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barcode.py
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# -*- coding: utf-8 -*-
"""
# Author : Ming
# File : {NAME}.py
# Time : 2019/3/29 0029 下午 3:21
"""
from PIL import Image
import numpy as np
import matplotlib.pyplot as plt
'source: https://habr.com/en/post/439768/'
'''首先，让我们加载图像，拉伸它的宽度，从中间裁剪一条水平线，将其转换为b/w颜色并保存为数组。'''
image_path = "barcode.jpeg"
img = Image.open(image_path)
width, height = img.size
basewidth = 4*width
img = img.resize((basewidth, height), Image.ANTIALIAS)
hor_line_bw = img.crop((0, int(height/2), basewidth, int(height/2) + 1)).convert('L')
hor_data = np.asarray(hor_line_bw, dtype="int32")[0]
print('hor_data:', hor_data)
'''条形码上的黑线对应于«1»，但是在RGB中黑线是相反的，0，所以数组需要倒排。我们还将计算平均值。'''
hor_data = 255 - hor_data
avg = np.average(hor_data)
# plt.plot(hor_data)
# plt.show()
print('hor_data:', hor_data)
print(avg)
'现在我们需要确定一个“位”的宽度。为此，我们将提取序列，保存平均线交叉的位置。'
pos1, pos2 = -1, -1
bits = ""
for p in range(basewidth - 2):
if hor_data[p] < avg and hor_data[p + 1] > avg:
bits += "1"
if pos1 == -1:
pos1 = p
if bits == "101":
pos2 = p
break
if hor_data[p] > avg and hor_data[p + 1] < avg:
bits += "0"
bit_width = int((pos2 - pos1)/3)
print('bits,bit_width, pos1,pos2', bits, bit_width, pos1, pos2)
'现在让我们自己解码。我们需要找到每一个平均线交叉，并找到最后一个区间的比特数。'
'数字将不匹配完美(代码可以拉伸或弯曲一点)，因此，我们需要将值四舍五入为integer。'
bits = ""
for p in range(basewidth - 2):
if hor_data[p] > avg and hor_data[p + 1] < avg:
interval = p - pos1
cnt = interval/bit_width
bits += "1"*int(round(cnt))
pos1 = p
if hor_data[p] < avg and hor_data[p + 1] > avg:
interval = p - pos1
cnt = interval/bit_width
bits += "0"*int(round(cnt))
pos1 = p
print('pos1, pos2, bits', pos1, pos2,bits)
'''
在我们的例子中，序列的开头是11010010000，它对应于«Code B»。
我懒得手动输入所有代码，所以我只是从Wikipedia页面复制粘贴了它。
这几行代码的解析也是在Python上进行的(提示——不要在生产环境中做这样的事情)。
'''
CODE128_CHART = """
0 _ _ 00 32 S 11011001100 212222
1 ! ! 01 33 ! 11001101100 222122
2 " " 02 34 " 11001100110 222221
3 # # 03 35 # 10010011000 121223
4 $ $ 04 36 $ 10010001100 121322
5 % % 05 37 % 10001001100 131222
6 & & 06 38 & 10011001000 122213
7 ' ' 07 39 ' 10011000100 122312
8 ( ( 08 40 ( 10001100100 132212
9 ) ) 09 41 ) 11001001000 221213
10 * * 10 42 * 11001000100 221312
11 + + 11 43 + 11000100100 231212
12 , , 12 44 , 10110011100 112232
13 - - 13 45 - 10011011100 122132
14 . . 14 46 . 10011001110 122231
15 / / 15 47 / 10111001100 113222
16 0 0 16 48 0 10011101100 123122
17 1 1 17 49 1 10011100110 123221
18 2 2 18 50 2 11001110010 223211
19 3 3 19 51 3 11001011100 221132
20 4 4 20 52 4 11001001110 221231
21 5 5 21 53 5 11011100100 213212
22 6 6 22 54 6 11001110100 223112
23 7 7 23 55 7 11101101110 312131
24 8 8 24 56 8 11101001100 311222
25 9 9 25 57 9 11100101100 321122
26 : : 26 58 : 11100100110 321221
27 ; ; 27 59 ; 11101100100 312212
28 < < 28 60 < 11100110100 322112
29 = = 29 61 = 11100110010 322211
30 > > 30 62 > 11011011000 212123
31 ? ? 31 63 ? 11011000110 212321
32 @ @ 32 64 @ 11000110110 232121
33 A A 33 65 A 10100011000 111323
34 B B 34 66 B 10001011000 131123
35 C C 35 67 C 10001000110 131321
36 D D 36 68 D 10110001000 112313
37 E E 37 69 E 10001101000 132113
38 F F 38 70 F 10001100010 132311
39 G G 39 71 G 11010001000 211313
40 H H 40 72 H 11000101000 231113
41 I I 41 73 I 11000100010 231311
42 J J 42 74 J 10110111000 112133
43 K K 43 75 K 10110001110 112331
44 L L 44 76 L 10001101110 132131
45 M M 45 77 M 10111011000 113123
46 N N 46 78 N 10111000110 113321
47 O O 47 79 O 10001110110 133121
48 P P 48 80 P 11101110110 313121
49 Q Q 49 81 Q 11010001110 211331
50 R R 50 82 R 11000101110 231131
51 S S 51 83 S 11011101000 213113
52 T T 52 84 T 11011100010 213311
53 U U 53 85 U 11011101110 213131
54 V V 54 86 V 11101011000 311123
55 W W 55 87 W 11101000110 311321
56 X X 56 88 X 11100010110 331121
57 Y Y 57 89 Y 11101101000 312113
58 Z Z 58 90 Z 11101100010 312311
59 [ [ 59 91 [ 11100011010 332111
60 \ \ 60 92 \ 11101111010 314111
61 ] ] 61 93 ] 11001000010 221411
62 ^ ^ 62 94 ^ 11110001010 431111
63 _ _ 63 95 _ 10100110000 111224
64 NUL ` 64 96 ` 10100001100 111422
65 SOH a 65 97 a 10010110000 121124
66 STX b 66 98 b 10010000110 121421
67 ETX c 67 99 c 10000101100 141122
68 EOT d 68 100 d 10000100110 141221
69 ENQ e 69 101 e 10110010000 112214
70 ACK f 70 102 f 10110000100 112412
71 BEL g 71 103 g 10011010000 122114
72 BS h 72 104 h 10011000010 122411
73 HT i 73 105 i 10000110100 142112
74 LF j 74 106 j 10000110010 142211
75 VT k 75 107 k 11000010010 241211
76 FF l 76 108 l 11001010000 221114
77 CR m 77 109 m 11110111010 413111
78 SO n 78 110 n 11000010100 241112
79 SI o 79 111 o 10001111010 134111
80 DLE p 80 112 p 10100111100 111242
81 DC1 q 81 113 q 10010111100 121142
82 DC2 r 82 114 r 10010011110 121241
83 DC3 s 83 115 s 10111100100 114212
84 DC4 t 84 116 t 10011110100 124112
85 NAK u 85 117 u 10011110010 124211
86 SYN v 86 118 v 11110100100 411212
87 ETB w 87 119 w 11110010100 421112
88 CAN x 88 120 x 11110010010 421211
89 EM y 89 121 y 11011011110 212141
90 SUB z 90 122 z 11011110110 214121
91 ESC { 91 123 { 11110110110 412121
92 FS | 92 124 | 10101111000 111143
93 GS } 93 125 } 10100011110 111341
94 RS ~ 94 126 ~ 10001011110 131141
103 Start Start A 208 SCA 11010000100 211412
104 Start Start B 209 SCB 11010010000 211214
105 Start Start C 210 SCC 11010011100 211232
106 Stop Stop - - - 11000111010 233111""".split()
SYMBOLS = [value for value in CODE128_CHART[6::8]]
VALUESB = [value for value in CODE128_CHART[2::8]]
CODE128B = dict(zip(SYMBOLS, VALUESB))
print('SYMBOLS',SYMBOLS)
'最后一部分很简单。首先，我们将序列分割为11位块:'
sym_len = 11
symbols = [bits[i:i+sym_len] for i in range(0, len(bits), sym_len)]
print('symbols', symbols)
'最后，生成输出字符串并显示:'
str_out = ""
for sym in symbols:
print(sym)
try:
if CODE128B[sym] == 'Start':
continue
elif CODE128B[sym] == 'Stop':
break
str_out += CODE128B[sym]
print(" ", sym, CODE128B[sym])
except:
str_out += '*'
print(" *")
print("Str:", str_out)
'pip install pyzbar '
from pyzbar.pyzbar import decode
img = Image.open(image_path)
decode = decode(img)
print(decode)