你的意思是这样的吗?
d = { 0 : (1, 2, 3) , 1 : (2, 3, 4), 2: (5, 6, 7)}
for key in d.keys():
for val in d[key]:
try:
d[key]+=d[val]
except KeyError:
pass
给了
>>> d
{0: (1, 2, 3, 2, 3, 4, 5, 6, 7), 1: (2, 3, 4, 5, 6, 7), 2: (5, 6, 7)}
如果您想要唯一值,请将 d[key] = tuple(set(d[key])) 添加到 for key in d.keys() 循环的末尾。
给予
>>> d
{0: (1, 2, 3, 4, 5, 6, 7), 1: (2, 3, 4, 5, 6, 7), 2: (5, 6, 7)}
ps:d = {[ 0 : 1, 2, 3], [1 : 2, 3, 4], [2: 5, 6, 7]} 不是有效的 python!
编辑:见 cmets。
d = { 0 : [1, 2, 3] , 1 : [2, 3, 4], 2: [5, 6, 7]}
for key in d.keys():
orig_vals=d[key]
new_vals=[]
for val in orig_vals:
try:
new_vals+=d[val]
except KeyError:
pass
d[key] = list(set(new_vals)-set(orig_vals))
给予
>>> d
{0: [4, 5, 6, 7], 1: [5, 6, 7], 2: []}
如果您想避免清除未链接到其他键的值,例如2中的[5,6,7],然后把最后一行改成
if new_vals:
d[key] = list(set(new_vals)-set(orig_vals))
这给了
>>> d
{0: [4, 5, 6, 7], 1: [5, 6, 7], 2: [5, 6, 7]}
编辑 2:见 cmets。
d = { 0 : [1, 2, 3] , 1 : [2, 3, 4], 2: [5, 6, 7]}
for key in d.keys():
orig_vals=d[key]
new_vals=[]
count = 0
for val in orig_vals:
try:
new_vals+=d[val]
count+=1
if count >= yournumberhere: break
except KeyError:
pass
d[key] = list(set(new_vals)-set(orig_vals))