发布网友
共3个回答
热心网友
1:
2:
name="xxx"
热心网友
workers.sort(key=lambda x:x['Name'])
name=raw_input("please input Name:").strip()
found=False
low=0
high=len(workers)-1
mid=(high+low)/2
while low<=high:
if workers[mid]['Name']<name:
low=mid+1
elif workers[mid]['Name']>name:
high=mid-1
else:
found=True;
break
mid=(low+high)/2
if found:
workman=workers[mid]
print workman['Name']+":\t",
for key ,value in workman.items():
print key+":"+value+"\t",
else:
print "Not Find this man %s"%name
热心网友
workers_dict = dict([(w["Name"], w) for w in workers])
for name in li:
print workers_dict.get(name)
将列表转换为字典,可以高效查找;且保持良好的代码可读性
追问大神,你的用不了啊