LCD Horizontal bargraph like one black rectangle

Projects Programming
3

Thanks!
From internet possible code:

/*
  Voltmeter with graph
  Coded by Ollie Jeffcoate
*/
#include <LiquidCrystal.h>    

LiquidCrystal lcd(2, 3, 4, 5, 6, 7); //initialize the library, numbers of the interface pins

  int analogInput = 1;     //analog pin A1
  float VoltageOut = 0.0;
  float VoltageIn = 0.0;
  int refresh = 50;        //refresh rate
  int i = 0;               //variable used for clearing our byte 'graph'
  int value = 0;           //variable to store value
  byte graph[8] = 
    {
      B11111,
      B11111,
      B11111,
      B11111,
      B11111,
      B11111,
      B11111,
      B11111,
    };

void setup()
{
  pinMode(analogInput, INPUT);  //setting analog pin mode to input
  
  lcd.begin(16, 2);             //setting up the LCD screens rows and colums
  lcd.print("Voltage=");        
}

void loop(){
  value = analogRead(analogInput);  //reading the value on A1

  //VoltageOut = (value * 5.0) / 1024.0;
  VoltageIn = (value * 5.0) / 1024.0; 
  //VoltageOut / (R2/(R1+R2));  

  lcd.setCursor(8, 0);               //printing the result to the LCD display
  lcd.print(VoltageIn);
  lcd.print(" V");
  delay(refresh);                    //refreshing the screen
  
  voltmetergraph();                  //calling my function voltmetergraph 
}

void voltmetergraph()
  { 
    if (VoltageIn >= 0.3125)         //setting up the bar graph
    {
    lcd.createChar(0, graph);
    lcd.setCursor(0, 2);
    lcd.write(byte(0));}
    else {LcdClearByte(0);}   
   
   if (VoltageIn >= 0.625)
    {
    lcd.createChar(0, graph);
    lcd.setCursor(1, 2);
    lcd.write(byte(0));}
    else {LcdClearByte(1);}  
   
    if (VoltageIn >= .9375)
    {
    lcd.createChar(0, graph);
    lcd.setCursor(2, 2);
    lcd.write(byte(0));}
    else {LcdClearByte(2);} 
    
    if (VoltageIn >= 1.25){
    lcd.createChar(0, graph);
    lcd.setCursor(3, 2);
    lcd.write(byte(0));}
    else {LcdClearByte(3);} 
   
   if (VoltageIn >= 1.5625)
    {
    lcd.createChar(0, graph);
    lcd.setCursor(4, 2);
    lcd.write(byte(0));}
    else {LcdClearByte(4);} 
   
   if (VoltageIn >= 1.875)
    {
    lcd.createChar(0, graph);
    lcd.setCursor(5, 2);
    lcd.write(byte(0));}
    else {LcdClearByte(5);} 
    
   if (VoltageIn >= 2.1875)
    {
    lcd.createChar(0, graph);
    lcd.setCursor(6, 2);
    lcd.write(byte(0));}
    else {LcdClearByte(6);} 
    
    if (VoltageIn >= 2.5)
    {
    lcd.createChar(0, graph);
    lcd.setCursor(7, 2);
    lcd.write(byte(0));}
    else {LcdClearByte(7);} 
    
    if (VoltageIn >= 2.8125)
    {
    lcd.createChar(0, graph);
    lcd.setCursor(8, 2);
    lcd.write(byte(0));}
    else {LcdClearByte(8);}
    
   if (VoltageIn >= 3.125)
    {
    lcd.createChar(0, graph);
    lcd.setCursor(9, 2);
    lcd.write(byte(0));}
    else {LcdClearByte(9);}
    
   if (VoltageIn >= 3.4375)
    {
    lcd.createChar(0, graph);
    lcd.setCursor(10, 2);
    lcd.write(byte(0));}
    else {LcdClearByte(10);}
    
   if (VoltageIn >= 3.75)
    {
    lcd.createChar(0, graph);
    lcd.setCursor(11, 2);
    lcd.write(byte(0));}
    else {LcdClearByte(11);}
    
   if (VoltageIn >= 4.0625)
    {
    lcd.createChar(0, graph);
    lcd.setCursor(12, 2);
    lcd.write(byte(0));}
    else {LcdClearByte(12);}
    
   if (VoltageIn >= 4.375)
    {
    lcd.createChar(0, graph);
    lcd.setCursor(13, 2);
    lcd.write(byte(0));}
    else {LcdClearByte(13);}
    
    if (VoltageIn >= 4.6875)
    {
    lcd.createChar(0, graph);
    lcd.setCursor(14, 2);
    lcd.write(byte(0));}
    else {LcdClearByte(14);}
    
    if (VoltageIn >= 4.99)
    {
    lcd.createChar(0, graph);
    lcd.setCursor(15, 2);
    lcd.write(byte(0));}
    else {LcdClearByte(15);}
  }
  
 void LcdClearByte(int c)        //clearbyte function
{
  lcd.setCursor(c,1);
  for (i = 0; i < 16; i = i + 1) {
    lcd.print(" ");
  }
}

But cannot delete left-handed unwanted black boxes. lcd.clear() wasting time? How to clear unwanted left-handed boxes?