我希望将以下字典转换为矩阵,其中字典的第一个和第二个值是列和行值.在矩阵为真的情况下,我希望有一个’1′,当它是假的时我想要一个’0′.
{0:[2,5.0],1:[6,7.0],2:[6,8.0],3:[5,6.0],4:[1,5.0],5:[3,4.0],6 :[4,5.0]}
期望的输出看起来像这样
1 2 3 4 5 6 7 8
1 0 0 0 0 1 0 0 0
2 0 0 0 0 1 0 0 0
3 0 0 0 1 0 0 0 0
4 0 0 1 0 1 0 0 0
5 1 1 0 1 0 1 0 0
6 0 0 0 0 1 0 1 1
7 0 0 0 0 0 1 0 0
8 0 0 0 0 0 1 0 0
谢谢堆,任何指针都会很棒!
丹妮尔.
最佳答案 这是一个匹配你想要的输出,虽然我认为使用@ Antimony的答案作为基础(numpy)可能是要走的路(至少比这个答案更多)
N = 8
d = {0: [2, 5.0], 1: [6, 7.0], 2: [6, 8.0], 3: [5, 6.0], 4: [1, 5.0], 5: [3, 4.0], 6: [4, 5.0]}
m = [0] * (N ** 2 + 1)
for x, y in d.values():
m[x + int(y - 1) * N] = 1
m[int(y) + (x - 1) * N] = 1
print " ".ljust(9),
print " ".join(map(str, range(1, N + 1)))
print
for i in range(1, N ** 2 + 1):
if i % N == 1:
print "%d ".ljust(10) % (i / N + 1),
val = m[i]
print "%d " % val,
if not i % N:
print
OUTPUT
1 2 3 4 5 6 7 8
1 0 0 0 0 1 0 0 0
2 0 0 0 0 1 0 0 0
3 0 0 0 1 0 0 0 0
4 0 0 1 0 1 0 0 0
5 1 1 0 1 0 1 0 0
6 0 0 0 0 1 0 1 1
7 0 0 0 0 0 1 0 0
8 0 0 0 0 0 1 0 0
根据@ DSM对@ Antimony的回答的评论,这里有一个使用numpy:
import numpy
N = 8
m = numpy.zeros((N,N))
d = {0: [2, 5.0], 1: [6, 7.0], 2: [6, 8.0], 3: [5, 6.0], 4: [1, 5.0], 5: [3, 4.0], 6: [4, 5.0]}
for i,j in d.itervalues():
m[i-1,j-1] = 1
m[j-1,i-1] = 1
print " ".ljust(9),
print " ".join(map(str, range(1, N + 1)))
print
for i, line in enumerate(m.tolist()):
print "%s %s" % (("%s".ljust(10) % (i+1)," ".join(map(str, map(int, line)))))
OUTPUT
1 2 3 4 5 6 7 8
1 0 0 0 0 1 0 0 0
2 0 0 0 0 1 0 0 0
3 0 0 0 1 0 0 0 0
4 0 0 1 0 1 0 0 0
5 1 1 0 1 0 1 0 0
6 0 0 0 0 1 0 1 1
7 0 0 0 0 0 1 0 0
8 0 0 0 0 0 1 0 0