find the closest value in table

Similar Questions

• How to find the index of the closest value in table

  Hello

  I have a double-table A and a double value B.

  I'm looking for on what number in the table A is the closest number to number B?

  Thanks for help

  If the array is sorted, built-in 'Threshold 1 D Array' might work for you.

• Find the minimum value of the table in time...

  Firstly that it is not "find the min value in the table.

  It is:

  I have a table that changes its values at each iteration, and I want to compare the new values with previous values and make a new array with the minimum values...

  In C++, it would be like this:

  int main() {}

  int array [8];

  int min [8];

  int i;

  for (i = 0; i<>

  min [i] = 1000000; Initialize minimum table with a high value

  }

  {while (1)}

  Array = update_array ();

  for (i = 0; i<>

  If (array [i]<>

  min [i] = array [i];

  }

  }

  do things...

  }

  return 0;

  }

  IM new to labview and it always confuses me a lot...

  Thanks in advance.

  Using the function Max & Min regular (not table Max & min as you know) with tables and a good will of employment registry change.  Max Min &, as most of the digital functions, is polymorphic and works witt tables as well as scalars.

  Lynn

  

• PL/SQL loop to find the minimum value

  I want to find the minimum value from a table using the pl/sql loop

  I have data format below.

  1 ABC 2

  1 ABC 20

  1 ABC 500

  1 ABC 2000

  XYZ 2 250

  JKL 3 300

  MNP 3 350

  3 KLM 400

  I wrote the code below. His does not work. Please suggest.

  Set serveroutput on
  declare
  x number: = 3;
  y VARCHAR2 (3): = 'ABC ';
  fmin number: = 0;
  number of Fmax: = 0;
  Start
  for rec in (select sal from test
  where id = name and x = y) loop

  
  If rec.sal > fmin then
  fmin: = rec.sal;
  on the other
  FMAX: = rec.sal;
  fmin: = fmax.
  end if;
  end loop;
  dbms_output.put_line (Fmax);
  end;
  /

  I want to output 2 on top of code.

  Thanks in advance

  Why PL/SQL? SELECT MIN (SAL) OF TEST will be more rapid and much less coding. In any case:

  declare

  number of v_min_sal;

  Start

  for v_rec in a loop (select sal from emp)

  If nvl (v_min_sal, v_rec.sal + 1) > v_rec.sal

  then

  v_min_sal: = v_rec.sal;

  end if;

  end loop;

  dbms_output.put_line (' lowest salaray is "' |") v_min_sal | '"');

  end;

  /

  The lowest Salaray is "800".

  PL/SQL procedure successfully completed.

  SQL >

  SY.

• Find the average value of a buffer

  Hello

  In my VI Panel, I have a digital control, a button and a digital display.

  When the user, enter a number in numerical order and press the button, the digital display should show the average value of all the past 5 values entered in the digital control, for example:

  1st entry: 10-online average = 10 (as sum = 10)

  2nd type: 0-online average = 5 (as sum = 10 + 0 = 10)

  3rd enter: 5-online average = 5 (you get the idea...)

  4 enter: 1-average online = 4

  5 enter: 14-online medium = 6

  6 enter: 2-online average = 4.4 (sum = 0 + 5 + 1 + 14 + 2 = 22, because the 1st value is thrown)

  How to write this VI?

  Thank you!!

  You can also use arrays to hold the data in a registry change and find the average value of the last five items.

  

  Good luck

• Find the max value

  Hi team,

  I have the following query where I need to find the max value

  [code]

  with t as)

  Select 'L1' R_nm, R_Data 'Data', 'Obj' R_Obj, r_prec 'Wd', '2' double val

  Union of all the

  Select 'L1', 'Data', 'Obj', 'No', '4' double

  Union of all the

  Select 'L2', 'Data', 'Obj', 'No', '4' double)

  Select t.*, max (val) on rk (partition r_nm, r_data, r_obj) of t

  [/ code]

  My results should be

  [code]

  r_nm r_data r_obj r_prec rk val

  L1 Data Obj 1 2 Wd

  L1 data Obj 4 No.2

  L2 Data Yes 1 1 Obj

  [/ code]

  Thank you

  
  

  Hello

  It seems that you are not interested in the MAX at all.  Looks like you want to classify the lines, such as one that has the lowest value is assigned the number 1, the 2nd line with the lowest values gets 2,... and line with him gets N-th lowest value N.  If is not serious if the MAX is 4, 2 or 420.

  Here's a way to do it:

  SELECT T.*

  , RANK () OVER (PARTITION BY r_nm, r_obj)

  ORDER BY val

  ) AS rk

  T

  ORDER BY r_nm, r_obj

  val

  ;

  Depending on how you want to deal with links, you can use ROW_NUMBER or DENSE_RANK instead of RANK.

• Find the closest CMYK Pantone spot colors?

  CS6 on a MacBookPro using

  Is there a way to find the closest matching Pantone shades of CMYK, I created in Illustrator?

  I know it's easy in Photoshop using the color picker, but there must be a way to do that in Illustrator.

  Help?

  Use Photoshop in tandem to do this.  You first need to know what book of Pantone color you plan to use.  As far as I know it is usually solid uncoated.

  1. Open your file IN Photoshop.
  2. Make sure you only choose CMYK as color mode (this is very important).
  3. Take the color eyedropper ( I )
  4. Click the first color
  5. Then click the shade 'picked' in your toolbar
  6. In the Picker dialog, choose "color libraries".
  7. You can now navigate PANTONE Solid not clothed with the top down in the next window
  8. At this point, Photoshop will match your selection with a color of the library of the PANTONE colors you have chosen.  Write it down.
  9. Jump to Illustrator.
  10. Choose a shape with the same color initially selected in Photoshop.
  11. In the window of your color chart, click on the small icon in the lower left (it says 'Menu libraries of nuances' bubble)
  12. Place the cursor of color books, and then click "Pantone Solid non-couche.
  13. Wow! Box of the Rainbow! But it's messy, so let's clean it up.
  14. In the palette options drown down (it's the icon at the top right that looks like an arrow with a bunch of lines), choose "sort by name".
  15. Now you can easily select the PANTONE color Photoshop guessed for you.
  16. The selected part of the work updates automatically to the PANTONE color, you clicked on on.j

  It is MUCH simpler than it seems.  I just explain like you MIKE.

• How to find the ASCII value of the TAB?

  Hi friends,
  
  I'm trying to find the ASCII value of the TAB character using ASCII standard as:
  
  SELECT ASCII ('< TAB >') FROM DUAL;
  
  I'm running the command above in SQL * more.
  
  Is not treated as TAB.
  It is considered space and give the ASCII value of the space - 32.
  
  Please let me know how?
  
  Version of Database - 9i
  
  Concerning
  Anthony Alix.

  Its only editor that gives you the problem; Try it on TOAD or sql developer, it works perfectly.

  Is there a particular reason you want in sql * more?

• How to find the closest higher value?

  Hi all
  
  I want to select the closest higher value in my data (test data are below):
  

  with test_data as
  ( select 0.21 as val1, 0.19 as val2 from dual
    union all
    select 0.21 as val1, 0.29 as val2 from dual
    union all
    select 0.21 as val1, 0.49 as val2 from dual
    union all
    select 0.67 as val1, 0.49 as val2 from dual
    union all
    select 0.67 as val1, 0.69 as val2 from dual
  )
  select * from 
  (
  select 
  val1,
  val2,
  rank() over (order by abs(val2 - val1)) nearest
  from test_data
  )
  where nearest = 1;

  output:

  VAL1    VAL2    NEAREST
  0.21     0.19     1
  0.67     0.69     1

  Output of this query is the NEAREST value to the value in the column of VAL1. But I would get the SUPERIOR value CLOSEST. As below the value at the higher value 0.21 closest isn't 0.29 0.19.
  
  desired output:

  VAL1    VAL2
  0.21     0.29
  0.67     0.69

  Is it possible to get the features you want using analytical functions?
  
  Thx for the help!
  
  Kind regards
  
  Jiri
  
  Published by: Jiri n. 10 October 2012 01:14

  with test_data as
  ( select 0.21 as val1, 0.19 as val2 from dual
    union all
    select 0.21 as val1, 0.29 as val2 from dual
    union all
    select 0.21 as val1, 0.49 as val2 from dual
    union all
    select 0.67 as val1, 0.49 as val2 from dual
    union all
    select 0.67 as val1, 0.69 as val2 from dual
  )
  
  select
   val1
  ,val2
  ,first_value(val2) over (partition by val1 order by
     case when val2 >= val1 then 1 else 2 end
     , val2
   ) nearest
  from test_data
  
  VAL1 VAL2 NEAREST
  .21 .29 .29
  .21 .49 .29
  .21 .19 .29
  .67 .69 .69
  .67 .49 .69 
  

• How to compare the 3 values in table

  Hi all

  This is my first post so be tolerant

  I need to find "bad values" in the table, it will be easier to explain with example:

  I have a table like this: 0 0 0 0 0 0 0 0 5 0 0 0 0<- so="" "5"="" is="" the="" bad="" value,="" my="" method="" is="" to="" check="" two="" of="" nearby="" values="" so="" x(i-1)="0," x(i)="5" x(i+1)="0," and="" if="" checked="" value="" is="" more="" than="" 25%="" bigger="" or="" smaller="" it="" should="" be="" replaced="" by="" average="" of="">

  example: 0 1 2 3 4 9 6 7 8<- bad="" value="" is="" 9,="" and="" result="" of="" my="" filter="" should="" be="" (4+6)/2="">

  I did it using the formula node (not 20% and no way, but the point is the same) and it works but now I want to do without C and I had a lot of problems because of the size of the array memory is 10000.

  I joined thah vi and signal, you can test that it works very well

  Thank you

  Mike

  MeeHow wrote:

  ... What is the problem?

  Because the first incorrect value is the point of 2780, because it is also disabled because the average of neighbour is biased by the incorrect value to 2781.

  You could jut blindly replace all values with the value that has the lowest absolute value among the three points that we have. Here's a quick project, seems pretty good, but it really depends what you want.

• How can I find the subset of a table that corresponds to a set of conditions/conditions?

  I have often encountered this problem in LabVIEW, probably the result of my Matlab based thinking on the tables. Suppose I have a 1 d array that consists of a real number between 0 and 2 (for example):

  1.3 0.1 1.9 0.7 1.4 1.5 1.5 0.8 1.3 0.3 1.4 1.4 0.1 0.6 0.1 0.2 1.6 1.6 1.4 0.6 0.1 1.9

  and I want to know the average of the subset of values greater than 0.5 and less than 1.2:

  average ([0.8 0.7 0.6 0.6]) = 0.6750.

  How can I find the subset of the array that satisfies the condition > 0.5 &<1.2>

  For your example (values between 0.5 and 2) If you wish:

  

  I would do something like this:

  

  I have generated just a table of 10 random numbers between 0.0 and 2.0

  The following loop sorts the data, using "range and force."
  It then displays the data in the selected range.

  The last step is to take the average

  I have attached the VI if you do not want to use the code.

• Find the same value in different columns

  Hello

  I'm trying to figure out how I would get the same value in the same row but in different columns. For example, in the table sometimes report the name is the same as the description. I would like to find all the lines where this is the case.

  Thank you

  Name of the report	Description	Date	Time
  Report 1	Report 1	Delalande	TTTT
  Report 2	Billing report	Delalande	TTTT
  Report 3	Report 3	Delalande	TTTT

  Hello

  Is that what you want?

  SELECT *- or whatever the columns that you want to

  FROM table_x

  WHERE description = report_name

  ;

  If not, post a small example data (CREATE TABLE and only relevant columns, INSERT statements) for all of the tables involved and also publish outcomes from these data.
  Explain, using specific examples, how you get these results from these data.
  Always say what version of Oracle you are using (for example, 11.2.0.2.0).

  See the FAQ forum: https://forums.oracle.com/message/9362002

• Take the minimum value for table data

  I'm storing test data in a table, with the hope that I can output the minimum value of the data. So far, I have used the table of max/min but can't make it work how I want. The test runs continuously in a loop of the holder. I get the minimum values in the table, but I would like to change the number of samples that the table contains, before it generates a minimum. The other option is that I run the test for a period of time (e.g. 3 seconds) and the table stores the data for this time then output 1 minimum value for the entire period.

  My current configuration VI is attached.

  Very appreciated

  Chris

  The size of the array in your VI is determined by the number of samples that you collect in your DAQ assistant, who is now 100 samples.  Never more, never less.

  How many samples do you want before you determine the minimum?

  Maybe what you really want to do is to build the table in the while loop and store it in a shift register, and when the loop has repeated enough times, then you take the minimum of built table.

• How to compare the original value of table size and the changed value

  juice I took a table and then took the function of the size of the array so that it shows me the number of the elements present in it. so it'll be the original table size value. If the items in the table even changes another value, then I want to compare the original table size value and the value of table size has changed. How to compare... Please help me. you are looking for a possible solution. Thank you

  Hi stara,

  the attached picture shows the ony solution.

  It will be useful.

  Mike

• Getting the closest value output

  I have the table as below

  with t as)

  Select 1 n double union

  Select 2.71 double Union

  Select 2.72 double Union

  Select 3 from dual union

  Select 4 double)

  Select * from t

  I have the input as 2 value. As there is no such thing as value 2 in my table, I ve to choose the nearest value max exists in my table.

  From the above table, I need to get the result as 2.71 only.

  Could you please suggest the same thing.

  Thank you.

  variable number of val

  exec: val: = 2;

  with t as)

  Select 1 n double union

  Select 2.71 double Union

  Select 2.72 double Union

  Select 3 from dual union

  Select 4 double

  ),

  x as)

  Select n,

  Rank() over (order of abs (n -: val)) rn

  t

  )

  Select n

  x

  where rn = 1

  /

  N
  ----------
  2.71

  SQL >

  SY.