我如何使用 Python 进行新加坡车牌识别的校验和

Question

I am able to do recognition of car license plate and extract the plate values. Sometimes, the results are inaccurate as i am using OCR to do the recognition. I uses a checksum to ensure only the correct results are being printed and viewed. After the calculation for the checksum, I need to use another formula to get the last letter of the plate. match with these 19 letter, A=0, Z=1, Y=2, X=3, U=4, T=5, S=6, R=7, P=8, M=9, L=10, K=11, J=12, H=13, G=14, E=15, D=16, C=17, B=18. Is there any way i can use a loop to like declare the values of this letters instead of doing it one by one manually? Please help out. Thank you.

Answer 1

You can use a list and perform the lookups according to your needs.

The list looks like this:

plate_letter_list = ['A', 'Z', 'Y', 'X', 'U', 'T', 'S', 'R', 'P', 'M', 'L', 'K', 'J', 'H', 'G', 'E', 'D', 'C', 'B']

Case 1: Lookup value from letter

If you need to find the numeric value associated with a letter, use the index method:

letter = 'T'
print(plate_letter_list.index(letter))
>> 5

Case 2: Lookup letter from value

If you need to find the letter associated with a numeric value, use it as index:

value = 13
print(plate_letter_list[value])
>> H

Answer 2

There are two ways to search for these values. One of it is to use the List datatype like @sal provided. The other way is to use Dictionary datatype

**

Solution using List Datatype

**

pl_vals_list = ['A', 'Z', 'Y', 'X', 'U', 'T', 'S', 'R', 'P', 'M', 'L', 'K', 'J', 'H', 'G', 'E', 'D', 'C', 'B']

You can then do a lookup either by position or by value

To search by position , will provide you the value you assigned to the alphabet.

print(pl_vals_list[0]) . This will result in A .

Alternatively, you can search by the alphabet itself. In this case, you have to use the index() function.

print(pl_vals_list.index('A')) . This will result in the assigned number you gave to the alphabet. The result will be 0

This provides you a means to look-up based on alphabet or value.

You can also check if the alphabet is inside the list using:

if 'A' in pl_vals_list:
    #then do something for value A
elif 'B' in pl_vals_list:
    #then do something for value B
else:
    #then do something

You can also iterate through the list using a for loop and enumerate. However, I don't know if that will be of value to you.

for i, v in enumerate(pl_vals_list):
    #do something with each value in the list
    #here i will store the index, and v will have the value - A, B, C, etc

You can get the value of each of these and determine what you want to do.

**

Solution using Dictionary Datatype

**

Similarly, you can do the same using a dictionary.

pl_vals_dict = {'A':0, 'Z':1, 'Y':2, 'X':3, 'U':4, 'T':5, 'S':6, 'R':7, 'P':8, 'M':9, 'L':10, 'K':11, 'J':12, 'H':13, 'G':14, 'E':15, 'D':16, 'C':17, 'B':18}

To look for alphabets within a dictionary, you can use

if 'A' in pl_vals_dict.keys():
    #then do something for value A
elif 'A' in pl_vals_dict.keys():
    #then do something for value B
else:
    #do something else

An alternate way to check for something would be:

x = True if 'F' in pl_vals_dict.keys() else False

In this case x will have a value of False

You can also use the get() function to get the value.

x = pl_vals_dict.get('A')  # OR

print (pl_vals_dict.get('A')

The simplest way to look up a dictionary value is:

print (pl_vals_dict['A'])

which will result in 0

However, you have to be careful if you try to print value of 'F', it will throw an error as 'F' is not part of the key value pair within the dictionary.

print (pl_vals_dict['F'])

this will give you the following error:

Traceback (most recent call last):
  File "<pyshell#48>", line 1, in <module>
    pl_vals_dict['F']
KeyError: 'F'

Similar to list, you can also iterate through the dictionary for keys and values. Not sure if you will need to use this but here's an example for you.

for k, v in pl_vals_dict.items():
    #do something with each pair of key and value
    #here k will have the keys A Z Y X ....
    #and v will have the values 1, 2, 3, 4, ....