How should I convert Wifi signal strength from a Quality in percentage, usually 0% to 100% into an RSSI value, usually a negative dBm number (i.e. -96db)?
Wifi Signal Strength Percentage to RSSI dBm
Microsoft defines Wifi signal quality in their WLAN_ASSOCIATION_ATTRIBUTES structure as follows:
wlanSignalQuality:
A percentage value that represents the signal quality of the network.
WLAN_SIGNAL_QUALITY is of type ULONG. This member contains a value
between 0 and 100. A value of 0 implies an actual RSSI signal strength
of -100 dbm. A value of 100 implies an actual RSSI signal strength of
-50 dbm. You can calculate the RSSI signal strength value for
wlanSignalQuality values between 1 and 99 using linear interpolation.
RSSI (or "Radio (Received) Signal Strength Indicator") are in units of 'dB' (decibel) or the similar 'dBm' (dB per milliwatt) (See dB vs. dBm) in which the smaller magnitude negative numbers have the highest signal strength, or quality.
Therefore, the conversion between quality (percentage) and dBm is as follows:
quality = 2 * (dBm + 100) where dBm: [-100 to -50]
dBm = (quality / 2) - 100 where quality: [0 to 100]
Pseudo Code (with example clamping):
// dBm to Quality:
if(dBm <= -100)
quality = 0;
else if(dBm >= -50)
quality = 100;
else
quality = 2 * (dBm + 100);
// Quality to dBm:
if(quality <= 0)
dBm = -100;
else if(quality >= 100)
dBm = -50;
else
dBm = (quality / 2) - 100;
Note:
Check the definition of Quality that you are using for your calculations carefully. Also check the range of dB (or dBm). The limits may vary.
Examples:
Medium quality: 50% -> -75dBm = (50 / 2) - 100
Low quality: -96dBm -> 8% = 2 * (-96 + 100)
In JS I prefer doing something like:
Math.min(Math.max(2 * (x + 100), 0), 100)
My personal opinion is that it's more elegant way to write it, instead of using if's.
From experience:
Less than -50dB (-40, -30 and -20) = 100% of signal strength
From -51 to -55dB= 90%
From -56 to -62dB=80%
From -63 to -65dB=75%
The below is not good enough for Apple devices
From -66 to 68dB=70%
From -69 to 74dB= 60%
From -75 to 79dB= 50%
From -80 to -83dB=30%
Windows laptops can work fine on -80dB however with slower speeds
Im glad I found this post cause I was looking for a way to convert the dbm to percentage. Using David's post, I wrote up a quick script in python to calculate the quality percentage.
#!/usr/bin/env python3
import os
import platform
system = platform.system()
if system == 'Linux':
cmd = "iwconfig wlan0 | grep Signal | /usr/bin/awk '{print $4}' | /usr/bin/cut -d'=' -f2"
elif system == 'Darwin':
cmd = "/System/Library/PrivateFrameworks/Apple80211.framework/Versions/Current/Resources/airport -I | grep CtlRSSI | awk '{ print $NF; }"
else:
print("Unsupported os: {}".format(system))
dbm = os.popen(cmd).read()
if dbm:
dbm_num = int(dbm)
quality = 2 * (dbm_num + 100)
print("{0} dbm_num = {1}%".format(dbm_num, quality))
else:
print("Wifi router connection signal strength not found")
In order to get the highest wifi quality from where my computer is located, I moved/rotated my antenna until I received the highest quality. To see real time quality, I ran the above script using:
watch -n0.1 "python getwifiquality.py"
I know this may be late but this may help someone in the future.
I took the value of dBm 30-90 for RSSI and correlated it to 100-0 %.
I used the basic linear equation to get the answer.
y = mx + b
We know our x values for dBm as 30 and 90.
We know our y values for % as 100 and 0.
We just need to find the slope. So we can make it linear.
m = 100-0/30-90
= 100/-60
= -5/3
b = y - mx
= 0 + 5/3*90
= 150
Final equation to put in code when you know the RSSI value.
% = 150 - (5/3) * RSSI
Note I did take the RSSI value that is normally negative and multiplied by the absolute value to get positive numbers.
quality = abs(RSSI)
% = 150 - (5/3) * quality
From RSSI vs RSS:
RSSI - Received Signal Strength Indicator
RSS - Received Signal Strength
RSSI is an indicator and RSS is the real value. Ok, now what do you mean by indicator, indicator mean it can be a relative value and RSSI is always a positive value and there is no unit for the RSSI.
We can say RSSI is for common man to understand. RF values are always told in dBm and the values are negative values most of the time. To make it easy for the people to understand these negative values are converted to positive values through scaling.
Say for example, if the maximum signal strength is 0 dBm and minimum is -100 dBm. We can scale it like as explained. We can put 0 dBm and more (RSS) as 100 RSSI (i. e. maximum RSSI) and -100 dBm (or less) as 0 RSSI (minimum RSS).
This is what i have done :
long rssi = WiFi.RSSI();
rssi=-rssi;
int WiFiperct;
if (rssi<27){
WiFiperct =100;
}
else if(rssi>=27&&rssi<33){
WiFiperct=150-(5/2.7)*rssi;
}
else if(rssi>=33&&rssi<36){
WiFiperct=150-(5/3)*rssi;
}
else if(rssi>=36&&rssi<40){
WiFiperct=150-(5/3.3)*rssi;
}
else if(rssi>=40&&rssi<80){
WiFiperct=150-(5/3.5)*rssi;
}
else if(rssi>=80&&rssi<90){
WiFiperct=150-(5/3.4)*rssi;
}
else if(rssi>=90&&rssi<99){
WiFiperct=150-(5/3.3)*rssi;
}
else{
WiFiperct=0;
}
This article is a more detailed explanation of mW, dBm and RSSI
http://madwifi-project.org/attachment/wiki/UserDocs/RSSI/Converting_Signal_Strength.pdf?format=raw
According to it RSSI do not have a unit. It's a value defined in 802.11 standard and calculated by nic card and sent to OS. The nic card vendor should provide a mapping table of dBm-RSSI values.
Sorry for the direct link, but I can not found the original page for the file link.
Mentioned pseudocode will not work all the ranges, the ranges example (-80dBm to 0, and -40dBm to 100).
Generic simple logic to map any range to 0 to 100.
Usage example, for below code ConvertRangeToPercentage(-80,-40,-50)
int ConvertRangeToPercentage (int a_value_map_to_zero, int a_value_map_to_100, int a_value_to_convert)
{
int percentage = 0;
if (a_value_map_to_zero < a_value_map_to_100)
{
if (a_value_to_convert <= a_value_map_to_zero)
{
percentage = 0;
}
else if (a_value_to_convert >= a_value_map_to_100)
{
percentage = 100;
}
else
{
percentage = (a_value_to_convert - a_value_map_to_zero) * 100 / (a_value_map_to_100 - a_value_map_to_zero);
}
}
else if (a_value_map_to_zero > a_value_map_to_100)
{
if (a_value_to_convert >= a_value_map_to_zero)
{
percentage = 0;
}
else if (a_value_to_convert <= a_value_map_to_100)
{
percentage = 100;
}
else
{
percentage = (a_value_to_convert - a_value_map_to_zero) * 100 / (a_value_map_to_100 - a_value_map_to_zero);
}
}
else
{
percentage = 0;
}
return percentage;
}
Ok.. I agree...but why is then:
Quality=29/100 Signal level=-78 dBm
Quality=89/100 Signal level=-55 dBm
Quality=100/100 Signal level=-21 dBm
this does not agree with the formula
percentage=quality/2 - 100.
Also, you can try inverse this Bash function which converts dBm to percentage:
#!/bin/bash
function dbmtoperc { # Convert dBm to percentage (based on https://www.adriangranados.com/blog/dbm-to-percent-conversion)
dbmtoperc_d=$(echo "$1" | tr -d -)
dbmtoperc_r=0
if [[ "$dbmtoperc_d" =~ [0-9]+$ ]]; then
if ((1<=$dbmtoperc_d && $dbmtoperc_d<=20)); then dbmtoperc_r=100
elif ((21<=$dbmtoperc_d && $dbmtoperc_d<=23)); then dbmtoperc_r=99
elif ((24<=$dbmtoperc_d && $dbmtoperc_d<=26)); then dbmtoperc_r=98
elif ((27<=$dbmtoperc_d && $dbmtoperc_d<=28)); then dbmtoperc_r=97
elif ((29<=$dbmtoperc_d && $dbmtoperc_d<=30)); then dbmtoperc_r=96
elif ((31<=$dbmtoperc_d && $dbmtoperc_d<=32)); then dbmtoperc_r=95
elif ((33==$dbmtoperc_d)); then dbmtoperc_r=94
elif ((34<=$dbmtoperc_d && $dbmtoperc_d<=35)); then dbmtoperc_r=93
elif ((36<=$dbmtoperc_d && $dbmtoperc_d<=38)); then dbmtoperc_r=$((92-($dbmtoperc_d-36)))
elif ((39<=$dbmtoperc_d && $dbmtoperc_d<=51)); then dbmtoperc_r=$((90-($dbmtoperc_d-39)))
elif ((52<=$dbmtoperc_d && $dbmtoperc_d<=55)); then dbmtoperc_r=$((76-($dbmtoperc_d-52)))
elif ((56<=$dbmtoperc_d && $dbmtoperc_d<=58)); then dbmtoperc_r=$((71-($dbmtoperc_d-56)))
elif ((59<=$dbmtoperc_d && $dbmtoperc_d<=60)); then dbmtoperc_r=$((67-($dbmtoperc_d-59)))
elif ((61<=$dbmtoperc_d && $dbmtoperc_d<=62)); then dbmtoperc_r=$((64-($dbmtoperc_d-61)))
elif ((63<=$dbmtoperc_d && $dbmtoperc_d<=64)); then dbmtoperc_r=$((61-($dbmtoperc_d-63)))
elif ((65==$dbmtoperc_d)); then dbmtoperc_r=58
elif ((66<=$dbmtoperc_d && $dbmtoperc_d<=67)); then dbmtoperc_r=$((56-($dbmtoperc_d-66)))
elif ((68==$dbmtoperc_d)); then dbmtoperc_r=53
elif ((69==$dbmtoperc_d)); then dbmtoperc_r=51
elif ((70<=$dbmtoperc_d && $dbmtoperc_d<=85)); then dbmtoperc_r=$((50-($dbmtoperc_d-70)*2))
elif ((86<=$dbmtoperc_d && $dbmtoperc_d<=88)); then dbmtoperc_r=$((17-($dbmtoperc_d-86)*2))
elif ((89<=$dbmtoperc_d && $dbmtoperc_d<=91)); then dbmtoperc_r=$((10-($dbmtoperc_d-89)*2))
elif ((92==$dbmtoperc_d)); then dbmtoperc_r=3
elif ((93<=$dbmtoperc_d)); then dbmtoperc_r=1; fi
fi
echo $dbmtoperc_r
}
Usage:
echo $(dbmtoperc -48)% # returns 81%
Airodump RXQ is really usefull in the real world conditions...
"Receive Quality as measured by the percentage of packets (management and data frames) successfully received over the last 10 seconds."
"Its measured over all management and data frames. The received frames contain a sequence number which is added by the sending access point. RXQ = 100 means that all packets were received from the access point in numerical sequence and none were missing. That's the clue, this allows you to read more things out of this value. Lets say you got 100 percent RXQ and all 10 (or whatever the rate) beacons per second coming in. Now all of a sudden the RXQ drops below 90, but you still capture all sent beacons. Thus you know that the AP is sending frames to a client but you can't hear the client nor the AP sending to the client (need to get closer)."
Related
Check out this question:
Swift probability of random number being selected?
The top answer suggests to use a switch statement, which does the job. However, if I have a very large number of cases to consider, the code looks very inelegant; I have a giant switch statement with very similar code in each case repeated over and over again.
Is there a nicer, cleaner way to pick a random number with a certain probability when you have a large number of probabilities to consider? (like ~30)
This is a Swift implementation strongly influenced by the various
answers to Generate random numbers with a given (numerical) distribution.
For Swift 4.2/Xcode 10 and later (explanations inline):
func randomNumber(probabilities: [Double]) -> Int {
// Sum of all probabilities (so that we don't have to require that the sum is 1.0):
let sum = probabilities.reduce(0, +)
// Random number in the range 0.0 <= rnd < sum :
let rnd = Double.random(in: 0.0 ..< sum)
// Find the first interval of accumulated probabilities into which `rnd` falls:
var accum = 0.0
for (i, p) in probabilities.enumerated() {
accum += p
if rnd < accum {
return i
}
}
// This point might be reached due to floating point inaccuracies:
return (probabilities.count - 1)
}
Examples:
let x = randomNumber(probabilities: [0.2, 0.3, 0.5])
returns 0 with probability 0.2, 1 with probability 0.3,
and 2 with probability 0.5.
let x = randomNumber(probabilities: [1.0, 2.0])
return 0 with probability 1/3 and 1 with probability 2/3.
For Swift 3/Xcode 8:
func randomNumber(probabilities: [Double]) -> Int {
// Sum of all probabilities (so that we don't have to require that the sum is 1.0):
let sum = probabilities.reduce(0, +)
// Random number in the range 0.0 <= rnd < sum :
let rnd = sum * Double(arc4random_uniform(UInt32.max)) / Double(UInt32.max)
// Find the first interval of accumulated probabilities into which `rnd` falls:
var accum = 0.0
for (i, p) in probabilities.enumerated() {
accum += p
if rnd < accum {
return i
}
}
// This point might be reached due to floating point inaccuracies:
return (probabilities.count - 1)
}
For Swift 2/Xcode 7:
func randomNumber(probabilities probabilities: [Double]) -> Int {
// Sum of all probabilities (so that we don't have to require that the sum is 1.0):
let sum = probabilities.reduce(0, combine: +)
// Random number in the range 0.0 <= rnd < sum :
let rnd = sum * Double(arc4random_uniform(UInt32.max)) / Double(UInt32.max)
// Find the first interval of accumulated probabilities into which `rnd` falls:
var accum = 0.0
for (i, p) in probabilities.enumerate() {
accum += p
if rnd < accum {
return i
}
}
// This point might be reached due to floating point inaccuracies:
return (probabilities.count - 1)
}
Is there a nicer, cleaner way to pick a random number with a certain probability when you have a large number of probabilities to consider?
Sure. Write a function that generates a number based on a table of probabilities. That's essentially what the switch statement you've pointed to is: a table defined in code. You could do the same thing with data using a table that's defined as a list of probabilities and outcomes:
probability outcome
----------- -------
0.4 1
0.2 2
0.1 3
0.15 4
0.15 5
Now you can pick a number between 0 and 1 at random. Starting from the top of the list, add up probabilities until you've exceeded the number you picked, and use the corresponding outcome. For example, let's say the number you pick is 0.6527637. Start at the top: 0.4 is smaller, so keep going. 0.6 (0.4 + 0.2) is smaller, so keep going. 0.7 (0.6 + 0.1) is larger, so stop. The outcome is 3.
I've kept the table short here for the sake of clarity, but you can make it as long as you like, and you can define it in a data file so that you don't have to recompile when the list changes.
Note that there's nothing particularly specific to Swift about this method -- you could do the same thing in C or Swift or Lisp.
This seems like a good opportunity for a shameless plug to my small library, swiftstats:
https://github.com/r0fls/swiftstats
For example, this would generate 3 random variables from a normal distribution with mean 0 and variance 1:
import SwiftStats
let n = SwiftStats.Distributions.Normal(0, 1.0)
print(n.random())
Supported distributions include: normal, exponential, binomial, etc...
It also supports fitting sample data to a given distribution, using the Maximum Likelihood Estimator for the distribution.
See the project readme for more info.
You could do it with exponential or quadratic functions - have x be your random number, take y as the new random number. Then, you just have to jiggle the equation until it fits your use case. Say I had (x^2)/10 + (x/300). Put your random number in, (as some floating-point form), and then get the floor with Int() when it comes out. So, if my random number generator goes from 0 to 9, I have a 40% chance of getting 0, and a 30% chance of getting 1 - 3, a 20% chance of getting 4 - 6, and a 10% chance of an 8. You're basically trying to fake some kind of normal distribution.
Here's an idea of what it would look like in Swift:
func giveY (x: UInt32) -> Int {
let xD = Double(x)
return Int(xD * xD / 10 + xD / 300)
}
let ans = giveY (arc4random_uniform(10))
EDIT:
I wasn't very clear above - what I meant was you could replace the switch statement with some function that would return a set of numbers with a probability distribution that you could figure out with regression using wolfram or something. So, for the question you linked to, you could do something like this:
import Foundation
func returnLevelChange() -> Double {
return 0.06 * exp(0.4 * Double(arc4random_uniform(10))) - 0.1
}
newItemLevel = oldItemLevel * returnLevelChange()
So that function returns a double somewhere between -0.05 and 2.1. That would be your "x% worse/better than current item level" figure. But, since it's an exponential function, it won't return an even spread of numbers. The arc4random_uniform(10) returns an int from 0 - 9, and each of those would result in a double like this:
0: -0.04
1: -0.01
2: 0.03
3: 0.1
4: 0.2
5: 0.34
6: 0.56
7: 0.89
8: 1.37
9: 2.1
Since each of those ints from the arc4random_uniform has an equal chance of showing up, you get probabilities like this:
40% chance of -0.04 to 0.1 (~ -5% - 10%)
30% chance of 0.2 to 0.56 (~ 20% - 55%)
20% chance of 0.89 to 1.37 (~ 90% - 140%)
10% chance of 2.1 (~ 200%)
Which is something similar to the probabilities that other person had. Now, for your function, it's much more difficult, and the other answers are almost definitely more applicable and elegant. BUT you could still do it.
Arrange each of the letters in order of their probability - from largest to smallest. Then, get their cumulative sums, starting with 0, without the last. (so probabilities of 50%, 30%, 20% becomes 0, 0.5, 0.8). Then you multiply them up until they're integers with reasonable accuracy (0, 5, 8). Then, plot them - your cumulative probabilities are your x's, the things you want to select with a given probability (your letters) are your y's. (you obviously can't plot actual letters on the y axis, so you'd just plot their indices in some array). Then, you'd try find some regression there, and have that be your function. For instance, trying those numbers, I got
e^0.14x - 1
and this:
let letters: [Character] = ["a", "b", "c"]
func randLetter() -> Character {
return letters[Int(exp(0.14 * Double(arc4random_uniform(10))) - 1)]
}
returns "a" 50% of the time, "b" 30% of the time, and "c" 20% of the time. Obviously pretty cumbersome for more letters, and it would take a while to figure out the right regression, and if you wanted to change the weightings you're have to do it manually. BUT if you did find a nice equation that did fit your values, the actual function would only be a couple lines long, and fast.
Check out this question:
Swift probability of random number being selected?
The top answer suggests to use a switch statement, which does the job. However, if I have a very large number of cases to consider, the code looks very inelegant; I have a giant switch statement with very similar code in each case repeated over and over again.
Is there a nicer, cleaner way to pick a random number with a certain probability when you have a large number of probabilities to consider? (like ~30)
This is a Swift implementation strongly influenced by the various
answers to Generate random numbers with a given (numerical) distribution.
For Swift 4.2/Xcode 10 and later (explanations inline):
func randomNumber(probabilities: [Double]) -> Int {
// Sum of all probabilities (so that we don't have to require that the sum is 1.0):
let sum = probabilities.reduce(0, +)
// Random number in the range 0.0 <= rnd < sum :
let rnd = Double.random(in: 0.0 ..< sum)
// Find the first interval of accumulated probabilities into which `rnd` falls:
var accum = 0.0
for (i, p) in probabilities.enumerated() {
accum += p
if rnd < accum {
return i
}
}
// This point might be reached due to floating point inaccuracies:
return (probabilities.count - 1)
}
Examples:
let x = randomNumber(probabilities: [0.2, 0.3, 0.5])
returns 0 with probability 0.2, 1 with probability 0.3,
and 2 with probability 0.5.
let x = randomNumber(probabilities: [1.0, 2.0])
return 0 with probability 1/3 and 1 with probability 2/3.
For Swift 3/Xcode 8:
func randomNumber(probabilities: [Double]) -> Int {
// Sum of all probabilities (so that we don't have to require that the sum is 1.0):
let sum = probabilities.reduce(0, +)
// Random number in the range 0.0 <= rnd < sum :
let rnd = sum * Double(arc4random_uniform(UInt32.max)) / Double(UInt32.max)
// Find the first interval of accumulated probabilities into which `rnd` falls:
var accum = 0.0
for (i, p) in probabilities.enumerated() {
accum += p
if rnd < accum {
return i
}
}
// This point might be reached due to floating point inaccuracies:
return (probabilities.count - 1)
}
For Swift 2/Xcode 7:
func randomNumber(probabilities probabilities: [Double]) -> Int {
// Sum of all probabilities (so that we don't have to require that the sum is 1.0):
let sum = probabilities.reduce(0, combine: +)
// Random number in the range 0.0 <= rnd < sum :
let rnd = sum * Double(arc4random_uniform(UInt32.max)) / Double(UInt32.max)
// Find the first interval of accumulated probabilities into which `rnd` falls:
var accum = 0.0
for (i, p) in probabilities.enumerate() {
accum += p
if rnd < accum {
return i
}
}
// This point might be reached due to floating point inaccuracies:
return (probabilities.count - 1)
}
Is there a nicer, cleaner way to pick a random number with a certain probability when you have a large number of probabilities to consider?
Sure. Write a function that generates a number based on a table of probabilities. That's essentially what the switch statement you've pointed to is: a table defined in code. You could do the same thing with data using a table that's defined as a list of probabilities and outcomes:
probability outcome
----------- -------
0.4 1
0.2 2
0.1 3
0.15 4
0.15 5
Now you can pick a number between 0 and 1 at random. Starting from the top of the list, add up probabilities until you've exceeded the number you picked, and use the corresponding outcome. For example, let's say the number you pick is 0.6527637. Start at the top: 0.4 is smaller, so keep going. 0.6 (0.4 + 0.2) is smaller, so keep going. 0.7 (0.6 + 0.1) is larger, so stop. The outcome is 3.
I've kept the table short here for the sake of clarity, but you can make it as long as you like, and you can define it in a data file so that you don't have to recompile when the list changes.
Note that there's nothing particularly specific to Swift about this method -- you could do the same thing in C or Swift or Lisp.
This seems like a good opportunity for a shameless plug to my small library, swiftstats:
https://github.com/r0fls/swiftstats
For example, this would generate 3 random variables from a normal distribution with mean 0 and variance 1:
import SwiftStats
let n = SwiftStats.Distributions.Normal(0, 1.0)
print(n.random())
Supported distributions include: normal, exponential, binomial, etc...
It also supports fitting sample data to a given distribution, using the Maximum Likelihood Estimator for the distribution.
See the project readme for more info.
You could do it with exponential or quadratic functions - have x be your random number, take y as the new random number. Then, you just have to jiggle the equation until it fits your use case. Say I had (x^2)/10 + (x/300). Put your random number in, (as some floating-point form), and then get the floor with Int() when it comes out. So, if my random number generator goes from 0 to 9, I have a 40% chance of getting 0, and a 30% chance of getting 1 - 3, a 20% chance of getting 4 - 6, and a 10% chance of an 8. You're basically trying to fake some kind of normal distribution.
Here's an idea of what it would look like in Swift:
func giveY (x: UInt32) -> Int {
let xD = Double(x)
return Int(xD * xD / 10 + xD / 300)
}
let ans = giveY (arc4random_uniform(10))
EDIT:
I wasn't very clear above - what I meant was you could replace the switch statement with some function that would return a set of numbers with a probability distribution that you could figure out with regression using wolfram or something. So, for the question you linked to, you could do something like this:
import Foundation
func returnLevelChange() -> Double {
return 0.06 * exp(0.4 * Double(arc4random_uniform(10))) - 0.1
}
newItemLevel = oldItemLevel * returnLevelChange()
So that function returns a double somewhere between -0.05 and 2.1. That would be your "x% worse/better than current item level" figure. But, since it's an exponential function, it won't return an even spread of numbers. The arc4random_uniform(10) returns an int from 0 - 9, and each of those would result in a double like this:
0: -0.04
1: -0.01
2: 0.03
3: 0.1
4: 0.2
5: 0.34
6: 0.56
7: 0.89
8: 1.37
9: 2.1
Since each of those ints from the arc4random_uniform has an equal chance of showing up, you get probabilities like this:
40% chance of -0.04 to 0.1 (~ -5% - 10%)
30% chance of 0.2 to 0.56 (~ 20% - 55%)
20% chance of 0.89 to 1.37 (~ 90% - 140%)
10% chance of 2.1 (~ 200%)
Which is something similar to the probabilities that other person had. Now, for your function, it's much more difficult, and the other answers are almost definitely more applicable and elegant. BUT you could still do it.
Arrange each of the letters in order of their probability - from largest to smallest. Then, get their cumulative sums, starting with 0, without the last. (so probabilities of 50%, 30%, 20% becomes 0, 0.5, 0.8). Then you multiply them up until they're integers with reasonable accuracy (0, 5, 8). Then, plot them - your cumulative probabilities are your x's, the things you want to select with a given probability (your letters) are your y's. (you obviously can't plot actual letters on the y axis, so you'd just plot their indices in some array). Then, you'd try find some regression there, and have that be your function. For instance, trying those numbers, I got
e^0.14x - 1
and this:
let letters: [Character] = ["a", "b", "c"]
func randLetter() -> Character {
return letters[Int(exp(0.14 * Double(arc4random_uniform(10))) - 1)]
}
returns "a" 50% of the time, "b" 30% of the time, and "c" 20% of the time. Obviously pretty cumbersome for more letters, and it would take a while to figure out the right regression, and if you wanted to change the weightings you're have to do it manually. BUT if you did find a nice equation that did fit your values, the actual function would only be a couple lines long, and fast.
Is there any algorithm for the peak finder in numeric signal? For example if we have:
x = [ 0.01 -0.5 0.02 0.5 0.003 0.8 1 0 0 0 -1 0 0.0001 0 0 0 ]
As you see we have 5 peak sample:
( -0.5 , 0.5 , 0.8 , 1 , -1 )
These peaks must capture 80 % of energy of signal x?
In the comments you finally mention you want to do this in Matlab.
Well, have this Pseudo-code instead:
power = x .^2 #element-wise squaring
total = sum(power)
sorted, indices = sort(power) # sort and note down permutation indices
max_indices = []
max_sum = 0
while sum < total * 0.8:
currentpower = pop(sorted) # get the highest power
currentindex = pop(indices)
max_sum+=currentpower
max_indices.append(currentindex)
return max_indices
I'm trying the second day to send a midi signal. I'm using following code:
int pitchValue = 8191 //or -8192;
int msb = ?;
int lsb = ?;
UInt8 midiData[] = { 0xe0, msb, lsb};
[midi sendBytes:midiData size:sizeof(midiData)];
I don't understand how to calculate msb and lsb. I tried pitchValue << 8. But it's working incorrect, When I'm looking to events using midi tool I see min -8192 and +8064 max. I want to get -8192 and +8191.
Sorry if question is simple.
Pitch bend data is offset to avoid any sign bit concerns. The maximum negative deviation is sent as a value of zero, not -8192, so you have to compensate for that, something like this Python code:
def EncodePitchBend(value):
''' return a 2-tuple containing (msb, lsb) '''
if (value < -8192) or (value > 8191):
raise ValueError
value += 8192
return (((value >> 7) & 0x7F), (value & 0x7f))
Since MIDI data bytes are limited to 7 bits, you need to split pitchValue into two 7-bit values:
int msb = (pitchValue + 8192) >> 7 & 0x7F;
int lsb = (pitchValue + 8192) & 0x7F;
Edit: as #bgporter pointed out, pitch wheel values are offset by 8192 so that "zero" (i.e. the center position) is at 8192 (0x2000) so I edited my answer to offset pitchValue by 8192.
If I am given a floating point number but do not know beforehand what range the number will be in, is it possible to scale that number in some meaningful way to be in another range? I am thinking of checking to see if the number is in the range 0<=x<=1 and if not scale it to that range and then scale it to my final range. This previous post provides some good information, but it assumes the range of the original number is known beforehand.
You can't scale a number in a range if you don't know the range.
Maybe what you're looking for is the modulo operator. Modulo is basically the remainder of division, the operator in most languages is is %.
0 % 5 == 0
1 % 5 == 1
2 % 5 == 2
3 % 5 == 3
4 % 5 == 4
5 % 5 == 0
6 % 5 == 1
7 % 5 == 2
...
Sure it is not possible. You can define range and ignore all extrinsic values. Or, you can collect statistics to find range in run time (i.e. via histogram analysis).
Is it really about image processing? There are lots of related problems in image segmentation field.
You want to scale a single random floating point number to be between 0 and 1, but you don't know the range of the number?
What should 99.001 be scaled to? If the range of the random number was [99, 100], then our scaled-number should be pretty close to 0. If the range of the random number was [0, 100], then our scaled-number should be pretty close to 1.
In the real world, you always have some sort of information about the range (either the range itself, or how wide it is). Without further info, the answer is "No, it can't be done."
I think the best you can do is something like this:
int scale(x) {
if (x < -1) return 1 / x - 2;
if (x > 1) return 2 - 1 / x;
return x;
}
This function is monotonic, and has a range of -2 to 2, but it's not strictly a scaling.
I am assuming that you have the result of some 2-dimensional measurements and want to display them in color or grayscale. For that, I would first want to find the maximum and minimum and then scale between these two values.
static double[][] scale(double[][] in, double outMin, double outMax) {
double inMin = Double.POSITIVE_INFINITY;
double inMax = Double.NEGATIVE_INFINITY;
for (double[] inRow : in) {
for (double d : inRow) {
if (d < inMin)
inMin = d;
if (d > inMax)
inMax = d;
}
}
double inRange = inMax - inMin;
double outRange = outMax - outMin;
double[][] out = new double[in.length][in[0].length];
for (double[] inRow : in) {
double[] outRow = new double[inRow.length];
for (int j = 0; j < inRow.length; j++) {
double normalized = (inRow[j] - inMin) / inRange; // 0 .. 1
outRow[j] = outMin + normalized * outRange;
}
}
return out;
}
This code is untested and just shows the general idea. It further assumes that all your input data is in a "reasonable" range, away from infinity and NaN.