[英]TypeError: Expected Ptr<cv::UMat> for argument 'src'?
[英]Python - tkinter; TypeError: Expected Ptr<cv::UMat> for argument 'src'
我正在嘗試為圖像處理構建圖形界面。 我在上傳 function 時遇到問題。 當我嘗試檢查文件/圖像是否存在並對其進行修改時,我收到此錯誤並且我不知道如何修復它。
TypeError: Expected Ptr<cv::UMat> for argument 'src'
這是我的代碼:
import cv2
import instructions as instructions
from PIL import Image,ImageTk
from tkinter.filedialog import askopenfile
root = tk.Tk()
logo = Image.open('logo.png')
logo = ImageTk.PhotoImage(logo)
logo_label = tk.Label(image = logo)
logo_label.image = logo
logo_label.grid(column=1,row=0)
def upload():
browse.set("loading...")
file = askopenfile(parent=root,mode="rb",title="Choose an image",filetypes =[("JPG file","*.jpg"),("PNG file","*.png"),("JPEG file","*.jpeg")])
if file:
gray = cv2.cvtColor(file, cv2.COLOR_RGB2GRAY)
gray = cv2.medianBlur(gray,5)
edges = cv2.adaptiveThreshold(gray,255,cv2.ADAPTIVE_THRESH_MEAN_C,cv2.THRESH_BINARY, 9, 9)
color = cv2.bilateralFilter(file,9,250,250)
cartoon =cv2.bitwise_and(color,color,mask=edges)
cv2.imshow("Cartoon", cartoon)
intructions = tk.Label(root,text= "Select an image",font = "Raleway")
instructions.grid(columnspan=3,column=0,row=1)
browse = tk.StringVar()
browse_button = tk.Button(root,textvariable = browse,command = lambda:upload(),font = "Raleway",bg="#20bebe",fg ="white",width=15,height =2)
browse.set("Browse")
browse_button.grid(column=1,row=2)
canvas = tk.Canvas(root,width = 600,height = 300)
canvas.grid(columnspan = 3)
root.mainloop()
謝謝!
您需要將圖像傳遞給cv2.cvtColor
,您當前正在傳遞一個字符串file
。
if file:
src = cv2.imread(file)
gray = cv2.cvtColor(src, cv2.COLOR_RGB2GRAY)
gray = cv2.medianBlur(gray,5)
在第 20 行:您正在使用 CvtColor 將圖像更改為灰度。
gray = cv2.cvtColor(file, cv2.COLOR_RGB2GRAY)
您需要傳遞文件指針來代替文件。 這就是為什么您收到錯誤TypeError: Expected Ptr<cv::UMat> for argument 'src'
你需要:
首先使用img=cv2.imread(file)
讀取文件,然后使用
使用gray = cv2.cvtColor(img, cv2.COLOR_RGB2GRAY)
對 img 數據進行 CvtColor。
所以刪除第20行的內容並添加
img=cv2.imread(file) gray = cv2.cvtColor(img, cv2.COLOR_RGB2GRAY)
編輯:
第 6 行的一件重要事情是導入 askopenfile,它返回圖像的二進制編碼路徑。 這是錯誤的實現。
相反,您應該導入 askopenfilename; 這將返回圖像文件的路徑。
我在這里分享整個更新的代碼:
import cv2 import tkinter as tk import instructions as instructions from PIL import Image,ImageTk from tkinter.filedialog import askopenfilename root = tk.Tk() logo = Image.open('logo.jpg') logo = ImageTk.PhotoImage(logo) logo_label = tk.Label(image = logo) logo_label.image = logo logo_label.grid(column=1,row=0) def upload(): browse.set("loading...") file = askopenfilename(parent=root,title="Choose an image",filetypes =[("JPG file","*.jpg"),("PNG file","*.png"),("JPEG file","*.jpeg")]) print(file) if file: img=cv2.imread(file) gray = cv2.cvtColor(img, cv2.COLOR_RGB2GRAY) gray = cv2.medianBlur(gray,5) edges = cv2.adaptiveThreshold(gray,255,cv2.ADAPTIVE_THRESH_MEAN_C,cv2.THRESH_BINARY, 9, 9) color = cv2.bilateralFilter(img,9,250,250) cartoon =cv2.bitwise_and(color,color,mask=edges) cv2.imshow("Cartoon", cartoon) intructions = tk.Label(root,text= "Select an image",font = "Raleway") instructions.grid(columnspan=3,column=0,row=1) browse = tk.StringVar() browse_button = tk.Button(root,textvariable = browse,command = lambda:upload(),font = "Raleway",bg="#20bebe",fg ="white",width=15,height =2) browse.set("Browse") browse_button.grid(column=1,row=2) canvas = tk.Canvas(root,width = 600,height = 300) canvas.grid(columnspan = 3) root.mainloop()
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