我有2种的节点 - 的信节点的(L)和'编号节点'(N)的曲线图。我有2个字典,一是表明由L边缘N和其他节目边从N到L;
I have a graph with 2 kinds of nodes- 'Letter nodes' (L) and 'Number nodes' (N). I have 2 dictionaries, one shows edges from L to N and the other shows edges from N to L.
A = {0:(b,), 1:(c,), 2:(c,), 3:(c,)}
B = {a:(3,), b:(0,), c:(1,2,3)}
有一个键,值对 C:(1,2,3)
意味着有来自 C
边缘 1,2,3
(3边)
A key,value pair c:(1,2,3)
means there are edges from c
to 1,2,3
(3 edges)
我想这些合并成一个字典 C
,这样的结果是一个新的字典:
I want to merge these to one dictionary C
so that the result is a new dictionary:
C = {(0,): (b,), (1, 2, 3): (a, c)}
或
C = {(b,):(0,), (a, c):(1, 2, 3)}
在生成的词典我想信节点和数字节点上的键和值的不同侧面。我不在乎它只是需要他们分开的键或值。我该如何去有效地解决这个?
In the resulting dictionary I want the letter nodes and numerical nodes to be on separate sides of keys and values. I don't care which is the key or value just need them separated. How can I go about solving this efficiently?
澄清:这与2类型的节点的图的 - 号节点,和信节点。字典Ç说,从信节点(A,C)可以(1,2,3),即A-> 3-> C-> 1,A-> 3-> C-> 2从而达到数量的节点就可以得到1,2,3从。虽然没有直接的EDGE从A到2或1。
CLARIFICATION: this of a graph with 2 types of nodes - number nodes, and letter nodes. the dictionary C says from letter nodes (a,c) you can reach the number nodes (1,2,3) i.e a->3->c->1, a->3->c->2 thus you can get to 1,2,3 from a. EVEN THOUGH THERE IS NO DIRECT EDGE FROM a to 2 or a to 1.
推荐答案
根据你的说法,我猜你正在努力寻找一个图算法。
According to your statement, I guess you are trying to find a graph algorithms.
import itertools
def update_dict(A, result): #update vaules to the same set
for k in A:
result[k] = result.get(k, {k}).union(set(A[k]))
tmp = None
for i in result[k]:
tmp = result.get(k, {k}).union(result.get(i, {i}))
result[k] = tmp
for i in result[k]:
result[i] = result.get(i, {i}).union(result.get(k, {k}))
A = {0:('b',), 1:('c',), 2:('c',), 3:('c',)}
B = {'a':(3,), 'b':(0,), 'c':(1,2,3)}
result = dict()
update_dict(A, result)
update_dict(B, result)
update_dict(A, result) #update to fix bugs
update_dict(B, result)
k = sorted([sorted(list(v)) for v in result.values()])
k = list( k for k, _ in itertools.groupby(k)) #sort and remove dumplicated set
final_result = dict()
for v in k: #merge the result as expected
final_result.update({tuple([i for i in v if isinstance(i, int)]):tuple([i for i in v if not isinstance(i, int)])})
print final_result
#output
{(0,): ('b',), (1, 2, 3): ('a', 'c')}
相关推荐
最新文章