Arduino map
Why it may not work exactly the way you think it does! The
map function is intended to change one range of values into another range
of values and a common use is to read an analogue input (10 bits long, so values range from 0 to 1023) and change the output to a byte so the output would be from 0 to 255.
You would then write the following code
Everything is fine 0 maps to 0, and 1023 maps to 255...
...with an even distribution (really?????).
The map() function is useful but it has a secret hiding within. Its
easy
enough to use (or so you think) until you look a bit deeper inside. An
even distribution is what you want but does it really do that?
Lets write a sketch to test out exactly what it does
void setup() {
Serial.begin(9600);
for (int adc = 0; adc < 1024; adc++) {
int mapped = map(adc, 0, 1023, 0, 255);
Serial.print(adc);
Serial.print(',');
Serial.println(mapped);
}
}
void loop() {
}
First part of the serial output (i/p value, o/p value) |
Final part of the serial output (i/p value, o/p value) |
0,0 1,0 2,0 3,0 |
1007,251 1008,251 1009,251 1010,251 |
4,0 5,1 6,1 7,1 |
1011,252 1012,252 1013,252 1014,252 |
8,1 9,2 10,2 11,2 |
1015,253 1016,253 1017,253 1018,253 |
12,2 13,3 14,3 15,3 |
1019,254 1020,254 1021,254 1022,254 |
1023,255 |
In the table above the value to the left of the comma is the adc value while to the right is the mapped output value.
You can see from the table above that input is mapped to output ranges in blocks of 4 (1 output value for a range of 4 input values). This is expected since 1024/4 = 256.
The problem is the final output value (highlighted in red) has only 1
output bin for one input value i.e. 1023 results in 255, while
1019~1022 results in 254 as output (4 input bins).
What you really want is an even spread of values across the whole range.
To get to this point some of the other outputs must have had 5 values
as inputs (you can see that for adc values 0~4 - all 5 inputs result in
output of a zero.
This sketch increases the value in bins array every time map returns a
value - so each bins[] holds the number of times an output was created.
#define SAMPLES 1024
#define OUTPUTS 256
int bins[OUTPUTS];
const int binsize = (SAMPLES-1)/(OUTPUTS-1);
void setup() {
Serial.begin(9600);
Serial.println("Arduino map output distribution");
Serial.print("Bin size: ");
Serial.println(binsize);
// Initialise bins
for (int i = 0; i < OUTPUTS; i++) bins[i]=0;
for (int adc = 0; adc < SAMPLES; adc++) {
int mapped = map(adc, 0, SAMPLES-1, 0, OUTPUTS-1);
bins[mapped] +=1;
}
for(int i=0;i<OUTPUTS;i++) {
Serial.print("Bin: ");
Serial.print(i);
Serial.print(" count ");
Serial.print(bins[i]);
// Show bin distribution error
if ((bins[i] != binsize)) {
Serial.println(" ***") ;
} else Serial.println();
}
// Find wrong bin count
for(int i=0;i<OUTPUTS;i++) {
if (bins[i]!=binsize) {
Serial.print("Bin error: ");
Serial.println(i);
}
}
}
void loop() {
}
The last part of the output shows four bins with wrong values :
Bin: 253 count 4 Bin: 254 count 4 Bin: 255 count 1 *** Bin error: 0 Bin error: 85 Bin error: 170 Bin error: 255
The arduino map() reference has this to say:
"As previously mentioned, the map() function uses integer math. So fractions might get suppressed due to this. For example, fractions like 3/2, 4/3, 5/4 will all be returned as 1 from the map() function, despite their different actual values. So if your project requires precise calculations (e.g. voltage accurate to 3 decimal places), please consider avoiding map() and implementing the calculations manually in your code yourself."
The problem is that the upper value of 1023 is not exactly divisible
by 255 so you get a slight error. So it is a fraction that is
suppressed.
The code for the function is:
long map(long x, long in_min, long in_max, long out_min, long out_max) {
return (x - in_min) * (out_max - out_min) / (in_max - in_min) + out_min;
}
The calculations are
0 *255/1023 = 0 1 *255/1023 = 0 2 *255/1023 = 0 3 *255/1023 = 0 4 *255/1023 = 0 5 *255/1023 = 1 1022 * 255/1023 = 254 1023 * 255/1023 = 255
These match the Arduino outputs.
The problem becomes even more apparent.
if you map 1023 to the output range 0 to 7 the problem becomes even
more apparent.
Change the code in the previous sketch redefining 'OUTPUTS' as 8.
The output then becomes:
Arduino map output distribution Bin size: 146 Bin: 0 count 147 *** Bin: 1 count 146 Bin: 2 count 146 Bin: 3 count 146 Bin: 4 count 146 Bin: 5 count 146 Bin: 6 count 146 Bin: 7 count 1 *** Bin error: 0 Bin error: 7
If you tried to map an ADC input using a potentiometer to control an 8
LED bargraph the only time the last LED would light is when the input is
1023!
To fix it make the values a multiple of a power of 2 so the fraction is not suppressed.
Note: For different ranges the difference between max and min must be a power of 2.
The following program uses power of 2 values as input and shows the corrected output.
//#define NUM_SAMPLES 1024 // Must be a power of 2 (Range 0-1023)
//#define NUM_OUTPUTS 256 // Must be a power of 2 (Range 0-255)
#define NUM_SAMPLES 1024 // Must be a power of 2 (Range 0-1023)
#define NUM_OUTPUTS 8 // Must be a power of 2 (Range 0-7)
#define NUM_BINS (NUM_OUTPUTS)
int bins[NUM_BINS];
const int binsize = (NUM_SAMPLES)/(NUM_OUTPUTS);
void setup() {
Serial.begin(9600);
Serial.println("Arduino map output distribution");
Serial.print("Bin size: ");
Serial.println(binsize);
// Initialise bins
for (int i = 0; i < NUM_BINS; i++) bins[i]=0;
for (int adc = 0; adc < NUM_SAMPLES; adc++) {
int mapped = map(adc, 0, NUM_SAMPLES, 0, NUM_OUTPUTS);
bins[mapped] +=1;
}
for(int i=0;i< NUM_BINS;i++) {
Serial.print("Bin: ");
Serial.print(i);
Serial.print(" count ");
Serial.print(bins[i]);
// Show bin distribution error
if ((bins[i] != binsize)) {
Serial.println(" ***") ;
} else Serial.println();
}
// Find wrong bin count
for(int i=0;i< NUM_BINS;i++) {
if (bins[i]!=binsize) {
Serial.print("Bin error: ");
Serial.println(i);
}
}
}
void loop() {
}
Output results: for NUM_SAMPLES = 1024, NUM_OUTPUTS = 8
Arduino map output distribution Bin size: 128 Bin: 0 count 128 Bin: 1 count 128 Bin: 2 count 128 Bin: 3 count 128 Bin: 4 count 128 Bin: 5 count 128 Bin: 6 count 128 Bin: 7 count 128
Note: The first example is also corrected with an even spread of 4 input values per output.
Another example
Here's another example using random samples and outputs i.e. non-power of 2 values.
#define SAMPLES 1976
#define OUTPUTS 153
Here 1976/153 = 12.91 and each bin has 13 values in it except the
last that has one. So the algorithm gets it approximately right.
First part of the serial output |
Final part of the serial output |
Bin: 0 count 13 *** Bin: 1 count 13 *** Bin: 2 count 13 *** Bin: 3 count 13 *** Bin: 4 count 13 *** Bin: 5 count 13 *** Bin: 6 count 13 *** Bin: 7 count 13 *** |
Bin: 145 count 13 *** Bin: 146 count 13 *** Bin: 147 count 13 *** Bin: 148 count 13 *** Bin: 149 count 13 *** Bin: 150 count 13 *** Bin: 151 count 12 Bin: 152 count 1 *** |
Given it is such a pain to get working exactly right, its probably better to do one of the following.
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