#bit messy to say the least will put cleaner version in separate space 
def imgOCR_img2text(imgFilename):
  import easyocr
  #reader = easyocr.Reader(['en'], gpu=True) #GPU inference - faster and more accurate but need GPU. Enable and try/excpet CPU users down to CPU
  reader = easyocr.Reader(['en'], gpu=False) #CPU inference - slower and less accurate

  '''
  try:
      reader = easyocr.Reader(['en'], gpu=True) #GPU inference - faster and more accurate but need GPU. Enable and try/except CPU users down to CPU
  except:
      reader = easyocr.Reader(['en'], gpu=False) #CPU inference - slower and less accurate
  '''

  # Create a reader to do OCR.
  # If you change to GPU instance, it will be faster. But CPU is enough.
  # (by MENU > Runtime > Change runtime type > GPU, then redo from beginning )
  #import easyocr
  #reader = easyocr.Reader(['en'], gpu=True)

  # Doing OCR. Get bounding boxes.
  bounds2 = reader.readtext(imgFilename) #'writing_demo1.png'
  #bounds2 = reader.readtext('writing_demo1.png', detail = 0) # detail = 0 turns off details, ie coordinates of bounding boxes and just returns the text 

  OCRbox = []
  for kk in range(len(bounds2)): #don't want to alter original with the operations below
    OCRbox.append( bounds2[kk] )

  def getX1ofBoundingBox(inputArray1): # inputArray1 = bounds2[kk]
    boundingX1 = (inputArray1[0])[0][0]
    return boundingX1

  def getY1ofBoundingBox(inputArray2): # inputArray2 = bounds2[kk]
    boundingY1 = (inputArray2[0])[0][1]
    return boundingY1

  def getX3ofBoundingBox(inputArray3): # inputArray3 = bounds2[kk]
    boundingX3 = (inputArray3[0])[2][0]
    return boundingX3

  def getY3ofBoundingBox(inputArray4): # inputArray4 = bounds2[kk]
    boundingY3 = (inputArray4[0])[2][1]
    return boundingY3


  def get_XcentroidCoordinate_ofBoundingBox(inputArray5): # inputArray5 = bounds2[kk]
    x1_0 = getX1ofBoundingBox(inputArray5)
    x3_0 = getX3ofBoundingBox(inputArray5)
    
    x_centroid0 = ( (x3_0 - x1_0) / 2 )  + x1_0
    return x_centroid0

  def get_YcentroidCoordinate_ofBoundingBox(inputArray6): # inputArray6 = bounds2[kk]
    y1_0 = getY1ofBoundingBox(inputArray6)
    y3_0 = getY3ofBoundingBox(inputArray6)
    
    y_centroid0 = ( (y3_0 - y1_0) / 2 )  + y1_0
    return y_centroid0



  for kk in range(len(OCRbox)):
    #bounds2[]
    #OCRbox.sort(key=getY1ofBoundingBox) #Sorts it by Y1 location, see here for use of function key in sort https://www.w3schools.com/python/ref_list_sort.asp
    OCRbox.sort(key=get_YcentroidCoordinate_ofBoundingBox) #Sorts it by Y centroid location

    # [ associatedText, boundingCoordinates ] = [ bounds2[kk][1] , [X1, X3, Y1, Y3] ]

  print( bounds2 )
  print( "Row sorted aka all Y_centroid (or Y1, Y3, whichever we chose to sort by) should be increasing in each new item   :   ", OCRbox )


  listOfRows = []
  minilist = []


  for kk in range(len(OCRbox) - 1):
    minilist.append( OCRbox[kk] )
    if get_YcentroidCoordinate_ofBoundingBox( OCRbox[kk] ) < getY1ofBoundingBox( OCRbox[kk + 1] ):
      listOfRows.append( minilist )
      #print( "this minilist aka row = " , minilist )
      minilist = []
    #minilist.append( OCRbox[kk] )

  print( "listOfRows = ", listOfRows)
  print( "len( listOfRows) = " , len( listOfRows) )
  print( "the final minilist aka row = " , minilist )
  print( "OCRbox[-1] = ", OCRbox[-1] )

  #boundary case for last row. If its a single box we append it as its own row. If not we append it to the last list. 
  if get_YcentroidCoordinate_ofBoundingBox( OCRbox[-2] ) < getY1ofBoundingBox( OCRbox[-1] ): #boundary case in case the last row also happens to be a single box
    listOfRows.append( [OCRbox[-1]] ) #tack on last one that for loop didnt AS ITS OWN LIST
  elif len(listOfRows) < 1: #basically no text or single row detected 
    listOfRows.append( [OCRbox[-1]] )
  else:
    listOfRows[-1].append( OCRbox[-1] ) #tack it onto the last row 


  #def readLeft2RightSort(): #aka English, for Japanese just do Right2Left; Really just an X-centroid sort on each element of list of rows SEPARATELY like we did Y-centroid sort above 

  listOfRows.append( [([[0, 0], [0, 0], [0, 0], [0, 0]], '', 1)] ) #preserve structure in empty case

  for kk in range(len(listOfRows)):
    listOfRows[kk].sort(key=get_XcentroidCoordinate_ofBoundingBox)

  print(listOfRows)
  print(listOfRows[0])
  print(listOfRows[1])
  print(listOfRows[0][0][1])

  rowOfTextList = []

  for kk in range(len(listOfRows)):
    for ii in range(len(listOfRows[kk])):
      rowOfTextString = ''.join(listOfRows[kk][ii][1])
      rowOfTextList.append(rowOfTextString)

  print(rowOfTextList)

  coordinateSortedText = ' '.join(rowOfTextList)

  print(coordinateSortedText)




  def cleanOCRtext(inputString2clean):
    inputString2clean = inputString2clean.replace("_", " ") #replace _ with space
    inputString2clean = inputString2clean.replace("  ", " ") #replace double space with single space
    inputString2clean = inputString2clean.lower()

    #import re #turn 0's that appear in the text into o's, this seems to be the major letter to number error 
    inputString2clean = re.sub("([a-z])[0]", "\\1o", inputString2clean)   #capture [a-z] with parentheses then reference the first capture as \\1 
    inputString2clean = re.sub("[0]([a-z])", "\\1o", inputString2clean)
    
    return inputString2clean

  cleanedText = cleanOCRtext(coordinateSortedText)


  print("============================== FINAL ==============================")
  print(cleanedText)

  return cleanedText