title: How to find numbers and alphabets in images and blur them(mask them)?
body: <p>I have to blur the identification number without doing a hard code. I want to find the identification number on its own and blur that number for example</p>
  <p><img src="https://i.stack.imgur.com/JYznL.jpg" alt="scanned image" /></p>
  <p>In this image I need to blur the Account number i.e ABCDE1234F. This is the code that I have done so far:</p>
  <pre><code># pylint: disable=import-error
  import matplotlib.pyplot as plt
  import cv2
  import easyocr
  from pylab import rcParams
  from IPython.display import Image
  # size of image
  rcParams['figure.figsize'] = 8, 16
  #language to read
  reader = easyocr.Reader(['en'])
  #read imgae
  ##image = cv2.imread('aadhar.jpg')
  image = cv2.imread('pancard.jpg')
  # saved pre trained model in variable called reader and extract text using readtext  and then saved the output in variable called output
  ##output = reader.readtext('aadhar.jpg')
  output = reader.readtext('pancard.jpg')
  print(output)
  # extracted the word which we need to blur and saved the it in the variable name cord
  ##cord = output[6][0]
  cord = output[8][0]
  print(cord)
  # catched up the min and max the cordinates of bounding box
  x_min, y_min = [int(min(idx)) for idx in zip(*cord)]
  x_max, y_max = [int(max(idx)) for idx in zip(*cord)]
  print(x_min)
  print(y_min)
  print(x_max)
  print(y_max)
  
  # Create ROI coordinates(region of interest)
  topLeft = (x_min, y_min)
  bottomRight = (x_max, y_max)
  x, y = topLeft[0], topLeft[1]
  w, h = bottomRight[0] - topLeft[0], bottomRight[1] - topLeft[1]
  
  # Grab ROI with Numpy slicing and blur
  ROI = image[y:y+h, x:x+w]
  blur = cv2.GaussianBlur(ROI, (51,51), 0)
  
  # Insert ROI back into image
  image[y:y+h, x:x+w] = blur
  
  #cv2.imshow('blur', blur)
  cv2.imshow('image', image)
  cv2.waitKey()
  </code></pre>
  <p>I hard code the coordinates of the number which I need to blur. I am able to identify the region but I am not sure how to identify the number on its own and blur the same.</p>