如何编辑单个像素？

Question

我决定只使用该方法提供给我的数组来制作一个图像编辑程序matplotlib.pyplot.imread，但是当我尝试将一个像素的内容分配给另一个像素时，它告诉我：can't assign to function call.import matplotlib.pyplot as pltf = plt.imread("GreenScreen.png")x1 = 250y1 = 350x2 = 750xCount = x2xCount2 = xCountxCountStart = x2xSet = 0for xCount in range(xCountStart, x1):    xCount2 = (x1 + xCount) / 2    xSet = xCount2    for y1 in range(350, 360):        for xCount2 in range(xSet, xCount):            f(xCount2, y1) = f(xSet, y1)            f(xCount2, y1) = f(xSet, y1)            f(xCount2, y1) = f(xSet, y1)            f(xCount2, y1) = f(xSet, y1)        xCount2 = xSet    y1 = 350plt.imshow(f)plt.show()

Answer 1

将括号更改为用于索引()的方括号。[]像这样的东西应该工作：

for xCount in range(xCountStart, x1):

    xCount2 = (x1 + xCount) / 2

    xSet = xCount2

    for y1 in range(350, 360):

        for xCount2 in range(xSet, xCount):

            f[xCount2, y1] = f[xSet, y1]

            f[xCount2, y1] = f[xSet, y1]

            f[xCount2, y1] = f[xSet, y1]

            f[xCount2, y1] = f[xSet, y1]

        xCount2 = xSet

    y1 = 350 

    ...

Answer 2

f(xCount2, y1) = f(xSet, y1)

f(xCount2, y1) = f(xSet, y1)

f(xCount2, y1) = f(xSet, y1)

f(xCount2, y1) = f(xSet, y1)

要在 中设置字段f，请使用f[xCount2, y1]。括号用于调用函数，方括号用于索引。