Oscar Benjamin於 2013年3月12日星期二UTC+8下午11時44分50秒寫道： > On 12 March 2013 14:59, Oscar Benjamin <oscar.j.benja...@gmail.com> wrote: > > > Numpy and matplotlib will do what you want: > > > > > > import numpy as np > > > import matplotlib.pyplot as plt > > > > > > def bits_to_ndarray(bits, shape): > > > abytes = np.frombuffer(bits, dtype=np.uint8) > > > abits = np.zeros(8 * len(abytes), np.uint8) > > > for n in range(8): > > > abits[n::8] = (abytes % (2 ** (n+1))) != 0 > > > > Whoops! The line above should be > > abits[n::8] = (abytes & (2 ** n)) != 0 > > > > > return abits.reshape(shape) > > > > > > # 8x8 image = 64 bits bytes object > > > bits = b'\x00\xff' * 4 > > > > > > img = bits_to_ndarray(bits, shape=(8, 8)) > > > plt.imshow(img) > > > plt.show() > > > > > > > > > Oscar
Now the dram is so cheap in the street. Please type in a tuple of all 8 bit inversions from the index to the result then just take a look up by the index to solve the problem. # there are ways to exchange the top 4 bits and the low 4bits, then swap inside the nibbles then swap the 4 2bit pairs in the old way. -- http://mail.python.org/mailman/listinfo/python-list
