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I have a price series and I'd like to know the indices where there has been a change of x bips. I worked out a very ugly way to accomplish this in a loop e.g.

q)bips:200
q)level:0.001*bips / 0.2
q)price: 1.0 1.1 1.3 1.8 1.9 2.0 2.3
q)ix:0
q)lastix:0
q)result:enlist lastix
q)do[count price;if[abs(price[ix]-price[lastix])>level;result,:ix;lastix:ix];ix:ix+1];
q)result
  0 2 3 6

This is a simple O(n) algo that walks through the price series and keeps a marked index (lastix) starts from the first element until it finds a price whose difference is greater than bips when found saves that index and updates lastix with the one found ... is there a more idiomatic way to do it?

My if condition inside the loop is somewhat flawed don't know exactly why if I check abs(price[lastix]-price[ix]) instead of abs(price[ix]-price[lastix]) it doesn't give correct results.

UPDATE: I was aware of deltasbut it compares consecutive elements only and that's not what I need in my OP. I apologize if the price series example in the OP was ambiguous and lead to correct results by simply using deltas. Here I have a counter example new prices series:

q)price: 1.0 1.1 1.21 1.42 1.4 1.32 1.63
q)where abs deltas price > level
,0

and this is not correct. The correct result which is produced by the accepted answer is still 

0 2 3 6
Thomas Smyth - Treliant
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SkyWalker
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4 Answers4

3

I'm not sure if this is exactly what you're looking for but deltas will give you the change between consectutive pairs:

q)deltas price
1 0.1 0.2 0.5 0.1 0.1 0.3

Checking for your condition returns a boolean list:

q)level<=deltas price
1011001b

Finally 'where' will return the indices:

q)where level<=deltas price
0 2 3 6

Thanks, Jamie

jomahony
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level:0.001*bips:200;
result:where level<=abs deltas price:1.0 1.1 1.3 1.8 1.9 2.0 2.3;

result
0 2 3 6

Is this close to what you're looking for?

Deltas checks the difference between the current and next value, abs will take the absolute value, and then you're comparing each difference against "level", which you have predefine, using where to find the associated indice.

Matthew McCann
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3

I think you're looking for something like this maybe:

f:{where differ{$[level<abs[y-x];y;x]}\[x]}

this carries forward the last value that satisfied your condition and uses if for comparison with the scan adverb, and then uses differ to pick out where the condition was satisfied and values were updated.

If I've understood your problem correctly, the same result should come from

newprice:1 1.1 1.3 1.8 1.9 2 2.1

since the final value is more than 0.2 greater than 1.8, the last value at which the level was updated.

q)f newprice
0 2 3 6

Thanks, Ryan

Ryan McCarron
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1

You've included index 0 in your answer but if you want to exclude it you can use the two argument form of deltas:

q)where level<=abs deltas[price 0;price]
2 3 6

Where the first argument sets the initial value to take away, in this case the first element of the price list.

An example of where this may be beneficial is if you were running the function for each date in a partitioned db you could pass in the last value from the previous day to ensure you didn't get the indices where there wasn't a significant difference of bips.

Thomas Smyth - Treliant
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