i'll call it a royal, armoured, cyclic avalanche matrix with trap elements...

avalanche is a stupid pluming martin :)

maybe a rebound will do the trick:)

 
sever29:

Yeah, let's have a laugh together.
you don't have a clear algorithm for determining the channel, in fact the programmer would have to build that algorithm himself, for 50, nah (only this made me smile nothing whiter)
 
sanyooooook:
you don't have a clear channel detection algorithm, in fact the programmer would have to build that algorithm himself, for 50, nah (only this made me smile nothing whiter)

You're right, I admit it, that's why I started this thread, so I wouldn't have to mooch off the question - "how do I do it?"
Reason: