构建 DeBruijn 图的算法给出了错误的结果

Algorithm to construct DeBruijn graph gives wrong results

我正在尝试编写一些代码,从 Python 中的一组 kmers(k 字母长字符串,DNA 测序读数)构建 DeBruijn 图,输出为边集合,连接同一个节点给其他人。

当我 运行 我的样本输入代码时:

['GAGG','CAGG','GGGG','GGGA','CAGG','AGGG','GGAG']

我得到:

CAG -> AGG
GAG -> AGG

而不是:

AGG -> GGG
CAG -> AGG,AGG
GAG -> AGG
GGA -> GAG
GGG -> GGA,GGG

有什么我做错的提示吗?
这是代码:

import itertools

inp=['GAGG','CAGG','GGGG','GGGA','CAGG','AGGG','GGAG']

y=[a[1:] for a in inp]
z=[b[:len(b)-1] for b in inp]

y.extend(z)
edjes=list(set(y))

w=[c[1:] for c in edjes]
v=[d[:len(d)-1] for d in edjes]

w.extend(v)

nodes=list(set(w))

graph={}


new=itertools.product(edjes,edjes)

for node in nodes:
    for edj in new:
        edje1,edje2=edj[0],edj[1]
        if edje1[1:]==node and edje2[:len(edje2)-1]==node:
            if edje1 in graph:
                graph[edje1].append(edje2)
            else:
                graph[edje1]=[edje2]

for val in graph.values():
    val.sort()


for k,v in sorted(graph.items()):
    if len(v)<1:
        continue
   else:
        line=k+' -> '+','.join(v)+'\n'
   print (line)

我认为你的算法太复杂了:你可以简单地先对输入执行唯一性过滤器:

inp=['GAGG','CAGG','GGGG','GGGA','CAGG','AGGG','GGAG']

edges=list(set(inp))

然后遍历 "edges" 这个列表。对于每条边,前三个字符是 from 节点,后三个字符是 to 节点:

for edge in edges:
    frm = edge[:len(edge)-1]
    to = edge[1:]
    #...

现在您只需将其添加到图表中:

for edge in edges:
    frm = edge[:len(edge)-1]
    to = edge[1:]
    if frm in graph:
        graph[frm].append(to)
    else:
        graph[frm]=[to]

最后像您自己一样进行排序和打印:

for val in graph.values():
    val.sort()

for k,v in sorted(graph.items()):
    print(k+' -> '+','.join(v))

这导致:

AGG -> GGG
CAG -> AGG
GAG -> AGG
GGA -> GAG
GGG -> GGA,GGG

如您所见,第 2 行略有不同:您的预期输出包含 AGG 两次,这没有多大意义。

所以完整的算法是这样的:

inp=['GAGG','CAGG','GGGG','GGGA','CAGG','AGGG','GGAG']

edges=list(set(inp))

graph={}

for edge in edges:
    frm = edge[:len(edge)-1]
    to = edge[1:]
    if frm in graph:
        graph[frm].append(to)
    else:
        graph[frm]=[to]

for val in graph.values():
    val.sort()

for k,v in sorted(graph.items()):
    print(k+' -> '+','.join(v))

你的算法

我认为的一个问题是,您将三个字母序列视为 "edjes"(可能是 edges)。边缘是四个序列字符。通过执行此转换,信息将丢失。接下来构造一组双字符项(nodes,它们根本不是节点)。他们似乎习惯"glue"把节点放在一起。但在那个阶段,你不再知道这些碎片是如何粘合在一起的。