Issue
I am trying to use the pytesseract
to detect the text from this image:
Specifically, I care more about detecting Commercial break in progress. I used the following code to achieve this:
from PIL import Image
import pytesseract
print(pytesseract.image_to_string(Image.open('/home/me/Desktop/dataset/my-dataset/Apex-Legends/loustreams_001.jpg')))
However, it returns:
nercial break in progress
I know I shouldn't expect SoTA result from one line of code but the text is very visible. How can I improve this?
Solution
You can use image_to_data
to get the "Commercial break in progress" string.
Convert image to rgb
Get output as a dictionary
If detected text pass the threshold confidence level, display.
Code:
import cv2
from pytesseract import *
img = cv2.imread("6cNav.jpg")
rgb = cv2.cvtColor(img, cv2.COLOR_BGR2RGB)
res = image_to_data(rgb, output_type=Output.DICT)
out_txt = ""
for i in range(0, len(res["text"])):
x = res["left"][i]
y = res["top"][i]
w = res["width"][i]
h = res["height"][i]
txt = res["text"][i]
cnf = int(res["conf"][i])
if cnf > 95:
text = "".join(txt).strip()
cv2.rectangle(img,
(x, y),
(x + w, y + h),
(0, 0, 255), 2)
cv2.putText(img,
text,
(x, y - 10),
cv2.FONT_HERSHEY_SIMPLEX,
1.2, (0, 255, 255), 3)
out_txt += " " + text
print(out_txt)
cv2.imshow("img", img)
cv2.waitKey(0)
Output:
Commercial break in progress
Please note that my pytesseract
version is 4.1.1
Updated Code
For the all the images above, you can apply adaptive-threshold
- (1st and the 3rd image is also similar to the above)
the result will be:
output 1:
Commercial loreak in progress
output 2:
Commercial break in progress
output 3:
Commercial break in progress
Code:
import cv2
import pytesseract
img1 = cv2.imread("6cNav.jpg")
img2 = cv2.imread("IKHKa.jpg")
img3 = cv2.imread("VTJse.jpg")
img_lst = [img1, img2, img3]
for i, img in enumerate(img_lst):
gry = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
thr = cv2.adaptiveThreshold(gry, 255, cv2.ADAPTIVE_THRESH_GAUSSIAN_C,
cv2.THRESH_BINARY_INV, 11, 2)
res = pytesseract.image_to_string(thr, config="--psm 6")
print("output {}:".format(i+1))
print(res)
Answered By - Ahx
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