using a LDR analogue read and converting it to percentage

hello,
I have set up a circuit which gets the analogue read from an LDR. I get the values ok but how would I convert these values to a percentage? I have seen calculations like (5v-LDR reading)/1023 what is this calculating? I know the light depends on the resister you use in the voltage divider but a percentage is easier to display and makes a bit more sense than a number from 0-1023.

here is my code for temp, humidity and light sensor.

#include <dht.h>   // including DHT11 libary

dht DHT;

int dht_dpin = A0;   // Data pin for DHT11is connected to pin A0
int LDRPin= A2;   // LDR pin for reADING IN A2

int x = 0;
int row = 0;
 
void setup(){
 
  Serial.begin(9600);
 
  delay(300);//Let system settle
 
  Serial.println("Humidity and temperature\n\n");   // Print to serial monitor
 
  delay(1000);   //Wait rest of 1000ms recommended delay before
  
  Serial.println("LABEL, Time, H %, T (C), Light");   // Print headers for serial monitor
 
 }
 
void loop(){
  
  int LDRReading = analogRead(LDRPin);   // Reading value from LDR
 
 DHT.read11(dht_dpin);
 
    Serial.print("DATA,TIME,");
 
     Serial.print(DHT.humidity); Serial.print(",");
 
     Serial.print(DHT.temperature);  Serial.print(",");
    
    Serial.println(LDRReading);
    
    
    row++;   // for PLX-DAQ
    x++;   // for PLX-DAQ
 
  delay(800);   // Read values every 800 ms
 
}

I know the light depends on the resister you use in the voltage divider but a percentage is easier to display and makes a bit more sense than a number from 0-1023.

In a voltage divider circuit, you won't get a range of values from 0 to 1023. You need to determine the upper limit of values that you do get with your circuit.

Then, the value that you get, times 100, divided by the upper limit will be the percentage. Do the multiplication first. Otherwise, n/upperLimit will be 0.

By upper limit do you mean but the brightest light on and the value I get is max and the darkest is min?