I have two fields
- amount (decimal (11, 2))
- gift_amount (decimal (11, 2))
When I do an update on either for a value equal to or below 999.99, it saves correctly.
However, if I go over that, then it drops the value right back to down 1 - 10.
Is this a known issue or am I going wrong using decimal?
Heres some PHP code of what I'm doing just to make it clearer (although I'm 100% its not the PHP's fault.
if ($total_balance >= $cost) {
if ($this->user->balance->gift_amount > 0) {
$total_to_be_paid = number_format($cost, 2) - number_format($this->user->balance->gift_amount, 2);//figure out how much is left after the gift total
$this->user->balance->gift_amount -= number_format($cost, 2); //deduct from the gift balance
$this->user->balance->gift_amount = (number_format($this->user->balance->gift_amount, 2) < 0) ? number_format(00.00, 2) : number_format($this->user->balance->gift_amount, 2); //if the gift balance went below 0, lets set it to 0
if ($total_to_be_paid > 0) {
$this->user->balance->amount = number_format($this->user->balance->amount, 2) - number_format($total_to_be_paid, 2);
}
} else {
$this->user->balance->amount = number_format($this->user->balance->amount, 2) - number_format($cost, 2);
}
if ($object = Model_ClipBought::create(array('clip_id' => $clip->id, 'user_id' => $this->user->id, 'currency_name' => $user_currency, 'cost' => $cost, 'downloads' => $clip->downloads, 'expires' => time() + ($clip->expires * 86400)))) {
$this->user->balance->save();
$download = new Model_Download(ROOT_PATH."/public/files/Clip/$clip->file_url");
$download->execute();
} else {
throw new exception('We could not finish the purchase, this has been reported, sorry for the inconvenience.');
}
} else {
throw new exception('You dont have enough money in your account todo this');
}
exit;
}
You should not be using number_format until it's time to actually output/display the number to the user. It inserts your system's default thousand seperator (a comma by default):
number_format(999.99, 2) -> 999.99
number_format(1234.56, 2) -> 1,234.56
If you use these values in subsequent calculations in PHP, or try to insert verbatim into MySQL, you'll get funky values:
2345.67 + 1.0 = 2346.67
but using number_format() gives you this parsing sequence:
number_format(2345.67) + 1.0 -> "2,345.67" + 1.0
"2,345.67" + 1.0 -> "2" + 1.0
2 + 1.0 -> 3
Notice how "2,345.67" was truncated down to just "2" - the comma turns your nice number into a string, and now you're bound by string->integer parsing rules, which drops everything after the first non-numeric character in the string.
If you're trying to keep everything down to 2 decimal places throughout the calculation sequence, consider using sprintf('%0.2f', $value) instead.
Related
I have a list of numbers, some of them have leading underscores, some of them don't.
A
B
_12
34
99
_42
Which is the best way of adding up these numbers?
Note: I tried this custom script formula which for some reason doesn"t work (only returns the first item passed in the range), and anyway I guess there should be an easier way just using native GoogleSheet formulas.
function sum_with_underscores(underscored_nums) {
let nums = underscored_nums.map( x => String(x).replace("_", ""))
return nums.reduce((pv, cv) => parseFloat(pv) + parseFloat(cv), 0);
}
=SUM(ARRAYFORMULA(VALUE(SUBSTITUTE(<your_range>, "_", ""))))
Assuming you would like to achieve this result using a script instead of a formula, the code below worked for me.
Make sure to flatten the array before manipulating the data. Hope this helps.
let values = sheet.getDataRange().getValues() // returns array of arrays
let nums = values.flat() // converts [[_12, 34.0], [99.0, _42]] to [_12, 34.0, 99.0, _42]
nums = nums.map( n => {
return typeof(n) === 'number' ? n : parseFloat(n.replace("_", ""));
})
let sum = nums.reduce((acc, num)=> acc + num, 0)
}
I'm trying to figure out what my options are here when I need to use a column number in a formula, and if I really need to write a column number to column letter method to accomplish what I'm trying to do.
See this method I have here:
createFormulas(lookupField, lookupColumns) {
// Iterate through the lookupColumn array
lookupColumns.forEach(value => {
let columnNumber = this.getColumn(this.headers, value);
let range = this.sheet.getRange(2, columnNumber, this.lastRow - 1, 1);
// range.setFormula('=$A2');
range.setFormula('=' + columnNumber + '2' ); // doesn't work obviously
})
}
I'm trying to add formulas in a column based on the column.
this.getColumn() returns the column number based on the column name being passed in.
let range sets the range I want to set the formula in
range.setFormula('=$A2') pastes this formula into range and updates the reference accordingly (i.e., $A3, $A4, etc.). This isn't the formula I ultimately want to use, just a simplified example.
I need to set the column in the reference dynamically, however.
What I have obviously won't work: range.setFormula('=' + columnNumber + '2' );. That would just result in something like 72 where 7 is the column number.
I know I can write a method that will convert the column number into a letter. I'm just surprised there isn't a built in method for doing that or some other native way of accomplishing this.
For example, in Excel VBA I think you can do something like "=" & Cells(2, columnNumber).Address or something like that (been a while, I could be wrong), which should equate to =A2, =A3, =A4, etc. in the range.
So before writing this column number to letter method, I just wanted to check: is that the only way to accomplish what I'm after or is there a native way of handling this that I'm just not seeing?
Actually, was able to do this using .getA1Notation().
Refactored to the following and it works as expected:
createFormulas(lookupField, lookupColumns) {
// Iterate through the lookupColumn array
lookupColumns.forEach(value => {
let columnNumber = this.getColumn(this.headers, value);
let formulaRange = this.sheet.getRange(2, columnNumber, this.lastRow - 1, 1);
let referenceRange = this.sheet.getRange(2, this.idColumn, this.lastRow - 1, 1);
formulaRange.setFormula("=" + referenceRange.getCell(1, 1).getA1Notation());
})
}
Column To Letters
I followed Yuri's path to the numbers to letter functions and I'm a bit baffled that we have forgotten that there are 26 letters in the alphabet and so after looking at the various functions at that reference none of them seem to have worked for me. So here's my replacement:
function colToletters(num) {
let a = " ABCDEFGHIJKLMNOPQRSTUVWXYZ";
if (num < 27) return a[num % a.length];
if (num > 26) {
num--;
let letters = '';
while (num >= 0) {
letters = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'[num % 26] + letters;
num = Math.floor(num / 26) - 1;
}
return letters;
}
}
This will calculate the column letters for 1 to 1000 and I check all the way to 703 where the letters go to AAA and they look good all the way.
Just in case. Based on https://stackoverflow.com/a/64456745/14265469
function numberToLetters(num) {
// num--; // if you need 1 --> A, 2 --> B, 26 --> Z
let letters = '';
while (num >= 0) {
letters = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'[num % 26] + letters;
num = (num - num % 26) / 26 - 1;
}
return letters;
}
console.log(numberToLetters(0)); // --> A
console.log(numberToLetters(25)); // --> Z
console.log(numberToLetters(26)); // --> AA
I managed to do it, some other way.
but I have a question, I had this code before
def jumphunt(start, mylist, count = 0):
if count < len(mylist):
place = mylist[start]
print(place)
if place == 0:
return True
elif start >= len(mylist) or start < 0:
return False
move_left = (start - place)
move_right = (start + place)
return jumphunt(move_right, mylist, count+1) or jumphunt(move_left, mylist, count+1)
else:
return False
but for some reason it's not trying both ways
to get to the last item on the list.
for example: [1,2,2,3,4,5,3,2,1,7,0] and ,start=mylist[0]
it supposed to jump like this (from 1-2-4-1-left to 2-left to 5-right to 0)
but it keeps trying to go right and then index is out of range etc.
I thought that if u use return of or this or that, it will try both until it reaches True, why won't it work here?
Thanks!
Include the value you want to keep as a default parameter for the method, like this:
def my_func(int, list, i=0):
a = (i + int)
if int == 0:
return True
elif a > len(list):
i -= int
else:
i += int
int = list[i]
my_func(int, list, i)
Bear in mind that it may not even always be possible to arrive at the end of the list doing the jumping pattern you describe, and even if it is possible, this method may choose the wrong branch.
A better algorithm would look like this:
def branching_search(list, start):
marks = [0]*len(list)
pos = start
while list[pos]!=0:
marks[pos]++
if marks[pos] % 2 == 0 and pos + list[pos] < len(list):
pos += list[pos]
elif marks[pos] % 2 == 1 and pos - list[pos] >= 0:
pos -= list[pos]
else:
return False
if all(item == 0 or item > 1 for item in list)
return False
return True
This way, if it comes to an item that it has already visited, it will decide to go the opposite direction that it went last time. Also, if it comes to an item that it can't leave without going out-of-bounds, or if there is not way to get to the end, it will give up and return.
EDIT: I realized there are a number of flaws in this algorithm! Although it is better than the first approach, it is not guaranteed to work, although the reasons are somewhat complicated.
Just imagine this array (the unimportant elements are left blank):
1, 2, , 5, , , , , 5, 0
The first two elements would get only one mark (thus the loop checking condition would not work), but it would still get stuck looping between the two fives.
Here is a method that will always work:
def flood_search(list):
marks = [[]]*len(list)
marks[0] = [0]
still_moving = True
while still_moving:
still_moving = False
for pos in range(0,len(list)):
if marks[pos]:
if pos + list[pos] < len(list) and not marks[pos + list[pos]]:
marks[pos + list[pos]] = marks[pos] + [list[pos]];
pos += list[pos]
still_moving = True
if pos - list[pos] >= 0 and not marks[pos - list[pos]]:
marks[pos - list[pos]] = marks[pos] + [-list[pos]];
pos -= list[pos]
still_moving = True
return marks[-1]
This works by taking every possible branch at the same time.
You can also use the method to get the actual route taken to get to the end. It can still be used as a condition, since it returns an empty list if no path is found (a falsy value), or a list containing the path if a path is found (a truthy value).
However, you can always just use list[-1] to get the last item.
This question relates to an animated map template which we have developed at the UKs Office for National Statistics. It has been applied to many datasets and geographies many uses without problem. For example,
http://www.ons.gov.uk/ons/interactive/vp3-census-map/index.html
http://www.statistica.md/pageview.php?l=ro&idc=390&id=3807
The .fla calls on a supporting .as file (see below) to introduce a thousand separator (in the UK a comma, in Germany a full stop (period) defined elsewhwere.
However, the dataset I am currently mapping has large negative values, and it tutrns out that the ORIGINAL HELPER FUNCTION below does not like negative values with 3, 6, 9 or 12 (etc) digits.
-100 to -999 for instance are rendered NaN,100 to NaN,999.
This is because such values are recognised as being 4 digits long. They are being split, the comma introduced, and the -ve sign is misunderstood.
I reckon the approach must be to use absolute values, add in the comma and then (for the negative values) add the -ve sign back in afterwards. But so far, trials of the ADAPTED HELPER FUNCTION have produced only error. :-(
Can anyone tell me how to put the -ve sign back in , please?
Many thanks.
Bruce Mitchell
==================================================================================
//ORIGINAL HELPER FUNCTION: ACCEPTS A NUMBER AND RETURNS A STRING WITH THOUSANDS SEPARATOR ATTACHED IF NECESSARY
function addThouSep(num) {
/*
a. Acquire the number - 'myTrendValue' or 'myDataValue' - from function calcValues
b. Record it (still as a number) to data precision.
1. Turn dataORtrend into a string
2. See if there is a decimal in it.
3. If there isn't, just run the normal addThouSep.
4. If there is, run addThouSep just on the first bit of the string - then add the decimal back on again at the end.
*/
var myNum:Number = correctFPE(num); // Create number variable myNum and populate it with 'num'
// (myTrendvalue or myData Value from calcValues function) passed thru 'correctPFE'
var strNum:String = myNum+""; // Create string version of the dataORtrend number - so instead of 63, you get '63'
var myArray = strNum.split("."); // Create array representing elements of strNum, split by decimal point.
//trace(myArray.length); // How long is the array?
if (myArray.length==1) { // Integer, no decimal.
if (strNum.length < 4)//999 doesn't need a comma.
return strNum;
return addThouSep(strNum.slice(0, -3))+xmlData.thouSep+strNum.slice(-3);
}
else { // Float, with decimal
if (myArray[0].length < 4)//999 doesn't need a comma
return strNum;
return (addThouSep(myArray[0].slice(0, -3))+xmlData.thouSep+myArray[0].slice(-3)+"."+myArray[1]);
}
}
==================================================================================
//ADAPTED HELPER FUNCTION: ACCEPTS A NUMBER AND RETURNS A STRING WITH THOUSANDS SEPARATOR ATTACHED IF NECESSARY
function addThouSep(num) {
/*
a. Acquire the number - 'myTrendValue' or 'myDataValue' - from function calcValues
b. Record it (still as a number) to data precision.
1. Turn dataORtrend into a string
2. See if there is a decimal in it.
3. If there isn't, just run the normal addThouSep.
4. If there is, run addThouSep just on the first bit of the string - then add the decimal back on again at the end.
*/
var myNum:Number = correctFPE(num); // Create number variable myNum and populate it with 'num'
// (myTrendvalue or myData Value from calcValues function) passed thru 'correctPFE'
var myAbsNum:Number = Math.abs(myNum); // ABSOLUTE value of myNum
var strNum:String = myAbsNum+""; // Create string version of the dataORtrend number - so instead of 63, you get '63'
var myArray = strNum.split("."); // Create array representing elements of strNum, split by decimal point.
//trace(myArray.length); // How long is the array?
if (myNum <0){ // negatives
if (myArray.length==1) { // Integer, no decimal.
if (strNum.length < 4)//999 doesn't need a comma.
return strNum;
return addThouSep(strNum.slice(0, -3))+xmlData.thouSep+strNum.slice(-3);
}
else { // Float, with decimal
if (myArray[0].length < 4)//999 doesn't need a comma
return strNum;
return (addThouSep(myArray[0].slice(0, -3))+xmlData.thouSep+myArray[0].slice(-3)+"."+myArray[1]);
}
}
else // positive
if (myArray.length==1) { // Integer, no decimal.
if (strNum.length < 4)//999 doesn't need a comma.
return strNum;
return addThouSep(strNum.slice(0, -3))+xmlData.thouSep+strNum.slice(-3);
}
else { // Float, with decimal
if (myArray[0].length < 4)//999 doesn't need a comma
return strNum;
return (addThouSep(myArray[0].slice(0, -3))+xmlData.thouSep+myArray[0].slice(-3)+"."+myArray[1]);
}
}
==================================================================================
If you're adding commas often (or need to support numbers with decimals) then you may want a highly optimized utility function and go with straightforward string manipulation:
public static function commaify( input:Number ):String
{
var split:Array = input.toString().split( '.' ),
front:String = split[0],
back:String = ( split.length > 1 ) ? "." + split[1] : null,
pos:int = input < 0 ? 2 : 1,
commas:int = Math.floor( (front.length - pos) / 3 ),
i:int = 1;
for ( ; i <= commas; i++ )
{
pos = front.length - (3 * i + i - 1);
front = front.slice( 0, pos ) + "," + front.slice( pos );
}
if ( back )
return front + back;
else
return front;
}
While less elegant it's stable and performant — you can find a comparison suite at my answer of a similar question https://stackoverflow.com/a/13410560/934195
Why not use something simple like this function I've made?
function numberFormat(input:Number):String
{
var base:String = input.toString();
base = base.split("").reverse().join("");
base = base.replace(/\d{3}(?=\d)/g, "$&,");
return base.split("").reverse().join("");
}
Tests:
trace( numberFormat(-100) ); // -100
trace( numberFormat(5000) ); // 5,000
trace( numberFormat(-85600) ); // -85,600
Explanation:
Convert the input number to a string.
Reverse it.
Use .replace() to find all occurrences of three numbers followed by another number. We use $&, as the replacement, which basically means take all of those occurences and replace it with the value we found, plus a comma.
Reverse the string again and return it.
Did you try using the built in Number formatting options that support localized number values:
Localized Formatting with NumberFormatter
In this example I'm trying to create an Array of length 5 where each ellement contains the number of times .3 can be summed without exceeding 1. i.e. 3 times. So each element should contain the number 3. Here is my code:
Array[(
workingCount = 0;
workingSum = 0;
done = false;
While[! done,
workingSum = workingSum + .3;
If[workingSum > 1, done = true; workingCount, workingCount++]
])
, 5]
In the 3rd to last line there I have workingCount without a ; after it because it seems like in Mathematica omitting the ; causes the value a statement resolves to to be returned.
Instead I get this:
{Null[1], Null[2], Null[3], Null[4], Null[5]}
Why does this happen? How can I get my program to do what I want it to do? i.e. In the context of the function passed to Array to initialize it's elements, how to I use complicated multi-line functions?
Thanks in advance.
Two things:
First, one way to be able to do that in Mathematica is
Array[
Catch[
workingCount = 0;
workingSum = 0;
done = False;
While[! done,
workingSum = workingSum + .3;
If[workingSum > 1,
done = True; Throw#workingCount,
workingCount++]]] &,
5]
Second, and most important: you never should do that in Mathematica! Really.
Please visit for example the Stack Exchange site for Mathematica, and read the questions an answers there to get some grip on the programming style.
Your problem comes from the fact that you are trying to initialize your array, but are trying to do so without an explicit function call - which is what you need to do.
See here for documentation on Arrays in Mathematica:
http://reference.wolfram.com/mathematica/ref/Array.html
That aside, and minor errors (True and False have to be capitalized), this is what you want to do:
f[x_] :=
(
workingCount = 0;
workingSum = 0;
done = False;
While[done != True, workingSum = workingSum + 0.3;
If[workingSum > 1, done = True, workingCount++]
];
Return[workingCount];
);
Array[f, 5] (* The array here is generating 5 values of the return value of f[x_] *)