Question

用于编写sage中的汉明鳕鱼（基于python的编译器）我需要创建一个矩阵，其中每列是数字的二进制表示说汉明（3）矩阵看起来应该是这样的

0  0  0  1  1  1  1
0  1  1  0  0  1  1
1  0  1  0  1  0  1

这是从1到7的数字的二进制表示。所以我到现在所做的是将任何给定数字转换为它的二进制表示：我做了这个小功能它需要两个值n和r 和r比特的反复意见

binrep(n,r)
    x=n
    L=[]
    LL=[]
    while (n>0):
        a=int(float(n%2))
        L.append(a)
        n=(n-a)/2
    while (len(L)<r):
        L.append(0)
    #print(L)
    LL=L[::-1]
    return LL

所以现在我想收集我得到的所有LL并将它们放在一个像上面的那个大矩阵中

Answer 1

您只需要将所有这些二进制表示转换为整数列表，将它们收集到列表列表中，最后transpose列表列表以获得所需的输出。

n = 3
binary = []
for i in range(1, 2**n):
    s = binrep(i, n)
    binary.append(map(int, s))
matrix = zip(*binary)

或者在一行中：matrix = zip(*(map(int, binrep(i, n)) for i in range(1, 2**n)))

结果是

[(0, 0, 0, 1, 1, 1, 1), 
 (0, 1, 1, 0, 0, 1, 1), 
 (1, 0, 1, 0, 1, 0, 1)]

另请注意，convert numbers to binary还有其他方式，例如使用str.format：

def binrep(n,r):
    return "{0:0{1}b}".format(n, r)

Answer 2

除了其他的，使用numpy的答案：

import numpy as np

# we're creating the binary representation for all numbers from 0 to N-1
N = 8

# for that, we need a 1xN matrix of all the numbers
a = np.arange(N, dtype=int)[np.newaxis,:]

# we also need a log2(N)x1 matrix, for the powers of 2 on the numbers.
# floor(log(N)) is the largest component that can make up any number up to N
l = int(np.log2(N))
b = np.arange(l, dtype=int)[::-1,np.newaxis]

# This step is a bit complicated, so I'll explain it below.
print np.array(a & 2**b > 0, dtype=int)

打印：

[[0 0 0 0 1 1 1 1]
 [0 0 1 1 0 0 1 1]
 [0 1 0 1 0 1 0 1]]

该行

print np.array(a & 2**b > 0, dtype=int)

一次完成一些事情。我将它分成更简单的步骤：

# this takes our matrix b and creates a matrix containing the powers of 2
# up to 2^log2(N) == N
# (if N is a power of 2; otherwise, up to the next one below)
powers = 2**b

# now we calculate the bit-wise and (&) for each combination from a and b.
# because a has one row, and b as one column, numpy will automatically
# broadcast all values, so the resulting array has size log2(N)xN.
u = a & powers

# this is almost what we want, but has 4's in the first row,
# 2's in the second row and 1's in the last one.
# one method of getting 1's everywhere is to divide by powers:
print u / powers

# another way is to check where u > 0, which yields an array of bools,
# which we then convert to numbers by turning it into an array of ints.
print np.array(u > 0, dtype=int)

Answer 3

就像提到的zoosuck一样，我建议使用bin（）函数代替你的代码。要按顺序打印它们，假设格式相似：

L = ['10101010', '10101010']

for i in L:
    print ' '.join([j for j in i])

以类似的方式，您可以将其附加到列表或字典中。

在python中创建一个二进制数字表示矩阵

3 个答案: