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- JSL Use Lag or Dif Functions in List?

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Jun 7, 2013 11:27 AM
(3547 views)

Hi Folks,

I have a list ( list = {1, 3, 5, 7, 5, 9, 11, 13, 10, 14, 15, 17}; ) I want to modify so that it is monotonically increasing. That is, I want to evaluate the list so that any current value < previous value is set to previous value, else, use current value. In a JSL formula it looks like this:

If( Dif( :List, 1 ) < 0,

Lag( :List, 1 ),

:List

);

Is it possible to conduct Dif and Lag-like functions on items in a matrix? Thanks.

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Jun 7, 2013 1:53 PM
(5549 views)

Matrices are fun! Loops can often be avoided if taking advantage of the matrix operations in jsl.

a = **[****1**, **3**, **5**, **7**, **5**, **9**, **11**, **13**, **10**, **14**, **15**, **17****]**;

n = N Row**(** a **)**;

pos = Loc**(** a**[****1** :: n - **1****]** > a**[****2** :: n**]** **)**;

a**[**pos + **1****]** = a**[**pos**]**;

Show**(** a **)**;

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Jun 7, 2013 11:34 AM
(3097 views)

I don't believe that there are DIF and LAG functions for a matrix, but you really don't need them. You can loop over your list/matrix and reference (i-1) or (i+1), where i is your looping index.

BTW I wouldn't use **list** as a variable, as it is a function in JSL.

Or, you can just sort your matrix:

a = **[****1**, **3**, **5**, **7**, **5**, **9**, **11**, **13**, **10**, **14**, **15**, **17]**;

b = sort ascending**(**a**)**;

**[1, 3, 5, 5, 7, 9, 10, 11, 13, 14, 15, 17]**

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Jun 7, 2013 11:38 AM
(3097 views)

Thanks PMroz,

I need the list to stay in the current order, just change the current value to the previous value if it is less than previous value. What I'm looking to produce is:

a = **[****1**, **3**, **5**, **7**, **7**, **9**, **11**, **13**,** 13**, **14**, **15**, **17]**;

from

a = **[****1**, **3**, **5**, **7**, **5**, **9**, **11**, **13**, **10**, **14**, **15**, **17]**;

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Jun 7, 2013 11:44 AM
(3097 views)

Ah OK. Then you want this:

a = **[****1**, **3**, **5**, **7**, **5**, **9**, **11**, **13**, **10**, **14**, **15**, **17]**;

for **(**i = **2**, i <= nrows**(**a**)**, i++,

if **(**a**[**i**]** < a**[**i-**1]**,

a**[**i**]** = a**[**i-**1]**;

**)**;

**)**;

**[1, 3, 5, 7, 7, 9, 11, 13, 13, 14, 15, 17]**

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Jun 7, 2013 1:53 PM
(5550 views)

Matrices are fun! Loops can often be avoided if taking advantage of the matrix operations in jsl.

a = **[****1**, **3**, **5**, **7**, **5**, **9**, **11**, **13**, **10**, **14**, **15**, **17****]**;

n = N Row**(** a **)**;

pos = Loc**(** a**[****1** :: n - **1****]** > a**[****2** :: n**]** **)**;

a**[**pos + **1****]** = a**[**pos**]**;

Show**(** a **)**;

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Jun 12, 2013 3:18 PM
(3097 views)

Thanks MS and PMroz for two interesting ways to think about and play with matrices.