Develop Reference

How to choose a weighted random array element in Javascript?

For example: There are four items in an array. I want to get one randomly, like this:
array items = [
"bike" //40% chance to select
"car" //30% chance to select
"boat" //15% chance to select
"train" //10% chance to select
"plane" //5% chance to select
]

Both answers above rely on methods that will get slow quickly, especially the accepted one.
function weighted_random(items, weights) {
var i;
for (i = 1; i < weights.length; i++)
weights[i] += weights[i - 1];
var random = Math.random() * weights[weights.length - 1];
for (i = 0; i < weights.length; i++)
if (weights[i] > random)
break;
return items[i];
}
I replaced my older ES6 solution with this one as of December 2020, as ES6 isn't supported in older browsers, and I personally think this one is more readable.
If you'd rather use objects with the properties item and weight:
function weighted_random(options) {
var i;
var weights = [options[0].weight];
for (i = 1; i < options.length; i++)
weights[i] = options[i].weight + weights[i - 1];
var random = Math.random() * weights[weights.length - 1];
for (i = 0; i < weights.length; i++)
if (weights[i] > random)
break;
return options[i].item;
}
Explanation:
I've made this diagram that shows how this works:
This diagram shows what happens when an input with the weights [5, 2, 8, 3] is given. By taking partial sums of the weights, you just need to find the first one that's as large as a random number, and that's the randomly chosen item.
If a random number is chosen right on the border of two weights, like with 7 and 15 in the diagram, we go with the longer one. This is because 0 can be chosen by Math.random but 1 can't, so we get a fair distribution. If we went with the shorter one, A could be chosen 6 out of 18 times (0, 1, 2, 3, 4), giving it a higher weight than it should have.

Some es6 approach, with wildcard handling:
const randomizer = (values) => {
let i, pickedValue,
randomNr = Math.random(),
threshold = 0;
for (i = 0; i < values.length; i++) {
if (values[i].probability === '*') {
continue;
}
threshold += values[i].probability;
if (threshold > randomNr) {
pickedValue = values[i].value;
break;
}
if (!pickedValue) {
//nothing found based on probability value, so pick element marked with wildcard
pickedValue = values.filter((value) => value.probability === '*');
}
}
return pickedValue;
}
Example usage:
let testValues = [{
value : 'aaa',
probability: 0.1
},
{
value : 'bbb',
probability: 0.3
},
{
value : 'ccc',
probability: '*'
}]
randomizer(testValues); // will return "aaa" in 10% calls,
//"bbb" in 30% calls, and "ccc" in 60% calls;

Here's a faster way of doing that then other answers suggested...
You can achieve what you want by:
dividing the 0-to-1 segment into sections for each element based on their probability (For example, an element with probability 60% will take 60% of the segment).
generating a random number and checking in which segment it lands.
STEP 1
make a prefix sum array for the probability array, each value in it will signify where its corresponding section ends.
For example:
If we have probabilities: 60% (0.6), 30%, 5%, 3%, 2%. the prefix sum array will be: [0.6,0.9,0.95,0.98,1]
so we will have a segment divided like this (approximately): [ | | ||]
STEP 2
generate a random number between 0 and 1, and find it's lower bound in the prefix sum array. the index you'll find is the index of the segment that the random number landed in
Here's how you can implement this method:
let obj = {
"Common": "60",
"Uncommon": "25",
"Rare": "10",
"Legendary": "0.01",
"Mythical": "0.001"
}
// turning object into array and creating the prefix sum array:
let sums = [0]; // prefix sums;
let keys = [];
for(let key in obj) {
keys.push(key);
sums.push(sums[sums.length-1] + parseFloat(obj[key])/100);
}
sums.push(1);
keys.push('NONE');
// Step 2:
function lowerBound(target, low = 0, high = sums.length - 1) {
if (low == high) {
return low;
}
const midPoint = Math.floor((low + high) / 2);
if (target < sums[midPoint]) {
return lowerBound(target, low, midPoint);
} else if (target > sums[midPoint]) {
return lowerBound(target, midPoint + 1, high);
} else {
return midPoint + 1;
}
}
function getRandom() {
return lowerBound(Math.random());
}
console.log(keys[getRandom()], 'was picked!');
hope you find this helpful.
Note:
(In Computer Science) the lower bound of a value in a list/array is the smallest element that is greater or equal to it. for example, array:[1,10,24,99] and value 12. the lower bound will be the element with value 24.
When the array is sorted from smallest to biggest (like in our case) finding the lower bound of every value can be done extremely quickly with binary searching (O(log(n))).

Here is a O(1) (constant time) algo to solve your problem.
Generate a random number from 0 to 99 (100 total numbers). If there are 40 numbers (0 to 39) in a given sub-range, then there is a 40% probability that the randomly chosen number will fall in this range. See the code below.
const number = Math.floor(Math.random() * 99); // 0 to 99
let element;
if (number >= 0 && number <= 39) {
// 40% chance that this code runs. Hence, it is a bike.
element = "bike";
}
else if (number >= 40 && number <= 69) {
// 30% chance that this code runs. Hence, it is a car.
element = "car";
}
else if (number >= 70 && number <= 84) {
// 15% chance that this code runs. Hence, it is a boat.
element = "boat";
}
else if (number >= 85 && number <= 94) {
// 10% chance that this code runs. Hence, it is a train.
element = "train";
}
else if (number >= 95 && number <= 99) {
// 5% chance that this code runs. Hence, it is a plane.
element = "plane";
}
Remember this, one Mathematical principle from elementary school? "All the numbers in a specified distribution have equal probability of being chosen randomly."
This tells us that each of the random numbers have equal probability of occurring in a specific range, no matter how large or small that range might be.
That's it. This should work!

I added my solution as a method that works well on smaller arrays (no caching):
static weight_random(arr, weight_field){
if(arr == null || arr === undefined){
return null;
}
const totals = [];
let total = 0;
for(let i=0;i<arr.length;i++){
total += arr[i][weight_field];
totals.push(total);
}
const rnd = Math.floor(Math.random() * total);
let selected = arr[0];
for(let i=0;i<totals.length;i++){
if(totals[i] > rnd){
selected = arr[i];
break;
}
}
return selected;
}
Run it like this (provide the array and the weight property):
const wait_items = [
{"w" : 20, "min_ms" : "5000", "max_ms" : "10000"},
{"w" : 20, "min_ms" : "10000", "max_ms" : "20000"},
{"w" : 20, "min_ms" : "40000", "max_ms" : "80000"}
]
const item = weight_random(wait_items, "w");
console.log(item);

ES2015 version of Radvylf Programs's answer
function getWeightedRandomItem(items) {
const weights = items.reduce((acc, item, i) => {
acc.push(item.weight + (acc[i - 1] || 0));
return acc;
}, []);
const random = Math.random() * weights[weights.length - 1];
return items[weights.findIndex((weight) => weight > random)];
}
And ES2022
function getWeightedRandomItem(items) {
const weights = items.reduce((acc, item, i) => {
acc.push(item.weight + (acc[i - 1] ?? 0));
return acc;
}, []);
const random = Math.random() * weights.at(-1);
return items[weights.findIndex((weight) => weight > random)];
}

Sure you can. Here's a simple code to do it:
// Object or Array. Which every you prefer.
var item = {
bike:40, // Weighted Probability
care:30, // Weighted Probability
boat:15, // Weighted Probability
train:10, // Weighted Probability
plane:5 // Weighted Probability
// The number is not really percentage. You could put whatever number you want.
// Any number less than 1 will never occur
};
function get(input) {
var array = []; // Just Checking...
for(var item in input) {
if ( input.hasOwnProperty(item) ) { // Safety
for( var i=0; i<input[item]; i++ ) {
array.push(item);
}
}
}
// Probability Fun
return array[Math.floor(Math.random() * array.length)];
}
console.log(get(item)); // See Console.

In javascript, how do I add a random amount to a user's balance while controlling how much gets given total?

I'm trying to make it to where when a user does a certain thing, they get between 2 and 100 units. But for every 1,000 requests I want it to add up to 3,500 units given collectively.
Here's the code I have for adding different amounts randomly to a user:
if (Math.floor(Math.random() * 1000) + 1 === 900) {
//db call adding 100
}
else if (Math.floor(Math.random() * 100) + 1 === 90) {
//db call adding 40
}
else if (Math.floor(Math.random() * 30) + 1 === 20) {
//db call adding 10
}
else if (Math.floor(Math.random() * 5) + 1 === 4) {
//db call adding 5
}
else {
//db call adding 2
}
If my math is correct, this should average around 4,332 units per 1,000 calls. But obviously it would vary and I don't want that. I'd also like it to add random amounts instead, as the units added in my example are arbitrary.
EDIT: Guys, Gildor is right that I simply want to have 3,500 units, and give them away within 1,000 requests. It isn't even entirely necessary that it always reaches that maximum of 3,500 either (I could have specified that). The important thing is that I'm not giving users too much, while creating a chance for them to win a bigger amount.
Here's what I have set up now, and it's working well, and will work even better with some tweaking:
Outside of call:
var remaining = 150;
var count = 0;
Inside of call:
count += 1;
if (count === 100) {
remaining = 150;
count = 0;
}
if (Math.floor(Math.random() * 30) + 1 === 20) {
var addAmount = Math.floor(Math.random() * 85) + 15;
if (addAmount <= remaining) {
remaining -= addAmount;
//db call adding addAmount + 2
}
else {
//db call adding 2
}
}
else if (Math.floor(Math.random() * 5) + 1 === 4) {
var addAmount1 = Math.floor(Math.random() * 10) + 1;
if (addAmount1 <= remaining) {
remaining -= addAmount1;
//db call adding addAmount1 + 2
}
else {
//db call adding 2
}
}
else {
//db call adding 2
}
I guess I should have clarified, I want a "random" number with a high likelihood of being small. That's kind of part of the gimmick, where you have low probability of getting a larger amount.

As I've commented, 1,000 random numbers between 2 and 100 that add up to 3,500 is an average number of 3.5 which is not consistent with random choices between 2 and 100. You'd have to have nearly all 2 and 3 values in order to achieve that and, in fact couldn't have more than a couple large numbers. Nothing even close to random. So, for this to even be remotely random and feasible, you'd have to pick a total much large than 3,500. A random total of 1,000 numbers between 2 and 100 would be more like 51,000.
Furthermore, you can't dynamically generate each number in a truly random fashion and guarantee a particular total. The main way to guarantee that outcome is to pre-allocate random numbers that add up to the total that are known to achieve that and then random select each number from the pre-allocated scheme, then remove that from the choice for future selections.
You could also try to keep a running total and bias your randomness if you get skewed away form your total, but doing it that way, the last set of numbers may have to be not even close to random in order to hit your total consistently.
A scheme that could work if you reset the total to support what it should be for actual randomness (e.g. to 51,000) would be to preallocated an array of 500 random numbers between 2 and 100 and then add another 500 numbers that are the complements of those. This guarantees the 51 avg number. You can then select each number randomly from the pre-allocated array and then remove it form the array so it won't be selected again. I can add code to do this in a second.
function RandResults(low, high, qty) {
var results = new Array(qty);
var limit = qty/2;
var avg = (low + high) / 2;
for (var i = 0; i < limit; i++) {
results[i] = Math.floor((Math.random() * (high - low)) + low);
//
results[qty - i - 1] = (2 * avg) - results[i];
}
this.results = results;
}
RandResults.prototype.getRand = function() {
if (!this.results.length) {
throw new Error("getRand() called, but results are empty");
}
var randIndex = Math.floor(Math.random() * this.results.length);
var value = this.results[randIndex];
this.results.splice(randIndex, 1);
return value;
}
RandResults.prototype.getRemaining = function() {
return this.results.length;
}
var randObj = new RandResults(2, 100, 1000);
// get next single random value
if (randObj.getRemaining()) {
var randomValue = randObj.getRand();
}
Working demo for a truly random selection of numbers that add up to 51,000 (which is what 1,000 random values between 2 and 100 should add up to): http://jsfiddle.net/jfriend00/wga26n7p/
If what you want is the following: 1,000 numbers that add up to 3,500 and are selected from between the range 2 to 100 (inclusive) where most numbers will be 2 or 3, but occasionally something could be up to 100, then that's a different problem. I wouldn't really use the word random to describe it because it's a highly biased selection.
Here's a way to do that. It generates 1,000 random numbers between 2 and 100, keeping track of the total. Then, afterwards it corrects the random numbers to hit the right total by randomly selected values and decrementing them until the total is down to 3,500. You can see it work here: http://jsfiddle.net/jfriend00/m4ouonj4/
The main part of the code is this:
function RandResults(low, high, qty, total) {
var results = new Array(qty);
var runningTotal = 0, correction, index, trial;
for (var i = 0; i < qty; i++) {
runningTotal += results[i] = Math.floor((Math.random() * (high - low)) + low);
}
// now, correct to hit the total
if (runningTotal > total) {
correction = -1;
} else if (runningTotal < total) {
correction = 1;
}
// loop until we've hit the total
// randomly select a value to apply the correction to
while (runningTotal !== total) {
index = Math.floor(Math.random() * qty);
trial = results[index] + correction;
if (trial >= low && trial <= high) {
results[index] = trial;
runningTotal += correction;
}
}
this.results = results;
}
This meets an objective of a biased total of 3,500 and all numbers between 2 and 100, though the probability of a 2 in this scheme is very high and the probably of a 100 in this scheme is almost non-existent.
And, here's a weighted random generator that adds up to a precise total. This uses a cubic weighting scheme to favor the lower numbers (the probably of a number goes down with the cube of the number) and then after the random numbers are generated, a correction algorithm applies random corrections to the numbers to make the total come out exactly as specified. The code for a working demo is here: http://jsfiddle.net/jfriend00/g6mds8rr/
function RandResults(low, high, numPicks, total) {
var avg = total / numPicks;
var i, j;
// calculate probabilities for each value
// by trial and error, we found that a cubic weighting
// gives an approximately correct sub-total that can then
// be corrected to the exact total
var numBuckets = high - low + 1;
var item;
var probabilities = [];
for (i = 0; i < numBuckets; i++) {
item = low + i;
probabilities[i] = avg / (item * item * item);
}
// now using those probabilities, create a steps array
var sum = 0;
var steps = probabilities.map(function(item) {
sum += item;
return sum;
});
// now generate a random number and find what
// index it belongs to in the steps array
// and use that as our pick
var runningTotal = 0, rand;
var picks = [], pick, stepsLen = steps.length;
for (i = 0; i < numPicks; i++) {
rand = Math.random() * sum;
for (j = 0; j < stepsLen; j++) {
if (steps[j] >= rand) {
pick = j + low;
picks.push(pick);
runningTotal += pick;
break;
}
}
}
var correction;
// now run our correction algorithm to hit the total exactly
if (runningTotal > total) {
correction = -1;
} else if (runningTotal < total) {
correction = 1;
}
// loop until we've hit the total
// randomly select a value to apply the correction to
while (runningTotal !== total) {
index = Math.floor(Math.random() * numPicks);
trial = picks[index] + correction;
if (trial >= low && trial <= high) {
picks[index] = trial;
runningTotal += correction;
}
}
this.results = picks;
}
RandResults.prototype.getRand = function() {
if (!this.results.length) {
throw new Error("getRand() called, but results are empty");
}
return this.results.pop();
}
RandResults.prototype.getAllRand = function() {
if (!this.results.length) {
throw new Error("getAllRand() called, but results are empty");
}
var r = this.results;
this.results = [];
return r;
}
RandResults.prototype.getRemaining = function() {
return this.results.length;
}

As some comments pointed out... the numbers in the question does not quite make sense, but conceptually there are two approaches: calculate dynamically just in time or ahead of time.
To calculate just in time:
You can maintain a remaining variable which tracks how many of 3500 left. Each time when you randomly give some units, subtract the number from remaining until it goes to 0.
In addition, to make sure each time at least 2 units are given, you can start with remaining = 1500 and give random + 2 units each time.
To prevent cases that after 1000 gives there are still balances left, you may need to add some logic to give units more aggressively towards the last few times. However it will result in not-so-random results.
To calculate ahead of time:
Generate a random list with 1000 values in [2, 100] and sums up to 3500. Then shuffle the list. Each time you want to give some units, pick the next item in the array. After 1000 gives, generate another list in the same way. This way you get much better randomized results.
Be aware that both approaches requires some kind of shared state that needs to be handled carefully in a multi-threaded environment.
Hope the ideas help.

recursive function breaking in if statement javascript

i'm in the process of trying to write a function that converts an amount into a specified number of coins. For this I'm calling a function on itself (I think this is recursive programming?). I'm having 2 problems.
When i call the function on itself in the else section of the if statement I get the error message. "Maximum call stack size exceeded" ? Not sure why this is as the call above is the same and works fine when this second call is commented out.
Also when the second call is commented out and the function runs the variable total should be increasing by 1 with each call. However it doesn't make it past 1. I thought this may be because the variable was being reset at the top to 0 with each call. However the remainder variable is also set to 0 decreases its value each time.
Can anyone explain what is happening here ? How is this problem best solved ?
Thanks
function amountCoins(amt, coins) {
var remainder = 0;
total = 0;
if(amt >= coins[0]) {
total += 1;
remainder = (amt - coins[0]);
return amountCoins(remainder, coins);
} else {
coins.shift();
//return amountCoins(remainder,coins);
}
alert(total);
}
amountCoins(121,[20,15,6,1]);

You can use .reduce() for this as an alternative.
And we don't really need loops where simple math can handle it.
var total = [20,15,6,1].reduce(function(obj, denomination) {
return {
total: obj.total + Math.floor(obj.amount / denomination),
amount: obj.amount % denomination
};
}, {amount:121, total:0}).total;
Or iterate the array.
var arr = [20,15,6,1], amount = 121, total = 0;
for (var i = 0; i < arr.length; ++i) {
total += Math.floor(amount / arr[i]);
amount %= arr[i];
}

I'd use a regular for loop instead, if I was you:
var amt = 121;
var coins = [20, 15, 6, 1];
var coinsUsed = {};
var totalCoins = 0;
for (var i = 0; i < coins.length; i++) {
if (coins[i] <= amt) {
coinsUsed[coins[i]] = Math.floor(amt / coins[i]);
amt -= coinsUsed[coins[i]] * coins[i];
totalCoins += coinsUsed[coins[i]];
} else {
coinsUsed[coins[i]] = 0;
}
}
console.log(coinsUsed, totalCoins);
Output:
Object {
1: 1,
6: 0,
15: 0,
20: 6
}
7

The problem is that your algorithm never ends. Every recursive function must have an end, if not it will produce a stack overflow (xD), because recursive calls are stored in a stack.
This would be a solution:
function amountCoins(amt, coins, total) {
var remainder = 0;
total = total || 0; //Default value 0 (for the first call)
if(coins.length == 0) return total; //This is the ending condition (no coins)
if(amt >= coins[0]) {
total += 1;
remainder = (amt - coins[0]);
return amountCoins(remainder, coins, total);
} else {
coins.shift();
return amountCoins(remainder, coins, total);
}
}
var coins = amountCoins(121,[20,15,6,1]); //Returns: 6
alert(coins);
As you can see, I turned total into a param so we can store it from call to call.
Hope this helps. Cheers

i think this is what you are trying to do:
function amountCoins(amt, coins) {
var remainder = 0;
if(coins.length == 0) {
return 0;
} else if(amt >= coins[0]) {
remainder = (amt - coins[0]);
return amountCoins(remainder, coins)+1;
} else {
coins.shift();
return amountCoins(remainder,coins);
}
}
var total = amountCoins(121,[20,15,6,1]);
alert(total)
There are a few things here
A recursive algorithm needs a terminating condition. ie. when the function is calling itself recursively it should stop the chain at some point. If it doesnt, the program will run out of memory to accomodate all the calls in the chain. Because this is a potentially dangerous condition, programming languages like javascript limit the depth of a recursive call. This is what is meant by the Maximum call stack size exceeded error.
In your program, logically, the terminating condition is when we run out of coins. so a
coins.length == 0 check that returns a 0 total (which in turn seeds the upward chain) will fix the problem
Usually in recursive algorithms the result is passed backwards, up the call chain and not stored in an external variable. So the incrementing total value is expressed with a return statement like return amountCoins(remainder, coins)+1;
Finally, this problem can be much easily implemented with for loops. Try to think by unwinding the recursive call chains and you'll figure out a loop solution.

You must return total not only if no coins left, but also if the last coin's value is bigger then the remaining value:
function amountCoins(amt, coins, total) {
total = total || 0;
if (coins.length == 0 || (amt < coins[0] && coins.length == 1)) return total;
if (amt >= coins[0]) {
total += 1;
remainder = (amt - coins[0]);
return amountCoins(amt - coins[0], coins, total);
} else {
coins.shift();
return amountCoins(remainder, coins, total);
}
}

This correction to your original code will work:
function amountCoins(amt, coins, total) {
total = total || 0;
if (amt === 0) return total;
// throw out all the coins that are too big
while (amt < coins[0]) coins.shift();
// use the largest coin possible, then recurse
total += 1;
remainder = amt - coins[0];
return amountCoins(remainder, coins, total);
}
alert(amountCoins(121,[20,15,6,1])); // alerts 7
It has the advantage of avoiding unnecessary if ... else statements, and the logic is bit clearer imo

Coming Up with a Good Algorithm for a Simple Idea

I need to come up with an algorithm that does the following:
Lets say you have an array of positive numbers (e.g. [1,3,7,0,0,9]) and you know beforehand their sum is 20.
You want to abstract some average amount from each number such that the new sum would be less by 7.
To do so, you must follow these rules:
you can only subtract integers
the resulting array must not have any negative values
you can not make any changes to the indices of the buckets.
The more uniformly the subtraction is distributed over the array the better.
Here is my attempt at an algorithm in JavaScript + underscore (which will probably make it n^2):
function distributeSubtraction(array, goal){
var sum = _.reduce(arr, function(x, y) { return x + y; }, 0);
if(goal < sum){
while(goal < sum && goal > 0){
var less = ~~(goal / _.filter(arr, _.identity).length); //length of array without 0s
arr = _.map(arr, function(val){
if(less > 0){
return (less < val) ? val - less : val; //not ideal, im skipping some!
} else {
if(goal > 0){ //again not ideal. giving preference to start of array
if(val > 0) {
goal--;
return val - 1;
}
} else {
return val;
}
}
});
if(goal > 0){
var newSum = _.reduce(arr, function(x, y) { return x + y; }, 0);
goal -= sum - newSum;
sum = newSum;
} else {
return arr;
}
}
} else if(goal == sum) {
return _.map(arr, function(){ return 0; });
} else {
return arr;
}
}
var goal = 7;
var arr = [1,3,7,0,0,9];
var newArray = distributeSubtraction(arr, goal);
//returned: [0, 1, 5, 0, 0, 7];
Well, that works but there must be a better way! I imagine the run time of this thing will be terrible with bigger arrays and bigger numbers.
edit: I want to clarify that this question is purely academic. Think of it like an interview question where you whiteboard something and the interviewer asks you how your algorithm would behave on a different type of a dataset.

It sounds like you want to subtract a weighted amount from each number. I.E you want to subtract X/sum * amount_to_subtract from each item. You would of course need to round the amount your subtracting. The problem is then making sure that you've subtracted the total correct amount. Also, this depends on your input: are you guaranteeing that that the amount you want to subtract can be subtracted? Here's a rough python implementation, (I think):
def uniform_array_reduction(inp, amount):
total = sum(inp)
if amount > total:
raise RuntimeError('Can\'t remove more than there is')
if amount == total: #special case
return [0] * len(inp)
removed = 0
output = []
for i in inp:
if removed < amount:
to_remove = int(round(float(i)/float(total)*float(amount)))
output.append(i - to_remove)
removed += to_remove
else:
output.append(i)
# if we didn't remove enough, just remove 1 from
# each element until we've hit our mark.
# shouldn't require more than one pass
while removed < amount:
for i in range(len(output)):
if output[i] > 0:
output[i] -= 1
removed += 1
if removed == amount:
break
return output
EDIT: I've fixed a few bugs in the code.

s = Sum(x) - required_sum
do:
a = ceil( s/number_of_non_zeros(x) )
For i=1 to length(x):
v = min(a, x[i], s)
x[i]-=v
s-=v
while s>0
This version needs no sorting.

Problem with JavaScript arithmetic

I have a form for my customers to add budget projections. A prominent user wants to be able to show dollar values in either dollars, Kila-dollars or Mega-dollars.
I'm trying to achieve this with a group of radio buttons that call the following JavaScript function, but am having problems with rounding that make the results look pretty crummy.
Any advice would be much appreciated!
Lynn
function setDollars(new_mode)
{
var factor;
var myfield;
var myval;
var cur_mode = document.proj_form.cur_dollars.value;
if(cur_mode == new_mode)
{
return;
}
else if((cur_mode == 'd')&&(new_mode == 'kd'))
{
factor = "0.001";
}
else if((cur_mode == 'd')&&(new_mode == 'md'))
{
factor = "0.000001";
}
else if((cur_mode == 'kd')&&(new_mode == 'd'))
{
factor = "1000";
}
else if((cur_mode == 'kd')&&(new_mode == 'md'))
{
factor = "0.001";
}
else if((cur_mode == 'md')&&(new_mode == 'kd'))
{
factor = "1000";
}
else if((cur_mode == 'md')&&(new_mode == 'd'))
{
factor = "1000000";
}
document.proj_form.cur_dollars.value = new_mode;
var cur_idx = document.proj_form.cur_idx.value;
var available_slots = 13 - cur_idx;
var td_name;
var cell;
var new_value;
//Adjust dollar values for projections
for(i=1;i<13;i++)
{
var myfield = eval('document.proj_form.proj_'+i);
if(myfield.value == '')
{
myfield.value = 0;
}
var myval = parseFloat(myfield.value) * parseFloat(factor);
myfield.value = myval;
if(i < cur_idx)
{
document.getElementById("actual_"+i).innerHTML = myval;
}
}

First of all, Don't set myfield.value = myval after doing the math. You'll accumulate rounding errors with each additional selection of one of the radio buttons. Keep myfield.value as dollars all the time, then calculate a display value. This will also reduce the number of cases in your if-else cascade, as you will always be converting from dollars.
Now calculate by division -- you'll never have to multiply by 0.001 or 1000 since you're always converting in the same direction.
Use Math.round() to control rounding errors. You can multiply your original value first, then divide by a larger factor to also help control rounding errors; e.g. both these are the same in pure mathematics:
newvalue = value / 1000;
newvalue = (value * 10) / 10000;
Added
switch (new_mode)
{
case 'd':
divisor = 1;
break;
case 'kd':
divisor = 1000;
break;
case 'md':
divisor = 1000000;
break;
}
var displayValue = myfield.value / divisor;
Also, don't assign a string of numbers divisor = "1000"; use actual numbers divisor = 1000 (without the quotes)

Javascript's floating point numbers are a constant source of irritation.
You should try adding toFixed() to reduce the inevitable absurd decimals, and I've actually had to resort to forcing it to do it's rounding down where I don't care about it by doing something like:
number_to_be_rounded = Math.round(number_to_be_rounded*1000)/1000;
number_to_be_rounded = parseFloat(number_to_be_rounded.toFixed(2));
It's ugly, but it's close enough for a non-financial system.