執行結果比較：

上面是自己做中斷，每6筆為一行。

下面是用 numpy 一行指令，把原本的資料切開。取得的結果是一樣的。

舊的寫法：

row = ""
idx=0
values_formated = []
for item in values:
    idx += 1
    flag = 1
    if item[0]==0:
        flag = 0
    row += " %d" % (flag)
    values_formated.append(flag)
    if idx % (diff_x+1) == 0:
        print((diff_x+1), row)
        row = ""

新的寫法：

np_re = np.array(values_formated)
np2 = np_re.reshape([5,6])
print("np2:", np2)

How to get pixel values inside of a rectangle within an image
https://stackoverflow.com/questions/58790535/how-to-get-pixel-values-inside-of-a-rectangle-within-an-image

You can do that with Python/OpenCV by first drawing a white filled polygon on black background as a mask from your four points. The use np.where to locate and then print all the points in the image corresponding to the white pixels in the mask.

Input:

import cv2
import numpy as np

# read image
image = cv2.imread('lena.png')

# create mask with zeros
mask = np.zeros((image.shape), dtype=np.uint8)

# define points (as small diamond shape)
pts = np.array( [[[25,20],[30,25],[25,30],[20,25]]], dtype=np.int32 )
cv2.fillPoly(mask, pts, (255,255,255) )

# get color values
values = image[np.where((mask == (255,255,255)).all(axis=2))]
print(values)

# save mask
cv2.imwrite('diamond_mask.png', mask)

cv2.imshow('image', image)
cv2.imshow('mask', mask)
cv2.waitKey()

Mask:

Results:

 [[108 137 232]
 [104 134 232]
 [108 136 231]
 [106 134 231]
 [109 133 228]
 [108 136 229]
 [109 137 230]
 [110 135 232]
 [103 126 230]
 [112 134 228]
 [114 136 228]
 [111 138 230]
 [110 137 233]
 [103 135 234]
 [103 126 230]
 [101 120 226]
 [108 137 230]
 [112 133 228]
 [114 136 227]
 [115 139 232]
 [112 137 232]
 [105 134 233]
 [102 128 232]
 [ 98 119 226]
 [ 93 105 220]
 [108 139 230]
 [110 137 230]
 [112 135 230]
 [113 135 230]
 [111 138 231]
 [112 139 232]
 [109 134 233]
 [101 128 232]
 [100 120 224]
 [ 90 104 221]
 [ 87  95 211]
 [111 138 229]
 [109 135 231]
 [109 136 230]
 [113 141 233]
 [110 139 233]
 [105 136 234]
 [101 127 232]
 [ 95 117 225]
 [ 90 107 220]
 [110 137 231]
 [110 138 231]
 [107 140 236]
 [110 139 233]
 [104 135 234]
 [105 130 231]
 [ 92 116 227]
 [114 141 234]
 [112 142 235]
 [111 140 235]
 [111 138 234]
 [110 132 232]
 [114 140 234]
 [108 140 233]
 [107 134 233]
 [107 140 235]]

換成 PIL 寫法：

How to read an image file as ndarray

from PIL import Image
import numpy as np

im = np.array(Image.open('data/src/lena_square.png'))

print(im.dtype)
# uint8

print(im.ndim)
# 3

print(im.shape)
# (512, 512, 3)

在 FontForge 裡的一個坐標，換成BMP格式的檔案，新的坐標值如何轉換？

滿有趣的一個數學問題，第一次遇到。上圖裡 Y 軸是 790，在 bmp 系統是對到 90. fontforge Y 軸向下，數值會減少，但在 bmp 系統，Y軸向下，數值會增加。

解法是，把 bmp 系統推換回原點，再乘一個負數，換回去 bmp=0 對到 fontforge 裡對到的頂點。再把 bmp 系統裡的異動值，以 fontforge 的基準點做減少，就可以對應了。

例如：

def ff_y_to_bmp_y(y):
    return 880 + (y * -1)

def average(lst): 
    return sum(lst) / len(lst) 

def is_same_direction_list(args,deviation=0):
    ret = True
    args_average=average(args)
    #print("args_average:", args_average)

    direction = -1
    if args[0] <= args_average:
        direction = 1
    
    idx=0
    args_count = len(args)
    for item in args:
        idx+=1
        if idx == args_count:
            break

        if direction==1:
            if (args[idx]+deviation)<item and (args[idx]-deviation)<item:
                ret = False
                break
        else:
            if (args[idx]+deviation)>item and (args[idx]-deviation)>item:
                ret = False
                break

    return ret

def is_same_direction(*args,deviation=0):
    return is_same_direction_list(args,deviation=deviation)


print(&quot;case 1:&quot;, is_same_direction(10,20,30,40))
print(&quot;case 2:&quot;, is_same_direction(10,20,40,30))
print(&quot;case 3:&quot;, is_same_direction(40,30,20,10))

# allow small mistake.
print(&quot;case 4:&quot;, is_same_direction(10,20,33,30,deviation=3))

執行結果：

 case 1: True
 case 2: False
 case 3: True
 case 4: True

成功讓 case 4 號變成 True, 允許小範圍的誤差.

技術原理：

Python不定長度參數args,*kwargs
https://stackoverflow.max-everyday.com/2020/03/python-function-args-kwargs/

換成python 寫法：

dist = sqrt( (x2 - x1)2 + (y2 - y1)2 )

較佳寫法：

from math import hypot
dist = hypot(x2-x1, y2-y1)

如果使用 (3,4) 會得到 5.0