Other than using a long string of if statements is there a way to program an acrobat text box to incrementally calculate based on variables entered into an other text box?
Here is what I have.
var v = this.getField("FixNum").value;
if (v == "1")
{
event.value = 84 ;
}
else if (v == "2")
{
event.value = 88 ;
}
else if (v == "3")
{
event.value = 92 ;
}
else
{
event.value = "";
}
As you can see this will get cumbersome because this goes from 1 - 9 in this pattern where 9 = 116, then 10 - 100 where 10 = 135. After 100 the pattern is 495 + 6 for every unit i.e. 101 = 501. I hope someone can understand this because I can't think of another way to ask!
Several possibilities:
• If you can come up with a formula, it would be simplest.
• If the value pairs are arbitrary, you might create a lookup table (in form of an array).
• If the input numbers are (more or less) consecutive integers, you can work with a simple array, where the index is the input number, and the element with that index number is the output number.
• If it is more complicated, you would have to create a 2-dimensional array, where each element is an array of input and output number.
• If you are looking for a simple replacement of if…else if statements, you can use the switch… statement, which would work well for totally arbitrary value pairs.
Update, following the comment by OP:
According to a comment by the OP, the input numbers are consecutive, which means that the third point above would be the method of choice. For that, proceed as follows:
Create a document-level script (the name does not matter):
var myPricesArr = [0, 12, 15, 17, 23, 27, 30, 33] ;
alternate possibility:
var myPricesArr = new Array() ;
myPricesArr[0] = 0 ;
myPricesArr[1] = 12 ;
myPricesArr[2] = 15 ;
myPricesArr[3] = 17 ;
myPricesArr[4] = 23 ;
myPricesArr[5] = 27 ;
myPricesArr[6] = 30 ;
myPricesArr[7] = 33 ;
Both variants are equivalent, but the second one is easier to maintain, but requires more typing…
Create a text field named "numFixts", formatted as number, no decimals, maybe with a maximum value (to protect the form from failing).
Create a text field named "fixtsAmt, formatted as number. In the Calculate event of this field add the following JavaScript:
event.value = myPricesArr[this.getField("numFixts").value] ;
And that should do it.
Related
I have searched for this everywhere but I couldn't find it anywhere, sorry if this questions has already been posted. Please refer me to it if this already exists somewhere else, thank you :)
Ok so here are the two rules to follow, for a number 'N' return the two Integers (x and y) that:
When added together is smaller than 'N'.
When multiplied together is bigger than 'N'
Example: for N = 15, X = 4 and Y = 4, when added together make 8, but make 16 when multiplied together.
Note: X = 4 and Y = 5 would be a valid answer as well, but for this specific question, the valid solution needs to add up to the smallest value possible. (4 + 5 = 9 > 8)
Hopefully this makes sense, all I am really looking for is the logic to approach this from, not necessarily a full on solution written in code but I would still take it if anyone feels like it :)
My biggest issue here is how to know that you have the smallest two ints (optimal solution), especially when we get to bigger and bigger number?
var N = 15;
var x, y = 0;
while(keepLooping) {
if((x+y < N) && (x*y > N)) {
//values found that respects the clauses
keepLooping = false;
break;
} else {
if((x < y) || (x==y)) {
x++
} else {
y++
}
}
}
My logic here is to start with both numbers at 0 and to slowly increment each of them by 1 but also test if the clauses are respected between each increment. To me this insures that I get the smallest values because I am starting from 0.
This is how see it;
Does is work with:
x=0, y=0... False
x=1, y=0... False
x=1, y=1... False
x=2, y=1... False
...
How can I create a fixed multidimensional array in Specman/e using varibles?
And then access individual elements or whole rows?
For example in SystemVerilog I would have:
module top;
function automatic my_func();
bit [7:0] arr [4][8]; // matrix: 4 rows, 8 columns of bytes
bit [7:0] row [8]; // array : 8 elements of bytes
row = '{1, 2, 3, 4, 5, 6, 7, 8};
$display("Array:");
foreach (arr[i]) begin
arr[i] = row;
$display("row[%0d] = %p", i, row);
end
$display("\narr[2][3] = %0d", arr[2][3]);
endfunction : my_func
initial begin
my_func();
end
endmodule : top
This will produce this output:
Array:
row[0] = '{'h1, 'h2, 'h3, 'h4, 'h5, 'h6, 'h7, 'h8}
row[1] = '{'h1, 'h2, 'h3, 'h4, 'h5, 'h6, 'h7, 'h8}
row[2] = '{'h1, 'h2, 'h3, 'h4, 'h5, 'h6, 'h7, 'h8}
row[3] = '{'h1, 'h2, 'h3, 'h4, 'h5, 'h6, 'h7, 'h8}
arr[2][3] = 4
Can someone rewrite my_func() in Specman/e?
There are no fixed arrays in e. But you can define a variable of a list type, including a multi-dimensional list, such as:
var my_md_list: list of list of my_type;
It is not the same as a multi-dimensional array in other languages, in the sense that in general each inner list (being an element of the outer list) may be of a different size. But you still can achieve your purpose using it. For example, your code might be rewritten in e more or less like this:
var arr: list of list of byte;
var row: list of byte = {1;2;3;4;5;6;7;8};
for i from 0 to 3 do {
arr.add(row.copy());
print arr[i];
};
print arr[2][3];
Notice the usage of row.copy() - it ensures that each outer list element will be a copy of the original list.
If we don't use copy(), we will get a list of many pointers to the same list. This may also be legitimate, depending on the purpose of your code.
In case of a field (as opposed to a local variable), it is also possible to declare it with a given size. This size is, again, not "fixed" and can be modified at run time (by adding or removing items), but it determines the original size of the list upon creation, for example:
struct foo {
my_list[4][8]: list of list of int;
};
I am reading through perl6intro on lazy lists and it leaves me confused about certain things.
Take this example:
sub foo($x) {
$x**2
}
my $alist = (1,2, &foo ... ^ * > 100);
will give me (1 2 4 16 256), it will square the same number until it exceeds 100. I want this to give me (1 4 9 16 25 .. ), so instead of squaring the same number, to advance a number x by 1 (or another given "step"), foo x, and so on.
Is it possible to achieve this in this specific case?
Another question I have on lazy lists is the following:
In Haskell, there is a takeWhile function, does something similar exist in Perl6?
I want this to give me (1 4 9 16 25 .. )
The easiest way to get that sequence, would be:
my #a = (1..*).map(* ** 2); # using a Whatever-expression
my #a = (1..*).map(&foo); # using your `foo` function
...or if you prefer to write it in a way that resembles a Haskell/Python list comprehension:
my #a = ($_ ** 2 for 1..*); # using an in-line expression
my #a = (foo $_ for 1..*); # using your `foo` function
While it is possible to go out of one's way to express this sequence via the ... operator (as Brad Gilbert's answer and raiph's answer demonstrate), it doesn't really make sense, as the purpose of that operator is to generate sequences where each element is derived from the previous element(s) using a consistent rule.
Use the best tool for each job:
If a sequence is easiest to express iteratively (e.g. Fibonacci sequence):
Use the ... operator.
If a sequence is easiest to express as a closed formula (e.g. sequence of squares):
Use map or for.
Here is how you could write a Perl 6 equivalent of Haskell's takewhile.
sub take-while ( &condition, Iterable \sequence ){
my \iterator = sequence.iterator;
my \generator = gather loop {
my \value = iterator.pull-one;
last if value =:= IterationEnd or !condition(value);
take value;
}
# should propagate the laziness of the sequence
sequence.is-lazy
?? generator.lazy
!! generator
}
I should probably also show an implementation of dropwhile.
sub drop-while ( &condition, Iterable \sequence ){
my \iterator = sequence.iterator;
GATHER: my \generator = gather {
# drop initial values
loop {
my \value = iterator.pull-one;
# if the iterator is out of values, stop everything
last GATHER if value =:= IterationEnd;
unless condition(value) {
# need to take this so it doesn't get lost
take value;
# continue onto next loop
last;
}
}
# take everything else
loop {
my \value = iterator.pull-one;
last if value =:= IterationEnd;
take value
}
}
sequence.is-lazy
?? generator.lazy
!! generator
}
These are only just-get-it-working examples.
It could be argued that these are worth adding as methods to lists/iterables.
You could (but probably shouldn't) implement these with the sequence generator syntax.
sub take-while ( &condition, Iterable \sequence ){
my \iterator = sequence.iterator;
my \generator = { iterator.pull-one } …^ { !condition $_ }
sequence.is-lazy ?? generator.lazy !! generator
}
sub drop-while ( &condition, Iterable \sequence ){
my \end-condition = sequence.is-lazy ?? * !! { False };
my \iterator = sequence.iterator;
my $first;
loop {
$first := iterator.pull-one;
last if $first =:= IterationEnd;
last unless condition($first);
}
# I could have shoved the loop above into a do block
# and placed it where ｢$first｣ is below
$first, { iterator.pull-one } … end-condition
}
If they were added to Perl 6/Rakudo, they would likely be implemented with Iterator classes.
( I might just go and add them. )
A direct implementation of what you are asking for is something like:
do {
my $x = 0;
{ (++$x)² } …^ * > 100
}
Which can be done with state variables:
{ ( ++(state $x = 0) )² } …^ * > 100
And a state variable that isn't used outside of declaring it doesn't need a name.
( A scalar variable starts out as an undefined Any, which becomes 0 in a numeric context )
{ (++( $ ))² } …^ * > 100
{ (++$)² } …^ * > 100
If you need to initialize the anonymous state variable, you can use the defined-or operator // combined with the equal meta-operator =.
{ (++( $ //= 5))² } …^ * > 100
In some simple cases you don't have to tell the sequence generator how to calculate the next values.
In such cases the ending condition can also be simplified.
say 1,2,4 ...^ 100
# (1 2 4 8 16 32 64)
The only other time you can safely simplify the ending condition is if you know that it will stop on the value.
say 1, { $_ * 2 } ... 64;
# (1 2 4 8 16 32 64)
say 1, { $_ * 2 } ... 3;
# (1 2 4 8 16 32 64 128 256 512 ...)
I want this to give me (1 4 9 16 25 .. )
my #alist = {(++$)²} ... Inf;
say #alist[^10]; # (1 4 9 16 25 36 49 64 81 100)
The {…} is an arbitrary block of code. It is invoked for each value of a sequence when used as the LHS of the ... sequence operator.
The (…)² evaluates to the square of the expression inside the parens. (I could have written (…) ** 2 to mean the same thing.)
The ++$ returns 1, 2, 3, 4, 5, 6 … by combining a pre-increment ++ (add one) with a $ variable.
In Haskell, there is a takeWhile function, does something similar exist in Perl6?
Replace the Inf from the above sequence with the desired end condition:
my #alist = {(++$)²} ... * > 70; # stop at step that goes past 70
say #alist; # [1 4 9 16 25 36 49 64 81]
my #alist = {(++$)²} ...^ * > 70; # stop at step before step past 70
say #alist; # [1 4 9 16 25 36 49 64]
Note how the ... and ...^ variants of the sequence operator provide the two variations on the stop condition. I note in your original question you have ... ^ * > 70, not ...^ * > 70. Because the ^ in the latter is detached from the ... it has a different meaning. See Brad's comment.
I've got a csv file that contain more than 60 columns and 2 000 000 lines, I'm trying to count the number of null value per variable (per column) then to do the sum of that new row to get the number total of null value in the entire csv. For example if we got this file in input:
We expect this other file in output:
I know how to count the number of null value per line but, I didn't figure out how to count the number of null value per column.
There has to be a better way to do this, but I made a really nasty JavaScript which does the job.
It has some problems for different column types, as it doesn't set the column type. (It should set all columns to integer, but I don't know if that is possible from JavaScript.)
You have to run Identify last row in a stream first, and save it to the column last (or change the script).
var nulls;
var seen;
if (!seen) {
// Initialize array
seen = 1;
nulls = [];
for (var i = 0; i < getInputRowMeta().size(); i++) {
nulls[i] = 0;
}
}
for (var i = 0; i < getInputRowMeta().size(); i++) {
if (row[i] == null) {
nulls[i] += 1;
}
// Hack to find empty strings
else if (getInputRowMeta().getValueMeta(i).getType() == 2 && row[i].length() == 0) {
nulls[i] += 1;
}
}
// Don't store any values
trans_Status = SKIP_TRANSFORMATION;
// Only store the nulls at the last row
if (last == true) {
putRow(nulls);
}
Please drag and drop below steps in to canvas.
step1: Add constants: create one variable called constant and value = 1
step2: Filter Rows: you have filter null values of all columns.
step3: Group by: here group by field constant variable
aggregates section we have to specify remaining columns like ct_inc.And type is Number of Values (N)
If you have any doubts feel free to ask.
skype_id : panabakavenkatesh
I'm looking for a way to generate combinations of objects ordered by a single attribute. I don't think lexicographical order is what I'm looking for... I'll try to give an example. Let's say I have a list of objects A,B,C,D with the attribute values I want to order by being 3,3,2,1. This gives A3, B3, C2, D1 objects. Now I want to generate combinations of 2 objects, but they need to be ordered in a descending way:
A3 B3
A3 C2
B3 C2
A3 D1
B3 D1
C2 D1
Generating all combinations and sorting them is not acceptable because the real world scenario involves large sets and millions of combinations. (set of 40, order of 8), and I need only combinations above the certain threshold.
Actually I need count of combinations above a threshold grouped by a sum of a given attribute, but I think it is far more difficult to do - so I'd settle for developing all combinations above a threshold and counting them. If that's possible at all.
EDIT - My original question wasn't very precise... I don't actually need these combinations ordered, just thought it would help to isolate combinations above a threshold. To be more precise, in the above example, giving a threshold of 5, I'm looking for an information that the given set produces 1 combination with a sum of 6 ( A3 B3 ) and 2 with a sum of 5 ( A3 C2, B3 C2). I don't actually need the combinations themselves.
I was looking into subset-sum problem, but if I understood correctly given dynamic solution it will only give you information is there a given sum or no, not count of the sums.
Thanks
Actually, I think you do want lexicographic order, but descending rather than ascending. In addition:
It's not clear to me from your description that A, B, ... D play any role in your answer (except possibly as the container for the values).
I think your question example is simply "For each integer at least 5, up to the maximum possible total of two values, how many distinct pairs from the set {3, 3, 2, 1} have sums of that integer?"
The interesting part is the early bailout, once no possible solution can be reached (remaining achievable sums are too small).
I'll post sample code later.
Here's the sample code I promised, with a few remarks following:
public class Combos {
/* permanent state for instance */
private int values[];
private int length;
/* transient state during single "count" computation */
private int n;
private int limit;
private Tally<Integer> tally;
private int best[][]; // used for early-bail-out
private void initializeForCount(int n, int limit) {
this.n = n;
this.limit = limit;
best = new int[n+1][length+1];
for (int i = 1; i <= n; ++i) {
for (int j = 0; j <= length - i; ++j) {
best[i][j] = values[j] + best[i-1][j+1];
}
}
}
private void countAt(int left, int start, int sum) {
if (left == 0) {
tally.inc(sum);
} else {
for (
int i = start;
i <= length - left
&& limit <= sum + best[left][i]; // bail-out-check
++i
) {
countAt(left - 1, i + 1, sum + values[i]);
}
}
}
public Tally<Integer> count(int n, int limit) {
tally = new Tally<Integer>();
if (n <= length) {
initializeForCount(n, limit);
countAt(n, 0, 0);
}
return tally;
}
public Combos(int[] values) {
this.values = values;
this.length = values.length;
}
}
Preface remarks:
This uses a little helper class called Tally, that just isolates the tabulation (including initialization for never-before-seen keys). I'll put it at the end.
To keep this concise, I've taken some shortcuts that aren't good practice for "real" code:
This doesn't check for a null value array, etc.
I assume that the value array is already sorted into descending order, required for the early-bail-out technique. (Good production code would include the sorting.)
I put transient data into instance variables instead of passing them as arguments among the private methods that support count. That makes this class non-thread-safe.
Explanation:
An instance of Combos is created with the (descending ordered) array of integers to combine. The value array is set up once per instance, but multiple calls to count can be made with varying population sizes and limits.
The count method triggers a (mostly) standard recursive traversal of unique combinations of n integers from values. The limit argument gives the lower bound on sums of interest.
The countAt method examines combinations of integers from values. The left argument is how many integers remain to make up n integers in a sum, start is the position in values from which to search, and sum is the partial sum.
The early-bail-out mechanism is based on computing best, a two-dimensional array that specifies the "best" sum reachable from a given state. The value in best[n][p] is the largest sum of n values beginning in position p of the original values.
The recursion of countAt bottoms out when the correct population has been accumulated; this adds the current sum (of n values) to the tally. If countAt has not bottomed out, it sweeps the values from the start-ing position to increase the current partial sum, as long as:
enough positions remain in values to achieve the specified population, and
the best (largest) subtotal remaining is big enough to make the limit.
A sample run with your question's data:
int[] values = {3, 3, 2, 1};
Combos mine = new Combos(values);
Tally<Integer> tally = mine.count(2, 5);
for (int i = 5; i < 9; ++i) {
int n = tally.get(i);
if (0 < n) {
System.out.println("found " + tally.get(i) + " sums of " + i);
}
}
produces the results you specified:
found 2 sums of 5
found 1 sums of 6
Here's the Tally code:
public static class Tally<T> {
private Map<T,Integer> tally = new HashMap<T,Integer>();
public Tally() {/* nothing */}
public void inc(T key) {
Integer value = tally.get(key);
if (value == null) {
value = Integer.valueOf(0);
}
tally.put(key, (value + 1));
}
public int get(T key) {
Integer result = tally.get(key);
return result == null ? 0 : result;
}
public Collection<T> keys() {
return tally.keySet();
}
}
I have written a class to handle common functions for working with the binomial coefficient, which is the type of problem that your problem falls under. It performs the following tasks:
Outputs all the K-indexes in a nice format for any N choose K to a file. The K-indexes can be substituted with more descriptive strings or letters. This method makes solving this type of problem quite trivial.
Converts the K-indexes to the proper index of an entry in the sorted binomial coefficient table. This technique is much faster than older published techniques that rely on iteration. It does this by using a mathematical property inherent in Pascal's Triangle. My paper talks about this. I believe I am the first to discover and publish this technique, but I could be wrong.
Converts the index in a sorted binomial coefficient table to the corresponding K-indexes.
Uses Mark Dominus method to calculate the binomial coefficient, which is much less likely to overflow and works with larger numbers.
The class is written in .NET C# and provides a way to manage the objects related to the problem (if any) by using a generic list. The constructor of this class takes a bool value called InitTable that when true will create a generic list to hold the objects to be managed. If this value is false, then it will not create the table. The table does not need to be created in order to perform the 4 above methods. Accessor methods are provided to access the table.
There is an associated test class which shows how to use the class and its methods. It has been extensively tested with 2 cases and there are no known bugs.
To read about this class and download the code, see Tablizing The Binomial Coeffieicent.
Check out this question in stackoverflow: Algorithm to return all combinations
I also just used a the java code below to generate all permutations, but it could easily be used to generate unique combination's given an index.
public static <E> E[] permutation(E[] s, int num) {//s is the input elements array and num is the number which represents the permutation
int factorial = 1;
for(int i = 2; i < s.length; i++)
factorial *= i;//calculates the factorial of (s.length - 1)
if (num/s.length >= factorial)// Optional. if the number is not in the range of [0, s.length! - 1]
return null;
for(int i = 0; i < s.length - 1; i++){//go over the array
int tempi = (num / factorial) % (s.length - i);//calculates the next cell from the cells left (the cells in the range [i, s.length - 1])
E temp = s[i + tempi];//Temporarily saves the value of the cell needed to add to the permutation this time
for(int j = i + tempi; j > i; j--)//shift all elements to "cover" the "missing" cell
s[j] = s[j-1];
s[i] = temp;//put the chosen cell in the correct spot
factorial /= (s.length - (i + 1));//updates the factorial
}
return s;
}
I am extremely sorry (after all those clarifications in the comments) to say that I could not find an efficient solution to this problem. I tried for the past hour with no results.
The reason (I think) is that this problem is very similar to problems like the traveling salesman problem. Until unless you try all the combinations, there is no way to know which attributes will add upto the threshold.
There seems to be no clever trick that can solve this class of problems.
Still there are many optimizations that you can do to the actual code.
Try sorting the data according to the attributes. You may be able to avoid processing some values from the list when you find that a higher value cannot satisfy the threshold (so all lower values can be eliminated).
If you're using C# there is a fairly good generics library here. Note though that the generation of some permutations is not in lexicographic order
Here's a recursive approach to count the number of these subsets: We define a function count(minIndex,numElements,minSum) that returns the number of subsets of size numElements whose sum is at least minSum, containing elements with indices minIndex or greater.
As in the problem statement, we sort our elements in descending order, e.g. [3,3,2,1], and call the first index zero, and the total number of elements N. We assume all elements are nonnegative. To find all 2-subsets whose sum is at least 5, we call count(0,2,5).
Sample Code (Java):
int count(int minIndex, int numElements, int minSum)
{
int total = 0;
if (numElements == 1)
{
// just count number of elements >= minSum
for (int i = minIndex; i <= N-1; i++)
if (a[i] >= minSum) total++; else break;
}
else
{
if (minSum <= 0)
{
// any subset will do (n-choose-k of them)
if (numElements <= (N-minIndex))
total = nchoosek(N-minIndex, numElements);
}
else
{
// add element a[i] to the set, and then consider the count
// for all elements to its right
for (int i = minIndex; i <= (N-numElements); i++)
total += count(i+1, numElements-1, minSum-a[i]);
}
}
return total;
}
Btw, I've run the above with an array of 40 elements, and size-8 subsets and consistently got back results in less than a second.