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假设我有以下列表:组元素的关键
mylist = [keyobj1, valobj1, valobj2, keyobj2, valobj1, valobj2, valobj3,...]
我想获得一个字典由包含它的:
mydict = {keyobj1: [valobj1, valobj2], keyobj2: [valobj1,valobj2,valobj3, ...}
是否有执行库这个任务?
下面是粗略的算法将被执行(我知道,相貌丑陋):
# convert mylist to iterator
mylist = iter(mylist)
n = mylist.next()
while True:
try:
if not iskey(n):
n = mylist.next()
value_list.append(n)
else:
mydict[key] = value_list
value_list = []
key = n
n = mylist.next()
except StopIteration:
mydict[key] = value_list
break
有没有必要手动迭代列表 - 只用于。 – jsbueno