python - 在 python 中打印前馈循环

Question

所以，我有一个看起来像这样的巨大输入文件:( you can download here )

 1. FLO8;PRI2
 2. FLO8;EHD3
 3. GRI2;BET2
 4. HAL4;AAD3
 5. PRI2;EHD3
 6. QLN3;FZF1
 7. QLN3;ABR5
 8. FZF1;ABR5
 ...

把它看成一个两列表，“;”之前的元素显示给“;”之后的元素

我想迭代地打印简单的字符串，以显示构成前馈循环的三个元素。上面的示例编号列表将输出:

"FLO8 PRI2 EHD3"
"QLN3 FZF1 ABR5"
...

将第一行输出解释为前馈循环:

A -> B  (FLO8;PRI2)
B -> C  (PRI2;EHD3)
A -> C  (FLO8;EHD3)

只有这个圈出来的link

所以，我有这个，但它非常慢......有什么建议可以加快实现速度吗？

import csv

TF = []
TAR = []

# READING THE FILE
with open("MYFILE.tsv") as tsv:
    for line in csv.reader(tsv, delimiter=";"):
        TF.append(line[0])
        TAR.append(line[1])

# I WANT A BETTER WAY TO RUN THIS.. All these for loops are killing me    
for i in range(len(TAR)):
    for j in range(len(TAR)):
        if ( TAR[j] != TF[j] and TAR[i] != TF[i] and TAR[i] != TAR[j] and TF[j] == TF[i] ): 
            for k in range(len(TAR )):
                    if ( not(k == i or k == j) and TF[k] == TAR[j] and TAR[k] == TAR[i]):
                        print "FFL: "+TF[i]+ " "+TAR[j]+" "+TAR[i]

注意:我不想要自循环...从 A -> A、B -> B 或 C -> C

Answer 1

我使用集合字典来实现非常快速的查找，如下所示:

编辑:防止自循环:

from collections import defaultdict

INPUT = "RegulationTwoColumnTable_Documented_2013927.tsv"

# load the data as { "ABF1": set(["ABF1", "ACS1", "ADE5,7", ... ]) }
data = defaultdict(set)
with open(INPUT) as inf:
    for line in inf:
        a,b = line.rstrip().split(";")
        if a != b:          # no self-loops
            data[a].add(b)

# find all triplets such that A -> B -> C and  A -> C
found = []
for a,bs in data.items():
    bint = bs.intersection
    for b in bs:
        for c in bint(data[b]):
            found.append("{} {} {}".format(a, b, c))

在我的机器上，这会在 0.36 秒内加载数据并在 2.90 秒内找到 1,933,493 个解决方案；结果看起来像

['ABF1 ADR1 AAC1',
 'ABF1 ADR1 ACC1',
 'ABF1 ADR1 ACH1',
 'ABF1 ADR1 ACO1',
 'ABF1 ADR1 ACS1',

---

Edit2: 不确定这是您想要的，但是如果您需要 A -> B 和 A -> C 和 B -> C 而不是 B -> A 或 C -> A 或C -> B，你可以试试

found = []
for a,bs in data.items():
    bint = bs.intersection
    for b in bs:
        if a not in data[b]:
            for c in bint(data[b]):
                if a not in data[c] and b not in data[c]:
                    found.append("{} {} {}".format(a, b, c))

但这仍然会返回 1,380,846 个解决方案。