I have two separate scripts in google sheets that are triggered by "TRUE" value checkboxes. The first script is set to uncheck all checkboxes from the sheet and the second is resetting a dropdown list. I'll post the code below for the two scripts. Ultimately I would like both conditions to run on one checkbox. If the code can be written in a more simplified manor that would not be a bad thing.
Best regards, Jon
Script 1:
function onEdit(e) {
if (e.range.getSheet().getName() == 'Audits') {
if (e.range.getA1Notation() == 'J27' && e.value == "TRUE") {
e.range.setValue("FALSE");
var ss = SpreadsheetApp.getActive();
var sh = ss.getActiveSheet();
var rg = sh.getDataRange();
var vA = rg.getDataValidations();
var cbA = [];
for (var i = 0; i < vA.length; i++) {
for (var j = 0; j < vA[i].length; j++) {
var rule = vA[i][j];
if (rule != null) {
var criteria = rule.getCriteriaType();
if (criteria == SpreadsheetApp.DataValidationCriteria.CHECKBOX) {
sh.getRange(i + 1, j + 1).setValue(null)
}
}
}
}
}
}
}
Script 2
function onEdit(e) {
if (e.range.getSheet().getName() == 'Audits') {
if (e.range.getA1Notation() == 'E27' && e.value == "TRUE") {
e.range.setValue("FALSE");
var ss = SpreadsheetApp.getActive();
var sh = ss.getActiveSheet();
var rg = sh.getDataRange();
var vA = rg.getDataValidations();
var cbA = [];
for (var i = 0; i < vA.length; i++) {
for (var j = 0; j < vA[i].length; j++) {
var rule = vA[i][j];
if (rule != null) {
var criteria = rule.getCriteriaType();
if (criteria == SpreadsheetApp.DataValidationCriteria.VALUE_IN_LIST) {
sh.getRange(i + 1, j + 1).setValue(null)
}
}
}
}
}
}
}
To combine and simplify your script, I extracted the variable declarations, which are common to both scripts, outside of the if statements, and then directly added the conditional statement for the criteria of the second script to the first criteria using an or statement (||) as shown below:
function onEdit(e) {
var ss = SpreadsheetApp.getActive();
var sh = ss.getActiveSheet();
var rg = sh.getDataRange();
var vA = rg.getDataValidations();
if (e.range.getSheet().getName() == 'Audits') {
if (e.range.getA1Notation() == 'E27' && e.value == "TRUE") {
e.range.setValue("FALSE");
for (var i = 0; i < vA.length; i++) {
for (var j = 0; j < vA[i].length; j++) {
var rule = vA[i][j];
if (rule != null) {
var criteria = rule.getCriteriaType();
if (criteria == SpreadsheetApp.DataValidationCriteria.CHECKBOX || criteria == SpreadsheetApp.DataValidationCriteria.VALUE_IN_LIST) {
sh.getRange(i + 1, j + 1).setValue(null)
}
}
}
}
}
}
}
To test the script's functionality, I created the following setup:
When the checkbox in cell E27 is toggled, all of the sample dropdown lists and checkboxes are reset.
I hope this helps.
I'm working on project euler problem number eight, in which ive been supplied this ridiculously large number:
7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450
and am supposed to "Find the thirteen adjacent digits in the 1000-digit number that have the greatest product." EG the product of the first four adjacent digits is 7 * 3 * 1 * 6.
My code is the following:
int main()
{
string num = /* ridiculously large number omitted */;
int greatestProduct = 0;
int product;
for (int i=0; i < num.length() -12; i++)
{
product = ((int) num[i] - 48);
for (int j=i+1; j<i+13; j++)
{
product = product * ((int) num[j] - 48);
if (greatestProduct <= product)
{
greatestProduct = product;
}
}
}
cout << greatestProduct << endl;
}
I keep getting 2091059712 as the answer which project euler informs me is wrong and I suspect its too large anyway. Any help would be appreciated.
EDIT: changed to unsigned long int and it worked. Thanks everyone!
In fact your solution is too small rather than too big. The answer is what was pointed out in the comments, that there is integer overflow, and the clue is in the fact that your solution is close to the largest possible value for an signed int: 2147483647. You need to use a different type to store the product.
Note that the answer below is still 'correct' in that your code does do this wrong, but it's not what is causing the wrong value. Try taking your (working) code to http://codereview.stackexchange.com if you would like the folks there to tell you what you could improve in your approach and your coding style.
Previous answer
You are checking for a new greatest product inside the inner loop instead of outside. This means that your maximum includes all strings of less or equal ton 13 digits, rather than only exactly 13.
This could make a difference if you are finding a string which has fewer than 13 digits which have a large product, but a 0 at either end. You shouldn't count this as the largest, but your code does. (I haven't checked if this does actually happen.)
for (int i=0; i < num.length() -12; i++)
{
product = ((int) num[i] - 48);
for (int j=i+1; j<i+13; j++)
{
product = product * ((int) num[j] - 48);
}
if (greatestProduct <= product)
{
greatestProduct = product;
}
}
9^13 ≈ 2.54e12 (maximal possible value, needs 42 bit to be represented exactly), which doesn't fit into signed int. You should use int64.
If you don't want to mess with BigNum libraries, you could just take logarithms of your digits (rejecting 0) and add them up. It amounts to the same comparison.
A faster way without internal loop, but works only where there isn't a 0 in the input:
long long greatest(string num)
{
if (num.length() < 13)
return 0;
// Find a product of first 13 numbers.
long long product = 1;
unsigned int i;
for (i=0; i<13; ++i) {
product *= (num[i]-'0');
}
long long greatest_product = product;
// move through the large number
for (i=0; i+13<num.length(); ++i) {
product = product/(num[i]-'0')*(num[i+13]-'0');
if (greatest_product < product)
greatest_product = product;
}
return greatest_product;
}
In order to make this work with inputs containing 0, split the input string into substrings:
int main()
{
string num = /* input value*/;
long long greatest_product = 0;
size_t start = -1;
// Iterate over substrings without zero
do {
++start;
size_t end = num.find('0', start);
long long product = greatest(num.substr(start, end-start));
if (greatest_product < product)
greatest_product = product;
start = end;
} while (start != string::npos);
cout << greatest_product << endl;
}
My functional programming solution
def product(number):
product_result = 1
for n in number:
product_result *= int(n)
return product_result
length = 13
indices = range(0, len(number) - 13 + 1)
values = indices
values = map(lambda index: {"index": index, "number": number}, values)
values = map(lambda value: value["number"][value["index"]:value["index"] + length], values)
values = map(product, values)
values = max(values)
print values
public static void main(String[] args) {
String val = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";
int Sum = 0;
String adjacent = null;
for (int i = 0; i < val.length()-13; i++) {
int total = 1;
for (int j = 0; j < 13; j++) {
total = total * Character.getNumericValue(val.charAt(i+j));
}
if(total > Sum){
Sum = total;
adjacent = val.substring(i, i+13);
}
}
System.out.println("Sum = " + Sum);
System.out.println("Adjsc = " + adjacent );
}
I had the same problem.
int product and int greatestproduct have maximum value it can store since they are 'int' type. They can only store values up to 2147483647.
Use 'long long' type instead of 'int'.
const thousandDigit =
`73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450`;
const productsOfAdjacentNths = num =>
[...thousandDigit].reduce((acc, _, idx) => {
const nthDigitAdjX = [...thousandDigit.substr(idx, num)].reduce(
(inAcc, inCur) => inCur * inAcc,
1
);
return acc.concat(nthDigitAdjX);
}, []);
console.log(Math.max(...productsOfAdjacentNths(13))); //5377010688
This is my O(n) approach
function findLargest(digits, n){
let largest = 0;
let j = 0;
let res = 1;
for(let i = 0; i< digits.length; i ++){
res = res * parseInt(digits[i]);
if(res == 0){
res = 1;
j = 0;
continue;
}
if(j === n-1){
if(res > largest)
largest = res;
res = res/parseInt(digits[i - j]);
j = j - 1;
}
j = j + 1;
}
return largest;
}
let val = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";
findLargest(val, 13);
Although above solutions are good, here is a python implementation which works perfectly:
def main():
num=7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450
prod,maxprod=1,0
numtostr=str(num)
length=len(numtostr)
for i in range(1,length-13+1):
prod=1
for z in range(i,i+13):
prod=prod*int(numtostr[z-1:z])
if prod>maxprod:
maxprod=prod
print "Largest 13 digit product is %d" %(maxprod)
if __name__ == '__main__':
main()
static long q8(){
long max_product = 1;
long product = 1;
String n = "LONG_INPUT";
for(int i=0;i<13;i++){
max_product *= Integer.parseInt(n.charAt(i)+"");
}
product = max_product;
for(int i=1;i<n.length()-13;i++){
int denom = Integer.parseInt(n.charAt(i-1)+"");
if(denom!=0)
product = product/denom * Integer.parseInt(n.charAt(i+12)+"");
else
product = product(i,n);
max_product = (max_product>product)?max_product:product;
}
return max_product;
}
static long product(int index,String n){
long pro = 1;
for(int i=index;i<index+13;i++){
pro *= Integer.parseInt(n.charAt(i)+"");
}
return pro;
}
Try this:
{
DateTime BeganAt = new DateTime();
BeganAt = DateTime.Now;
Int64 result = 0;
string testNumber = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";
StringBuilder StringBuilder = new StringBuilder(13);
Int64 maxNumber = 0;
try
{
char[] numbers = testNumber.ToCharArray();
int tempCounter = 13;
for (int i = 0;i < numbers.Length;i++)
{
if(i < tempCounter)
{
StringBuilder.Append(numbers[i]);
}
else if (i == tempCounter)
{
if (maxNumber < Convert.ToInt64(StringBuilder.ToString()))
{
maxNumber = Convert.ToInt64(StringBuilder.ToString());
}
StringBuilder.Clear();
StringBuilder.Append(numbers[i]);
tempCounter = tempCounter + n;
}
}
result = maxNumber;
}
catch
{
throw;
}
DateTime EndedAt = new DateTime();
EndedAt = DateTime.Now;
TimeSpan TimeItTook = (EndedAt - BeganAt);
return Convert.ToString(result) + " - Time took to execute: " + Convert.ToString(TimeItTook.TotalMilliseconds);
}
long long int sum = 1,high=0;
for (int i = 0; i < 988; i++)
{
sum = sum*(arr[i] - 48);
for (int j = i + 1; j < i+13; j++)
{
sum = sum*(arr[j]-48);
}
if (sum >= high)
{
high = sum;
}
sum = 1;
}
My solution to above problem if someone is still watching this in 2018.Although there are lots of solution to this question here my solution pre-checks the individual 13 digits which have a 0 in them.As something multiplied with 0 is always 0 so we can remove this useless computation
int j = 0, k = 12;
long long int mult = 1;
long long int max = 0;
char *digits = /*Big number*/
while(k < 1000){
for (int i = j; i <= k; ++i)
{
/* code */
long long int val = digits[i] -'0';
/* check if any number in series contains 0 */
if(val == 0){
break;
}
mult = mult * val;
}
printf("mult is %lld\n",mult );
/* check for max value */
if(max < mult){
max = mult;
}
printf("the new max is %lld\n", max);
j += 1;
k += 1;
mult = 1;
printf("%d iteration finished\n", k);
}
printf("%lld\n", max);
You should use 'int64_t' at the place of 'int'
I have solve this problem using python 're' module
Finding all possible 13 digits number without having zeros (total 988 with zero and 283 without zero) then find the product of each of these digits and check for max
here i have used look ahead regex
Note: string must not contain any newline char
s = '731671765...'
import re
def product_13(d):
pro = 1
while d:
pro *= d % 10
d //= 10
return pro
pat = re.compile(r'(?=([1-9]{13}))')
all = map(int, re.findall(pat, s))
pro = -1
for i in all:
v = product_13(i)
if pro < v:
pro = v
print(pro)
**This is JavaScript solution.**
greatestProductOfAdjacentDigit = (givenNumber, noOfDigit) => {
stringNumberToNumbers = givenNumber => givenNumber.split('').map(each => parseInt(each, 10))
let numbers = mathematicalProblems.stringNumberToNumbers(givenNumber);
return numbers.map(function (each, ind, arr) {
return mathematicalProblems.productOfAll(arr.slice(ind, ind + noOfDigit));
}).reduce(function (accumulator, currentValue) {
return accumulator > currentValue ? accumulator : currentValue;
});
};
Testing
const givenNumber = '73167176531330624919225119674426574742355349' +
'194934969835203127745063262395783180169848018694788518438586' +
'156078911294949545950173795833195285320880551112540698747158' +
'523863050715693290963295227443043557668966489504452445231617' +
'318564030987111217223831136222989342338030813533627661428280' +
'644448664523874930358907296290491560440772390713810515859307' +
'960866701724271218839987979087922749219016997208880937766572' +
'733300105336788122023542180975125454059475224352584907711670' +
'556013604839586446706324415722155397536978179778461740649551' +
'492908625693219784686224828397224137565705605749026140797296' +
'865241453510047482166370484403199890008895243450658541227588' +
'666881164271714799244429282308634656748139191231628245861786' +
'645835912456652947654568284891288314260769004224219022671055' +
'626321111109370544217506941658960408071984038509624554443629' +
'812309878799272442849091888458015616609791913387549920052406' +
'368991256071760605886116467109405077541002256983155200055935' +
'72972571636269561882670428252483600823257530420752963450';
console.log(greatestProductOfAdjacentDigit(givenNumber, 4)); //5832
console.log(greatestProductOfAdjacentDigit(givenNumber, 13)); //23514624000
My 2 cents ... Javascript
Largest_product_in_series_13 = g => {
return [...g.matchAll(/(?=([1-9]{13}))/g)]
.reduce((a,c)=>(p=eval(c[1].split``.join`*`),p>a?p:a),0)
}
First it matches all 13 digit series, then it reduces to the biggest product.
console.log(Largest_product_in_series_13(`7316717653133062491922....
> 23514624000
start: 14.292ms
I solved it using ruby. if
x = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450"
then used below logic to solve the problem
def print_lps(x)
p = 1
0.upto(x.length-13) do |i|
j = i
xy = 1
while j < i+13 do
xy *= x[j].to_i
j += 1
end
p = xy if xy > p
end
puts p
end
print_lps(x) # 23514624000
I solve the problem using this approach
1.Run two loops first outer loop which iterate over numbers in the string.
2.Inner loop reach to next 13 numbers afterwards and store its product in a variable
3.As inner loop ends we only use maximum value.(that what we need) :>
Have a look at my code in C++
#include <bits/stdc++.h>
using namespace std;
int main()
{
string s="//thestring is there";
long long unsigned mainans=1,ans=1; //initialize the value of mainans and ans to 1 (not 0 as product also become 0)
for(int i=0;i<s.length()-13;i++) //run the loop from 0 upto 13 minus string length
{
ans=1;
for(int j=i;j<i+13;j++) //now from that particular digit we check 13 digits afterwards it
{
int m=s[j]-'0'; //convert the number from string to integer
ans*=m; //taking product in every interation for 13 digits
}
mainans=max(mainans,ans); //now taking only max value because that what we need
}
cout<<mainans;
return 0;
}
This will find the answer in O(n) time and it handles zero. Written in Dart.
/**
* Main function
*
* Find the thirteen adjacent digits in the 1000-digit number that have the greatest product.
* What is the value of this product?
*
*/
const String DIGITS = "73167176531330624919225119674426574742355349194934"
"96983520312774506326239578318016984801869478851843"
"85861560789112949495459501737958331952853208805511"
"12540698747158523863050715693290963295227443043557"
"66896648950445244523161731856403098711121722383113"
"62229893423380308135336276614282806444486645238749"
"30358907296290491560440772390713810515859307960866"
"70172427121883998797908792274921901699720888093776"
"65727333001053367881220235421809751254540594752243"
"52584907711670556013604839586446706324415722155397"
"53697817977846174064955149290862569321978468622482"
"83972241375657056057490261407972968652414535100474"
"82166370484403199890008895243450658541227588666881"
"16427171479924442928230863465674813919123162824586"
"17866458359124566529476545682848912883142607690042"
"24219022671055626321111109370544217506941658960408"
"07198403850962455444362981230987879927244284909188"
"84580156166097919133875499200524063689912560717606"
"05886116467109405077541002256983155200055935729725"
"71636269561882670428252483600823257530420752963450";
main(List<String> args) {
const int MAX_DIGITS = 13;
int len = DIGITS.length;
int digits = 0;
int largest = 0;
int current = 0;
int index = 0;
while (index < len) {
int value = int.parse(DIGITS[index]);
// if we get a zero then rebuild the MAX_DIGITS consecutive digits
if (value == 0) {
current = 0;
digits = 0;
} else {
// Multiply consecutive digits up to target set
if (digits < MAX_DIGITS) {
if (current == 0) {
current = value;
} else
current *= value;
digits++;
} else {
int divisor = int.parse(DIGITS[index - MAX_DIGITS]);
if (current == 0) {
current = value;
} else {
current = (current ~/ divisor);
current *= value;
}
}
if (current > largest) {
largest = current;
}
}
index++;
}
print('Num $largest');
}
I'm extracting numbers from web page, and i want compare them, if
number is more than [previous one, then show it in iimDispaly(), otherwise if is it lower then skip, i trued here to do but cannot find a solution.
for(var i = 1; i <= 13; i++) {
for(var j = 3; j<=33; j+=3 ) {
iimPlayCode('TAG POS='+j+' TYPE=DIV ATTR=CLASS:"group_row_labeled" EXTRACT=TXT')
var res = iimGetLastExtract();
var result = res.replace(/[а-z]/g, '');
if(j==3) {
var firstRes = result;
}
if(result => firstRes) {
iimDisplay("Highest Number: " + result)
}
}
}
Copy & paste the following code and try to play it:
for(var i = 1; i <= 13; i++) {
for(var j = 3; j<=33; j+=3 ) {
iimPlayCode('TAG POS='+j+' TYPE=DIV ATTR=CLASS:"group_row_labeled" EXTRACT=TXT')
var res = iimGetLastExtract();
var result = parseFloat(res.replace(/[a-z]/g, ''));
if(j==3) {
var firstRes = result;
}
if(result >= firstRes) {
iimDisplay("Highest Number: " + result)
}
}
}
Given two strings (s1, s2), Levenshtein Distance is the minimum number of operations needed to change s1 to s2 or vice versa.
I want to show the result of changing s1 to s2. For example, changing Sunday to Saturday needs 3 operations. I need to show S++u+day. "+" is for each operations needed.
Here is a javascript snippet that returns what you want. If you are familiar with the dynamic programming algorithm you should be able follow this code. All the string operations/manipulation of the return string r and handling of min/curMin are what's changed from the original version.
function edits(t, s) {
var r = "";
if (s === t) {
return s;
}
var n = s.length, m = t.length;
if (n === 0 || m === 0) {
return "+".repeat(n + m);
}
var x, y, a, b, c, min = 0;
var p = new Array(n);
for (y = 0; y < n;)
p[y] = ++y;
for (x = 0; x < m; x++) {
var e = t.charCodeAt(x);
c = x;
b = x + 1;
var currMin = min;
min = n + 1;
for (y = 0; y < n; y++) {
a = p[y];
if (a < c || b < c) {
b = (a > b ? b + 1 : a + 1);
}
else {
if (e !== s.charCodeAt(y)) {
b = c + 1;
}
else {
b = c;
}
}
if (b < min) {
min = b;
}
p[y] = b;
c = a;
}
if (min > currMin) {
r += "+";
}
else {
r += t[x];
}
}
return r;
}
EDIT: The implementation above is an version optimized for speed and space so might be harder to read. The implemetation below is a modified version of the JS version from Wikipedia and should be easier to follow.
function getEditDistance(a, b) {
if(a.length === 0) return "+".repeat(b.length);
if(b.length === 0) return "+".repeat(a.length);
var matrix = [];
// increment along the first column of each row
var i;
for(i = 0; i <= b.length; i++){
matrix[i] = [i];
}
// increment each column in the first row
var j;
for(j = 0; j <= a.length; j++){
matrix[0][j] = j;
}
var r = "", min = 0;;
// Fill in the rest of the matrix
for(i = 1; i <= b.length; i++){
var currMin = min;
min = a.length + 1;
for(j = 1; j <= a.length; j++){
if(b.charAt(i-1) == a.charAt(j-1)){
matrix[i][j] = matrix[i-1][j-1];
} else {
matrix[i][j] = Math.min(matrix[i-1][j-1] + 1, // substitution
Math.min(matrix[i][j-1] + 1, // insertion
matrix[i-1][j] + 1)); // deletion
}
if (matrix[i][j] < min) {
min = matrix[i][j];
}
}
if (min > currMin) {
r += "+";
}
else {
r += b[i-1];
}
}
return r;
}
EDIT2: Added explanation of the algorithm and example output
Below is the levenshtein matrix from the input strings "kitten" and "sitting". What I changed in the original algorithm is that I added a check if the current rows minimum value is larger then the previous rows minimum, and if it is, there is an edit in the current row and thus adding a "+". If not and the " edit cost" for the current row is the same as the previous. Then there is no edit necessary and we just add the current character to the result string. You can follow the algorithm row by row in the image (starting at row 1, and column 1)
Examples:
> getEditDistance("kitten", "sitting");
'+itt+n+'
> getEditDistance("Sunday", "Saturday");
'S++u+day'
I'm using ASGALLANT'S Hide/Show method shown here:
http://jsfiddle.net/asgallant/6gz2Q/
Except I have fourth data series called average. When a series is hidden or shown, I recalculate the average... by adjusting the fiddle above:
function showHideSeries () {
var sel = chart.getChart().getSelection();
var view = chart.getView() || {};
// if selection length is 0, we deselected an element
if (sel.length > 0) {
// if row is undefined, we clicked on the legend
if (sel[0].row == null) {
var col = sel[0].column;
if (typeof(columns[col]) == 'number') {
var src = columns[col];
// hide the data series
columns[col] = {
label: datatable.getColumnLabel(src),
type: datatable.getColumnType(src),
sourceColumn: src,
calc: function () {
return null;
}
};
//record as hidden
hiddenSeries[col] = true;
// grey out the legend entry
series[columnsMap[src]].color = '#CCCCCC';
chart.setOption('series', series);
//If Exists a columnLabel called "Average", check if last column (average column is currently shown), recalculate average
if ((datatable.getColumnLabel(datatable.getNumberOfColumns() - 1) == 'Average') && (hiddenSeries[datatable.getNumberOfColumns() - 1] == false)){
for(var r = 0; r < datatable.getNumberOfRows(); ++r) {
sum = 0;
k = 0
for(var c = 1; c < datatable.getNumberOfColumns()-1; ++c) {
if(hiddenSeries[c] == false){
if(datatable.getValue(r, c) > 0){
sum = sum + datatable.getValue(r, c);
k = k + 1;
}
}
}
if (k == 0) k = 1;
datatable.setValue(r, datatable.getNumberOfColumns() - 1, sum/k);
}
}
var tmpColumn = new Array();
// Add each data value to the array with push()
for(var i = 0; i < datatable.getNumberOfRows(); ++i) {
tmpColumn.push(datatable.getValue(i, col));
}
}
else {
var src = columns[col].sourceColumn;
// show the data series
columns[col] = src;
series[columnsMap[src]].color = null;
//record as shown
hiddenSeries[col] = false;
//If Exists a columnLabel called "Average", check if last column (average column is currently shown), recalculate average
if ((datatable.getColumnLabel(datatable.getNumberOfColumns() - 1) == 'Average') && (hiddenSeries[datatable.getNumberOfColumns() - 1] == false)){
for(var r = 0; r < datatable.getNumberOfRows(); ++r) {
sum = 0;
k = 0;
for(var c = 1; c < datatable.getNumberOfColumns() - 1; ++c) {
if(hiddenSeries[c] == false){
if(datatable.getValue(r, c) > 0){
sum = sum + datatable.getValue(r, c);
k = k + 1;
}
}
}
if (k == 0) k = 1;
datatable.setValue(r, datatable.getNumberOfColumns() - 1, sum/k);
}
}
//chart.setOption('series.' + i + '.type', 'bars');
if(datatable.getColumnLabel(src) == 'Average' || datatable.getColumnLabel(src) == 'Peak Demand [kVA]'){
chart.getOptions().series[columnsMap[src]].type = 'line';
if(c.second_axis == true){
chart.getOptions().series[columnsMap[src]].targetAxisIndex = 1;
}
}
}
chart.setDataTable(datatable);
view.columns = columns;
chart.setView(view);
google.visualization.events.addListener(chart, 'ready', function() {
document.getElementById('chartImg').href = chart.getChart().getImageURI();
});
chart.draw();
}
}
}
It works great, except that the Standard tooltip is not being updated on the re-draw of the chart. I would like to avoid using custom tooltips as a solution. I would think that the ToolTip values should automatically update on each draw of the chart... if not, there must be a way to force it?
The issue here is that I was using formatted numbers as shown below:
var formatter = new google.visualization.NumberFormat(
{negativeColor: 'red', negativeParens: true, pattern: '###,###'});
formatter.format(datatable, i);
Therefore, not only did I have to use datatable.setValue(row, column, value) but I also have to use datatable.setFormattedValue(row, column, *String* value).