父)作为csv(700,000行)输入
Child Parent
fA00 f0
fA9 fA0
fA31 fA0
fA30 fA0
fA1 fA00
dccfA1 fA00
fA2 fA00
fA3 fA00
fA01 fA00
fA4 fA00
fA5 fA00
fA6 fA00
fA7 fA00
fA0 fA00
fA142149 fA00
fA02 fA00
fA8 fA00
qA1 fA10
fA22 fA10
fA23 fA10
fA11 fA10
qA2 fA10
fA15 fA11
fA13 fA11
fA12 fA11
fA14 fA13
fA17 fA16
fA18 fA17
fA19 fA17
fA20 fA17
fA21 fA19
etc....
它上升到14级。顶级父母是f0
我想迭代子父关系来确定路径
预期结果
f0 --- top
f0\fa00
f0\fa00\.Child
f0\fa00\.Child2etc
f0\fA0
f0\fA0\.Child
f0\fA0\.Child2etc
我怎么能用Python做到这一点?
我开始考虑复杂的tree structures递归构造,但基本上它非常简单。创建一个子到父的映射,然后从子列表开始,父子列表,父节点的父节点到顶部。递归例程很容易提取孩子的祖先。
'''
This is the family tree:
------------------------
f0:
a0:
b0
b1:
b2:
a1:
b3:
b4:
a2:
b5:
c0
c1
'''
ancestry = [
('b1', 'a0'),
('c1', 'b5'),
('b2', 'a0'),
('b3', 'a1'),
('b4', 'a1'),
('b5', 'a2'),
('a0', 'f0'),
('a1', 'f0'),
('a2', 'f0'),
('b0', 'a0'),
('c0', 'b5'),
]
代码是:
parents = set()
children = {}
for c,p in ancestry:
parents.add(p)
children[c] = p
# recursively determine parents until child has no parent
def ancestors(p):
return (ancestors(children[p]) if p in children else []) + [p]
# for each child that has no children print the geneology
for k in (set(children.keys()) - parents):
print '/'.join(ancestors(k))
输出是:
f0/a1/b4
f0/a0/b0
f0/a0/b1
f0/a0/b2
f0/a1/b3
f0/a2/b5/c1
f0/a2/b5/c0
我将把它作为练习留给read the csv file,并且可能更好地对输出进行排序。